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Non-Archimedean Geometry
Lectures from a course by Matt Baker at U.C. Berkeley
Scribes
Sarah Brodsky
Melody Chan
Michael Daub
Andrew Dudzik
Will Johnson
Shelly Manber
Maria Monks
Ralph Morrison
Andrew Niles
Doosung Park
Qingchun Ren
Lecture Notes: Math 274, Jan 18, 2012
Shelly Manber
Non-Archimedean Fields
Definition 1. An valued field is a field k together with a non-zero absolutevalue | · | : k ! R�0
, satisfying
1. |0| = 0, |1| = 1
2. |xy| = |x||y|
3. |x+ y| |x|+ |y|
A non-archimedean valued field is a valued field that additionally satisfiesthe stronger condition
3’ |x+ y| max(|x|, |y|)
An archimedean valued field is a valued field that is not non-archimedean.
We will almost always assume that k is complete, i.e. if we define ametric d on k by d(x, y) := |x � y| then k is a complete metric space withrespect to d.
Example 1. R and C are complete archimedean valued fields. In fact, it isa non-trivial theorem that they are the only complete archimedean valuedfields. (This is a consequence of a theorem by Gelfand and Mazur; seeTheorem 1.2.3 of Engler and Prestel, “Valued Fields”.)
Example 2. Qp, the field of p-adic numbers, is a complete non-archimedeanvalued field.
Aside: if you don’t know what Qp is, chances are this course is not atyour level and you should consider not taking it. However, the definition isreviewed here nonetheless.
1
Definition of Qp: Let p be a prime and 0 < ✏ < 1 (definition isindependent of choice of ✏; by convention we usually take ✏ = 1/p).
For n 2 Z, define ordp(n) := k i↵ pk||n (|| denotes “exactly divides”).
For ↵ = ab 2 Q define ordp(↵) := ordp(a)� ordp(b).
Let | · |p : Q ! R�0
be the valuation defined by |↵|p := ✏ordp(↵) for ↵ 2 Q.
Let Qp be the completion of Q with respect to | · |p.
Then Qp is a complete non-archimedean valued field and Q is dense inQp.
Example 3. Let F be any field and denote by F ((t)) the field of Laurantseries in t over F , i.e. elements of F ((t)) are of the form
P1n=N ant
n withN 2 Z, aN 6= 0.
Then we have an absolute value |·| : F ((t)) ! R�0
defined by |P1
n=N antn| :=
✏N , and F ((t)) is complete with respect to | · |.
Example 4. It is a non-trivial theorem that Cp, the completion of Qp
(algebraic closure of Qp), is complete and algebraically closed.Aside: later we will see that Cp is “not big enough” and we will bring in
the notions of Berkovich a�ne line and spherical closure to remedy this
Example 5. For a field F , the field of Puiseux Series over F is defined by
[
n2NF ((t1/n))
so that the elements of the field of Puiseux Series over F are series of theform
1X
k=k0
aktk/n where k
0
, n 2 Z, n > 0, and ak 2 F
If F is algebraically closed and characteristic zero, then the field of PuiseuxSeries over F is also algebraically closed. However, the field of Puiseux Seriesover F is not complete.
2
Geometry over k
Let X be an algebraic variety over k, an algebraically closed, complete,non-archimedean valued field with non-trivial absolute value (the trivialabsolute value is that defined by |0| = 0 and |x| = 1 for all x 6= 0)
We assign to X(k) the weakest topology such that for every Zariskiopen subset U of X and every regular function f in OX(U), the functionU(k) ! R defined by P 7! |f(P )| is continuous.
Problem: with this topology, X(k) is totally disconnected and not lo-cally compact
Solution: Two options:
1. Generalize the notion of topology and covering (Grothendick topology)
2. Add more points to “locally compactify” (Berkovich’s idea)
Sketch of solution 2: we will construct an object Xan with the followingproperties:
• Xan is locally compact and Hausdor↵
• X is path connected i↵ X is connected
• X(k) ,! Xan and X(k) is dense in Xan
• Xan is compact i↵ X is proper
• The topological dimension of Xan is equal to the Krull dimension ofX
We will define Xan in future lectures; for now we provide a picture of(A1)an:
3
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Figure 1: The Berkovich a�ne line (illustration courtesy of Joe Silverman)
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Lecture Notes: Math 274, Jan 20, 2012
Shelly Manber
Analogy to Understand Berkovich A�ne Line
ConsiderX := [0, 1] \Q = {x 2 Q | 0 x 1}
with the subspace topology inherited from the metric topology on the inter-val [0, 1]. It is totally disconnected and not locally compact.
We want to compactify X by first defining the “correct” notion of con-tinuous functions on X. For example, following Gelfand we can define thecompletion of X to be the spectrum of the algebra of continuous functionson X, given the “right” definition of continuous.
Naive Definition. Let f : X ! Q be continuous at x 2 X if for all✏ > 0 there exists a � > 0 such that for all x0 2 X, we have |f(x0)�f(x)| < ✏
whenever |x� x
0| < � .
Problem With Naive Definition: We get continuous functions thatcan’t be extended to continuous functions on the reals.
For example, let z 2 [0, 1] \Q and define a function f : X ! R by
f(x) :=
(0 x < z
1 x > z
Then there is no way to assign a value f(z) to z such that f becomes acontinuous function on [0, 1].
Solutions:
1. Tate-Grothendick: restrict the notion of open set and open cover
In this analogy, define open sets to be intervals of the form (a, b) witha, b 2 Q and open covers to be finite unions of open sets. With thisdefinition, X becomes compact and connected.
1
2. Berkovich: complete X
In this analogy, replace X with the interval [0, 1] and the usual notionof open set and open cover.
Berkovich A�ne Line
Given a complete valued field k, we would like construct a notion of a�neline A1
k such that
• A1C = C
• The ring of algebraic functions on A1k is k[T ].
To this end, we consider all multiplicative semi-norms on k[T ] that extendthe value | · | given on k.
Recall: a multiplicative semi-norm on a ring A is a function | · | : A !R�0, satisfying
1. |0| = 0, |1| = 1
2. |xy| = |x||y|
3. |x+ y| |x|+ |y|
If k is a valued field and A is a k-algebra, a k-semi-norm on A is asemi-norm on A whose restriction to k is the given absolute value on k.
Suppose k is a complete, algebraically closed, non-archimedean valuedfield (Cp is a good example to keep in mind). For every z 2 k, we have theevaluation semi-norm defined by:
| · |z : k[T ] ! R�0
f 7! |f(z)|
Then | · |z 6= | · |y when z 6= y, so identifying z 2 k with | · |z gives aninjection k ,! {multiplicative k-semi-norms on k[T ]}.
Definition 1. Let A1Berk be the set of multiplicative k-semi-norms on k[T ]
endowed with the weakest topology such that for all f 2 k[T ], the map
A1Berk ! R�0
|| · || 7! ||f ||
is continuous.
2
Note that when k = C, the Gelfand-Mazur theorem tells us that allmultiplicative semi-norms on k[T ] are of the form |f | = |f(z)| for somez 2 C.
Exercise 1. Prove that if k = C then the injection k ,! A1Berk is a homeo-
morphism k onto its image.
If k = Cp then A1Berk has many more points than those arising from k.
For example, given a 2 k and r 2 R�0, let D(a, r) := {z 2 k | |z � a| r}be a closed disk in k. Define
|| · ||D : k[T ] ! R�0
f 7! supz2D
|f(z)|
Exercise 2. Show that || · ||D is a multiplicative semi-norm on k[T ]
Hint: If f =Pn
i=0 ai(T � a)i then ||f ||D = max0in |ai|ri since | · | isnon-archimedean.
Note: The norm || · ||D(0,1) is often called the Gauss norm.
Exercise 3. Show that the map D(a, r) 7! || · ||D(a,r) yields an imbeddingfrom the set of generalized closed disks of k into A1
Berk that extends theinjection k ,! A1
Berk by associating an element a of k with the generalizeddisk D(a, 0).
Now we can begin to understand the tree-like picture of the Berkovicha�ne line (below). Each point on the tree is a point of A1
Berk, i.e. a mul-tiplicative semi-norm on k[T ]. The vertical axis measures the radius ofthe closed disk associated to the point. On the horizontal axis are thepoints of A1
Berk corresponding to disks of radius zero; as we saw theseare in bijection with elements of k. If a, b 2 k and |b � a| r then{z 2 k | |z � a| r} = {z 2 k | |z � b| r}. Thus the line segmentsstarting at a and b indicate points corresponding to disks of increasing ra-dius centered at a and b respectively, and at some radius r = |a � b|, thedisks become the same disk ; here the two segments meet.
3
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Figure 1: The Berkovich a�ne line (illustration courtesy of Joe Silverman)
Theorem 0.1. The Berkovich a�ne line is uniquely path-connected.
Proof. See Lemma 2.10 of “Potential Theory and Dynamics on the BerkovichProjective Line” (Baker and Rumley).
Note that what we have drawn is not locally compact as we have not yetidentified all the points of A1
Berk. In fact, there are multiplicative semi-normson k[T ] that are not of the form || · ||D for a closed disk D of k.
For example, let k = Cp and let D = D1 � D2 � D3 . . . be a sequenceof nested closed disks of k. If D1 \ D2 \ · · · = ; then || · ||D , defined by||f ||D := limn!1 ||f ||Dn , is a new point of A1
Berk, i.e. a multiplicative semi-norm on k[T ] that is not equal to || · ||D for any closed disk D. We call pointsarising in the this fashion “type 4 points”. Types 1 � 3 are points comingfrom generalized closed disks D(a, r) and are characterized as follows:
Type 1: r = 0 (equivalently, these are point of k)
Type 2: r 2 |k⇥|
Type 3: r /2 |k|
Theorem 0.2. (Berkovich) When k is algebraically closed, complete, and
non-archimedean, all points of A1Berk are of types 1� 4
4
Lecture Notes: Math 274, Jan 23, 2012
Ralph Morrison
(Throughout this day’s notes we’ll assume that k is a complete, alge-braically closed, non-Archimedian field. Also, a point in A1
Berk,k will some-times be written as x and sometimes as | · |
x
.)
Points of A1Berk,k
Recall 1. For k a complete, algebraically closed, non-Archimedian field, theBerkovich a�ne line (A1
k
)an (sometimes denoted A1Berk,k or A1
Berk), is defined(as a set) as
{multiplicative seminorms on k[T ] extending | · | on k}
.If D = D(a, r) := {x 2 k
�� |x� a| r}, then
|f |D
:= supz2D
|f(z)|
is a multiplicative seminorm.
Theorem 0.1 (Berkovich). Every point x 2 (A1k
)an can be realized as
limn!1
| · |Dn
for some nested sequence D1 ◆ D2 ◆ ... of closed disks in k.
Lemma 0.2. If k is a non-Archimedean field, A is a k-algebra, and | · | isa multiplicative seminorm on A extending the absolute value on k, then for
all f, g 2 A,
|f + g| max{|f |, |g|},
with |f + g| = |g| if |f | < |g|.
1
Proof of Lemma. This proof is essentially a clever application of the bino-mial theorem. In particular, we know that |m| = |1 + ... + 1| |1| = 1 forall integers m.
|f + g|n =|(f + g)n| =
�����
nX
k=0
✓n
k
◆f
k
g
n�k
����� nX
k=0
����
✓n
k
◆f
k
g
n�k
����
nX
k=0
���fk
g
n�k
��� (n+ 1)max{|f |, |g|}n.
Taking n
th roots and letting n ! 1, we have the desired result.For the case of |f | < |g|, we know that |f + g| |g|. On the other hand,
|g| = |g + f � f | max{|f + g|, |f |}. Since it is not the case |g| |f |, weknow that |g| |f + g|. Thus |g| = |f + g|.
Proof of Theorem. We will prove the theorem in the case of plugging in anylinear polynomial T � a of k[T ] into the seminorm, rather than an arbitrarypolynomial. This result, combined with k = k and the multiplicativity ofour seminorms, will give the desired result.
Fix x 2 A1Berk. Consider the family F of closed disks
F := {D(a, |T � a|x
) | a 2 k}.
Claim 1: F is totally ordered by inclusion.To establish this claim, let a, b 2 k. Without loss of generality, |T�a|
x
�|T � b|
x
. We have
|a� b| = |a� b|x
= |(T � b)� (T � a)|x
max{|T � a|x
, |T � b|x
} (⇤)
by the lemma, with equality if |T � a|x
> |T � b|x
. Thus b 2 D(a, |T � a|x
),implying that D(b, |T � b|
x
) ⇢ D(a, |T � a|x
), establishing Claim 1.
Now, let r = infa2k |T �a|
x
Choose a sequence of points an
such that rn
decreases to r, with r
n
= |T � a
n
|x
. Let Dn
= D(an
, r
n
).
Claim 2: For all a 2 k, |T � a|x
= limn!1 |T � a|
Dn .To prove this claim, choose any a 2 k. By definition of r, |T � a|
x
� r.We will deal with two cases.
(1) Assume |T � a|x
= r. Then r
n
= |T � a
n
|x
� |an
� a| by (⇤) (takinga = a
n
and b = a). Hence a 2 D
n
, so |T �a|Dn = sup
z2Dn|z�a| = r
n
.It follows that
|T � a|x
= r = lim r
n
= lim |T � a|Dn ,
as desired.
2
(2) Assume |T � a|x
> r. For n � 0, |T � a|x
> T � |an
|x
= r
n
. Applyingthe strict case of (⇤), this implies that |T � a|
x
= |a � a
n
| for n � 0,and therefore that |a � a
n
| > |T � a
n
|x
= r
n
for n � 0. Hence|T � a|
Dn = supz2Dn
|z � a| = |an
� a| = |T � a|x
for su�ciently largen. This lets us conclude that lim |T � a|
Dn = |T � a|x
, as desired.
This establishes Claim 2, as well as the theorem.
Remark 1. In the preceding proof, we can choose the radii of the disks Dn
to lie in the value group of k.
The four types of points in A1Berk To describe the four types of points in
the Berkovich a�ne line, for a point | · |x
we’ll let {Dn
} be the correspondingdescending sequence of closed disks, writing D
n
= D(an
, r
n
) and r = lim r
n
.
• Type I: These points, which correspond to points of k, have |f |x
=|f(a)| for some a 2 k. (That is, r = 0.)
• Type II: These points have
1\
n=1
D
n
= D(a, r)
with r 2 |k⇥| (the value group of k).
• Type III: These points have
1\
n=1
D
n
= D(a, r)
with r /2 |k|.
• Type IV: These points have
1\
n=1
D
n
= ;
with r > 0 by the completeness of k.
3
Local description of the points For x 2 A1Berk, let
T
x
= {tangent directions at x} = {connected components of A1Berk � {x}}.
• If x is a Type I point, then |Tx
| = 1.
• If x is a Type II point, then there are infinitely many branches at thatpoint. In particular, the branches are in noncanonical bijection with
P1(k) (to be made more precise later), so |Tx
| =���P1(k)
���.
• If x is a Type III point, then there is no branching, and |Tx
| = 2.
• Type IV points are sort of “dead ends”, giving |Tx
| = 1.
The topology on this infinitely branched tree, the observer’s topology, isa very weak topology which makes the tree locally compact. For a basis offundamental open neighborhoods, take the connected components of A1
Berk�{a finite set of points}.
4
Lecture Notes: Math 274, Jan 25, 2012
Ralph Morrison
Facts about non-Archimedean fields
1. If k is a complete non-Archimedean field and K/k is an algebraicextension, then there exists a unique extension to K of | · | on k. IfK/k is a finite extension, the K is complete.
2. If k is a non-Archimedean field which is algebraically closed, then itscompletion b
k is also algebraically closed (and complete, of course).Moreover, k is dense in b
k and bk/k is an immediate extension (meaning
that |bk| = |k| and bk and k have the same residue field).
Theorem 0.1. Qp
is not complete.
Sketch of Proof. Define X
n
= {x 2 Qp
�� [Qp
(x) : Qp
] = n}. Let X = Qp
, sothat
S1n=1Xn
= X.One may check that each X
n
is closed and has empty interior. Recall theBaire Category Theorem: in a complete metric space, any countable unionof closed sets with empty interior has empty interior (and so can’t be thewhole space). This implies that X is not complete.
Definition 1. We say k is spherically complete if every descending nested
sequence of closed disks has nonempty intersection.
Remark 1. Spherically complete implies complete. However, the converse isnot true, by the following theorem.
Theorem 0.2. Cp
:= bQp
is complete and algebraically closed but not spher-ically complete.
Proof. Let r > 0, and let r
n
be a descending sequence with r
n
! r, sothat r0 > r1 > ... > lim r
n
= r, where r
n
2 |Qp
|. Consider D0 = D(0, r0).Choose disjoint disks D1 and D
01 of radius r1 contained in D. Continue
this fashion, giving us a tree T that keeps branching o↵ in pairs. For each
1
infinite (rooted) path in T , we get a nested sequence D�
of closed disks. LetD
�
=T
D2D�D.
We claim that each D
�
is either empty or is a closed disk of radius r > 0.Assume D
�
is nonempty, and let x 2 D
�
. Since x is in each D 2 D�
, it is thecenter of each disk (by the ultrametric property), and so each D containsD(x, r), meaning D(x, r) ⇢ D
�
. However, if |x� y| = r
0> r, there is some
D 2 D of radius r
i
< r
0, meaning y /2 D, so y /2 D
�
. Thus D
�
is a disk ofradius r, or it is empty.
In particular, each D
�
is open (since r > 0). By construction the D
�
’sare disjoint. Since C
p
contains Q as a countable dense subset, only countablymany D
�
’s can be nonempty by disjointness. There were uncountably manypaths, so uncountably many D
�
’s are empty. We conclude that Cp
is notspherically complete.
Construction of a spherically complete extension of k Our con-struction is based on ultrafilters.
Definition 2. Let X be a set. A filter on X is a nonempty collection F ofsubsets of X such that
1. ; /2 F .
2. If A,B 2 F , then A \B 2 F .
3. If A 2 F and A ⇢ B ⇢ X, then B 2 F .
Example 1. Choose x 2 X. Define Fx
= {A ⇢ X |x 2 A}.
Example 2. If X is infinite, let F = {A ⇢ X
�� |AC | < 1}. This is calledthe Frechet filter.
Filters are ordered by refinement.
Definition 3. We say that F 0refines F , denoted F 0 � F , if F 0 � F in
P (X).
Definition 4. A maximal filter with respect to this relation is called anultrafilter.
Recommended reading for ultrafilters: Terry Tao’s blog.
Lemma 0.3. A filter F is an ultrafilter if and only if for all A ⇢ X, exactly
one of A and A
C
belongs to F .
2
By Zorn’s lemma, every filter is contained in an ultrafilter.
Definition 5. A filter of the form Fx
is called principal. Such a filter isalways an ultrafilter.
Remark 2. Arrow’s Impossibility Theorem says that every filter of a finiteset is principal (i.e., of the form F
x
).
On any infinite set, there exist non-principal ultrafilters. For instance,take a maximal filter containing the Frechet filter.
Fix a nonprincipal ultrafilter U on N. Then any bounded sequence {an
}of real numbers has a unique “U -limit.” This can be constructed by doing abinary search, dividing some bounded interval containing {a
n
} and zoomingin on the half with a
i
’s with indices forming an element of U .Given a non-Archimedean field k, we o↵er a construction of a field ⌦
k
that is algebraically closed and spherically complete.Fix U . Let R be the normed ring `
1(k) (i.e., bounded infinite sequencesin k with the `
1 norm). Define ' : R ! R�0 by '((an
)) = limU |an
|. LetJ = '
�1(0), which is an ideal in R. Let ⌦k
= R/J .
Theorem 0.4. 1. (⌦k
, | · |) is a non-Archimedean field containing k.
2. ⌦k
is spherically complete.
3. If k is algebraically closed, then so is ⌦k
.
4. If |k| is dense in R, then |⌦k
| = R.
Reference: Chapter 3 of Robert’s “A Course in p-adic Analysis.”
3
Lecture Notes: Math 274, Jan 27, 2012
Ralph Morrison
Recall: spherically complete extensions of k Given k, a non-Archimedeanfield, we constructed a non-Archimedean field ⌦ extending k such that ⌦is spherically complete (as well as algebraically closed if k is algebraicllyclosed).
The ⌦ we constructed is in general not an immediate extension of k.If k = k with nontrivial valuation, when the value group |k| is dense inR, meaning that |⌦
k
| = R. For instance, in the case of k = Cp
, we have|k⇥| = p
Q, but |⌦⇥k
| = R⇥. Immediate extensions preserve value groups, sothis extension is not immediate.
There is a minimal spherically closed field extension of k inside of ⌦. Allsuch extensions are isomorphic, but not canonically; in particular, the iso-morphisms might not preserve k. (Reference: van Rooij, “Non-archimedeanFunctional Analysis.”)
Analytification of A�ne Varieties over k We’ll take k to be completeand non-Archimedean, but maybe not algebraically closed.
An a�ne variety X/k corresponds to a reduced finitely generated k-algebra A. From this we will construct a topological space X
an, which we’llalso write as M(A).
Definition 1. As a set, we define
X
an :={multiplicative seminorms on A extending | · | on k}.(For concision we will call such seminorms k-multiplicative seminorms.)
The topology is the weakest such that all maps f
: X
an ! R arecontinuous, where
f
is defined for each f 2 A by x 7! |f |x
.Equivalently, this is the subspace topology induced by X ,! RA, x 7!
(f 7! |f |x
) (where RA has the product topology).
Some facts about Xan:
• The analytification X ! X
an is functorial.
1
• One can glue analytifications and thereby analytify any algebraic va-riety.
• Later on, we will put more structure on X
an and will be able to an-alytify more general objects, such as schemes of finite type over k, ormore generally quasiseparated and quasicompact rigid analytic spacesover k.
A motivating recollection from algebraic geometry Usually we de-fine
Spec A := {prime ideals in A}.We can instead define
Spec A := {k-algebra homomorphisms A! K, where K/k is an extension}/ ⇠,
where the equivalence relation ⇠ is given by ' ⇠ '
0 if and only if we havean injection K ,! K
0 such that A! K ,! K
0 agrees with A! K
0.This is equivalent to the prime ideal definition by the maps
p! [A! Frac(A/p)]
ker' (A'! K)
Say we’re in the case A = k[x1, ..., xn]/(f1, ..., fm). Then we can thinkof the points of Spec A as “solutions” to f1, ..., fm with values in some fieldK/k.
(One interpretation of this is in terms of transcendental versus algebraicelements. For instance, in Spec k[x, y]/(y), our points look like (x, 0). Maybex is algebraic, or maybe it’s transcendental. All transcendetal extensionsk(T ) are equivalent, so the generic point classifies the transcendental points.)
An alternate definition of X
an. We’ll consider
�(A) := {k-algebra homomorphisms A! K}/ ⇠
where K is a complete valued field extending | · | on k, and where equivalenceis the same as for the algebraic geometry case, except with the requirementthat K ,! K
0 is compatible with | · |.Claim 0.1. �(A) is naturally in bijection with M(A) = X
an.
2
This bijection is given by the maps
| · |x
! (A! H(x))
|f | := |'(f)| (A'! K),
where we define H(x) to be the completion of Frac(A/ ker(| · |x
)); we callH(x) the completed residue field of x.
Classification of Points on A1Berk in Terms of H(x)’s (See Proposition
2.3 in Baker-Rumely.)Assume that k is algebraically closed for simplicity. Let A = k[T ], and
x 2 A1Berk. Let k(x) = Frac(A/ ker(| · |
x
)), the algebraic residue field of x.
By definition, H(x) = dk(x).
• For Type I points, k(x) = H(x) = k.
• For points of Type II, III, and IV, k(x) ⇠= k(T ), and H(x) is a weirdcompletion. It can be described explicitly for Type III, but not forTypes II and IV.
Type Value Group |H(x)⇥| Residue field ]H(x)
I |k⇥| ek
II |k⇥| ek(T )
III h|k⇥|, ri , r /2 |k⇥| ek
IV |k⇥| ek
(Note that we have |H(x)⇥| = |k(x)⇥| and ]H(x) = gk(x).)
3
Classification of Berkovich points in terms of H(x)
Recall that K(x) is defined to be the fraction field of k[T ]/ ker(| · |x
), and H(x) is defined to be
its completion [K(x), with respect to the absolute value on K(x) induced by the multiplicativeseminorm | · |
x
on k[T ].
Theorem. Let k be a complete and algebraically closed non-Archimedean field. Then for each
point x 2 A1Berk,k, its completed residue field H(x) has one of the following descriptions:
Type Value group |H(x)⇤| Residue field
]H(x)
I |k⇤| ekII |k⇤| ek(t)III h|k⇤|, ri ekIV |k⇤| ek
where r 2 R \ |k⇤| in the value group of Type III above. Moreover, x is of Type I if and only if
H(x) ⇠= k.
Remark. A point x is of Type IV if and only if H(x) is a nontrivial immediate extension of k,that is, it is not equal to k but it has the same value group and residue field as k.
Proof. We give a sketch of the proof. For more details, see Proposition 2.3 in [1].We consider the four types of points separately.Type I: In this case, x corresponds to a point a 2 k, and |f |
x
= |f(a)|. Then ker | · |x
=m
a
= (T � a), and so H(x) = K(x) = k.Type II: In this case, x corresponds to a disk D(a, r) where r = |c| for some c 2 k⇤. Then
K(x) ⇠= k(T ) since the kernel of | · |x
must be 0.Now, H(x) is the completion of k(T ) with respect to the norm | · |
x
on k[T ]. By thenon-Archimedean maximum principle, for each f 2 k[T ] there exists a p 2 D(a, r) such that|f |
x
= |f(p)|. Hence, we have |k(T )⇤|x
= |k⇤|x
.
Let t be the reduction of (T � a)/c in ]H(x). (Exercise: Prove that t is transcendental overek.)
Then we have ]H(x) = ]K(x) = ek(t).Type III: In this case, the residue field is ek because all polynomials in T with | · | 1 have
constant reduction to ek. We have |T � a|x
= r 62 |k⇤|. This implies that |H(x)⇤| = h|k|⇤, ri.Type IV: In this case, x corresponds to a nested collection of disks D(a
i
, ri
) with emptyintersection. If f 2 k(T ) is nonzero, there exists an N such that D
N
:= D(aN
, rN
) contains nozeros or poles of f . This implies that |f | is constant on D
N
, and so |f |x
= |f |D(aN ,rN ) 2 |k⇤|. It
follows that |H(x)⇤| = |k⇤|.Now, let f = g/h, |f |
x
= 1. For N >> 0, we have |g|x
= |g|Dn = |h|
x
= |h|Dn . It follows
that
f ⌘ g(aN
)
h(aN
)(mod m
x
).
It follows that ef 2 ek.
1
References
[1] M. Baker, R. Rumley, Potential Theory on the Berkovich Projective Line, American MathSociety, 2010.
2
Inequalities in non-Archimedean field extensions
Algebraic extensions:
Let K/k be a finite extension of non-Archimedean valued fields. Let
e = [v(K
⇤) : v(k
⇤)]
and
f = [
eK :
ek].
Then
ef [K : k]. (1)
We sketch a proof below. Equality holds if k is discretely valued.
Proof. Choose x1, . . . , xe 2 K
⇤such that v(x1), . . . , v(xe) represent distinct cosets in v(K
⇤)/v(k
⇤).
Choose y1, . . . , yf 2 K
0such that ey1, . . . ,fyf in
eK are linearly independent over
ek. It su�ces to
prove the following:
Claim: The elements xiyj , where 1 i e and 1 j f , are linearly independent over k.
To prove the claim, assume Xaijxiyj = 0 (2)
for some aij 2 k not all zero. Choose (i0, j0) such that v(ai0j0xi0yj0) is minimal. We show that
any other monomial term with v(aijxiyj) minimal must also have i = i0. To see this, note that
v(yj) = 0, and if v(ai0j0xi0yj0) = v(ai1j1xi1yj1) then using the relation v(xy) = v(x) + v(y), we
obtain
v(xi0)� v(xi1) = v(ai1j1)� v(ai0j0) 2 v(k
⇤).
Since the xi were chosen so that v(x1), . . . , v(xe) represent e distinct cosets in v(K
⇤)/v(k
⇤), it
follows that i1 = i0.
Finally, multiplying (2) by (ai0j0xi0)�1
, we get a relation of the form
Pbiyj + c = 0 where
the bj ’s are in k with |bj | 1, c 2 K with |c| < 1, and |bj0 | = 1. Passing to the residue field, we
get a nontrivial relation in the eyj ’s, which contradicts the assumption that the eyj ’s are linearly
independent over
ek.
We can now prove a similar inequality for the transcendental case.
Transcendental extensions:
Suppose K/k is a finitely generated extension of non-Archimedean valued fields. Let
s = dimQ
✓v(K
⇤)
v(k
⇤)
⌦Z Q◆
and
t = tr. deg(
eK/
ek).
Then we have
s+ t tr. deg(K/k). (3)
This is known as Abhyankar’s Inequality.
1
Proof. (Sketch) Choose x1, . . . xs 2 K
⇤such that the [v(xi)] ⌦ 1 are linearly independent over
Q. Choose y1, . . . , yt 2 K
0such that ey1, . . . , eyt are algebraically independent over
ek. It su�ces
to show:
Claim: The elements x1, . . . , xs, y1, . . . , yt are algebraically independent over k.
To prove the claim, assume for the sake of contradiction that p is not identically zero. Then
p(x, y) = 0 for some p 2 k[X1, . . . , Xs, Y1, . . . , Yt].
We can then expand this relation into a sum of monomial terms, and consider their valuations.
Choosing a term of minimal valuation, by the same argument as above, we find that all of the
subscripts of x must be equal, and we can again divide out by that. We then get a polynomial
relation in the eyi’s, a contradiction.
Classification of points of A1
Berk
We can make the following table of values of s and t, for K(x), in the one-dimensional case.
Type s t tr. deg(K(x)/k)
1 0 0 0
2 0 1 1
3 1 0 1
4 0 0 1
Remark 1. There is a similar classification of points for nonsingular analytic curves.
We now take a look at the two-dimensional case, A2Berk.
• Every nested sequence of (possibly degenerate) closed polydisks gives a k-multiplicative
seminorm on R = k[X,Y ].
• Assume k is trivially valued. Every algebraic curve C in A2k gives a discrete valuation on R
given by ordC(f). (This is the “divisorial valuation.”) For such valuations we have s = 1
and t = 1.
• When k is trivially valued, we also get a di↵erent sort of point of A2Berk from germs of
arbitrary analytic curves: for example, let ⌧ 2 bR be transcendental over R, and write
⌧ = x+
P1i=1 ciy
iwhere ci 2 k
⇤for all i. Then the map
� : R ! k[[y]]
sending y 7! y and
x 7! x� ⌧ = �X
ciyi
is an injective homomorphism. In this case, we have s = 1 and t = 0.
We can then restrict the y-adic valuation on k[[y]] to R.
2
MATH 274: NON-ARCHIMEDIAN GEOMETRY
SARAH B. BRODSKY
Today we are going to talk about the spectrum of a Banach ring.
Let A be a commutative ring with identity (as all rings in this class, unless otherwisespecified). We then define a norm on A to be a function k · k : A ! R�0 such that:
(1) k0k = 0 and k1k = 1.(2) kfk = 0 i↵ f = 0.(3) kfgk kfk · kgk.(4) kf + gk kfk+ kgk.It is interesting to note that the spectrum of the zero ring is empty but that the zero ring
is still considered a normed ring.
A Banach ring is a pair (A, k · k) such that A is a commutative right with identity, k · kis a norm on A, and A is complete with respect to this norm.
Examples of Banach rings:
(1) Every field k which is complete with respect to some absolute value | · | is a Banachring.
(2) (Z, | · |1) is a Banach ring.(3) (Most important example for this course) Let k be a complete non-archimedean
valued field. Define the Tate algebra to be
Tn
:= khT1, ..., Tn
i := {X
⌫2Nn
a
⌫
T
⌫ : lim|⌫|!1
|a⌫
| = 0} ✓ k[[T1, ..., Tn
]]
and the Gauss norm to be kP
⌫2Nn a⌫
T
⌫k = max⌫2Nn |a
⌫
|. The Tate algebratogether with the Gauss norm is a Banach ring.
Note that Berkovich denotes the Tate algebra by k{T1, ..., Tn
} and uses the notationkhT1, ..., Tn
i for something di↵erent... which we will get to next lecture, but will not needfor the rest of the course. We are using the notation khT1, ..., Tn
i to denote the Tate algebraas this is the notation used in Bosch’s notes.
1
2 SARAH B. BRODSKY
The (Berkovich) spectrum M(A) of a Banach ring A is the set of all nonzero boundedmultiplicative seminorms on A, where by a bonded multiplicative seminorm we mean anonzero function | · | : A ! R�0 such that:
(1) |fg| = |f | · |g|.(2) |f + g| |f |+ |g|.(3) There exists C > 0 such that |f | Ckfk for all f 2 ANote that condition 3 for a seminorm is equivalent to f being continuous on A. The topol-
ogy on M(A) is the weakest one such that | · |x
7! |f |x
2 R is continuous for all f 2 A.
It should be clear from the context if A is a Banach ring or a k-algebra, where k is avalued field, as our definition of M(A) is slightly di↵erent in these two contexts. We nowhave the following theorem about M(A), whose proof we will cover in this lecture and thenext:
Theorem 0.1. (Berkovich) If A is a nonzero Banach ring, then M(A) is a non-emptycompact Hausdor↵ space.
In order to prove this theorem, we will need a couple of lemmas:
Lemma 0.2. If x 2 M(A) then |f |x
kfk for all f 2 A.
Proof. We have that (|f |x
)n = |fn|x
Ckfnk Ckfkn. So we have |f |x
npCkfk for all
n. Taking the limit as n ! 1 gives the desire result. ⇤In order to understand the statements of the next two lemmas, we need to know the
definition of a net (which is di↵erent than Berkovich’s notion of a net). First, we need thenotion of a directed set; a directed set I is a set together with a reflexive, transitive, binaryrelation such that for all ↵, � 2 I there exists � 2 I such that ↵ �, � �.
Examples of directed sets:
(1) (N,).(2) Let X be a topological space. For x 2 X, the set I = {Open nbds of x} with the
relation U V i↵ V ✓ U is a directed set.
A net in a set X is a map I ! X, where I is a directed set. We write hX↵
i instead of↵ 7! X
↵
. Note that a sequence is the same thing as a net indexted by the natural numbers.We say that X
↵
converges to X if for every neighborhood U of X, there exists ↵0 2 I suchthat X
↵
2 U whenever ↵ � ↵0.
Lemma 0.3. A subset Y of a topological space X is closed i↵ whenever hY↵
i is a net in Y
converging to x 2 X, we have x 2 Y .
Lemma 0.4. A map f : X ! Y , where X, Y are topological spaces, is coninuous i↵ f(X↵
) !f(X) whenever X
↵
! X.
MATH 274: NON-ARCHIMEDIAN GEOMETRY 3
Let us now begin to prove our theorem by showing that M(A) is compact. Consider thefollowing map:
i : M(A) ! T :=Y
f2A
[0, kfk], x 7! (|f |x
)f
By definition(s), i is injective and M(A) is homeomorphic to its image. By Tychono↵’stheorem, T is compact and Hausdor↵. Thus, it su�ce to prove thatM := i(M(A)) is closed.
So suppose we have a net hX↵
i in M(A) converging to some x 2 T . We want to show thatx 2 M. This is equivalent to showing that if | · |
↵
is a net in M(A) such that |f |↵
! |f |x
for all f 2 A, then | · |x
is a bounded multiplicative seminorm. This is clear.
Next lecture, we will show that M(A) 6= ;, discuss M(Z), and A1Berk,k
, where k is triviallyvalued.
Sarah B. Brodsky, Department of Mathematics, University of California, Berkeley, Cal-
ifornia 94720, USA
E-mail address: [email protected]
Math 274: Nonarchimedean Geometry
Will Johnson
February 6, 2012
Let (A, || · ||) be a Banach ring. We defined the Berkovich spectrum M(A) to be the set of all boundedmultiplicative seminorms on A, i.e., functions | · | : A ! R�0 such that
• |0| = 0 and |1| = 1,
• |fg| = |f ||g| and |f + g| |f |+ |g| for every f, g 2 A, and
• |f | ||f || for every f 2 A.
Our goal is to show that M(A) is nonempty if A is nonzero. This will require some preparation.
Definition 1. If (A, || · ||) is a Banach ring and I is an ideal, the residue seminorm on A/I is the normgiven by
|⇡(a)| = infa
02A, ⇡(a0)=⇡(a)||a0|| = inf
x2I
||a� x||,
for a 2 A, where ⇡ : A ⇣ A/I is the canonical map.
Lemma 1. If I is a proper ideal (1 /2 I), then the residue seminorm | · | on A/I is a seminorm. If I is closed,then the residue seminorm is a norm on A/I, making it into a Banach ring.
Proof. For any a 2 A, we have |⇡(a)| ||a||. Therefore, |0| = 0 and |1| ||1|| = 1. If |1| 6= 1, then||1 � x|| < 1 for some x 2 I. But then ||(1 � x)k|| ||1 � x||k by submultiplicativity of || · ||, and thereforeP1
k=0 ||(1� x)k|| < 1. Then the series
1 + (1� x) + (1� x)2 + · · ·converges absolutely to some value y 2 A. Multiplying by (1�x), we see that (1�x)y = (1�x)+(1�x)2+· · · =y � 1. Rearranging (1� x)y = y � 1 gives xy = 1. Then x 2 I implies that xy = 1 2 I, a contradiction. So|1| = 1.
To check submultiplicativity and the triangle law, let f and g be two elements of A/I. Let f1, f2, . . . andg1, g2, . . . be sequences of elements of A with ⇡(f
i
) = f and ⇡(gi
) = g for every i, and limi!1 ||f
i
|| = |f |and lim
i!1 ||gi
|| = |g|. Then ⇡(fi
g
i
) = fg and ⇡(fi
+ g
i
) = f + g, so
|fg| lim infi!1
||fi
g
i
|| limi!1
||fi
|| · ||gi
|| = |f | · |g|
and|f + g| lim inf
i!1||f
i
+ g
i
|| limi!1
||fi
||+ ||gi
|| = |f |+ |g|.
Finally, suppose that I is closed. If ⇡(a) 2 A/I has |⇡(a)| = 0, then by definition there are x 2 I with||a � x|| arbitrarily small. Since I is closed, a 2 I and ⇡(a) = 0. Thus | · | is a norm. Completeness of
1
A/M can be verified as follows. Let a1, a2, . . . be a Cauchy sequence. We only need to show that somesubsequence converges. Choose n1 < n2 < · · · such that |a
ni � a
ni+1 | 2�i for every i. Then we caninductively choose b
i
2 A such that ⇡(bi
) = a
ni and ||bi
� b
i+1|| 21�i. Then limi!1 b
i
converges sincePi>0 ||bi � b
i+1|| < 1. Let b = limi!1 b
i
. Then |ani � ⇡(b)| ||b
i
� b||, so limi!1 a
ni = ⇡(b). This in turnmakes lim
i!1 a
i
= ⇡(b) 2 A/M . Thus A/M is complete.
Lemma 2. Let (A, || · ||) be a Banach ring. Then every maximal ideal M ✓ A is closed.
Proof. Any nonzero element a 2 A/M has a multiplicative inverse, because A/M is a field. But then1 = |1| = |aa�1| |a| · |a�1|, which implies that |a| 6= 0. This shows that the residue seminorm on A/M is anorm. Therefore the topology on A/M from the residue seminorm is Hausdor↵, and {0} is a closed subsetof A/M . Now since |⇡(a)| ||a|| for any a 2 A, the map A
⇡! A/M is uniformly continuous.1 Since ⇡ iscontinuous and {0} is closed, ⇡�1({0}) = M is also a closed set.
Alternatively, we could note that the closure of any ideal is again an ideal. As noted in the proof of Lemma ??,every x 2 A with ||1 � x|| < 1 is invertible. Therefore, 1 is in the interior of the set of invertible elements,so the closure of any proper ideal is again a proper ideal. Therefore, if M is maximal, then M ✓ M impliesthat M = M .
Lemma 3. If A is a normed ring, and f 2 A is invertible, then for m,n � 0
||fn||�m ||f�m||n.
Proof. By submultiplicativity,||(fn)m|| ||fn||m (1)
and||fnm|| · ||f�nm|| � ||fnm
f
�nm|| = ||1|| = 1. (2)
Therefore,||fn||�m ||fnm||�1 ||f�nm|| ||f�m||n.
The first inequality is the reciprocal of (??), the second inequality follows from (??), and the third inequalityfollows directly from submultiplicativity.
Lemma 4. IfP
i
a
i
< 1 and the a
i
are positive numbers, thenP
i
a
n
i
< 1 for any n � 1.
Proof. Since limi!1 a
i
= 0, 0 a
i
1 for all but finitely many i. Then 0 a
n
i
a
i
for all but finitelymany i. Thus
Pi
a
n
i
converges by the comparison test.
Definition 2. If A is a Banach ring and r is a positive number, then Ahhr�1T ii is defined to be the Banach
ring
Ahhr�1T ii = {
1X
i=0
a
i
T
i 2 A[[T ]] :1X
i=0
||ai
||ri < 1},
with ring structure induced from A[[T ]] and with norm
||X
i
a
i
T
i|| =X
i
||ai
||ri.
Note that A injects into Ahhr�1T ii and that the norm on Ahhr�1
T ii extends the norm on A. Berkovich usesAhr�1
T i for what we denote as Ahhr�1T ii. We have been using AhT1, . . . , Tn
i to denote the Tate algebra,which Berkovich denotes as A{T1, . . . , Tn
}.1Given any ✏ > 0, ||a� b|| < ✏ implies |⇡(a)� ⇡(b)| ||a� b|| < ✏.
2
Lemma 5. Ahhr�1T ii is a Banach ring.
Proof. The triangle law is easy (and shows that Ahhr�1T ii is closed under addition). To check submulti-
plicativity, letP
i
a
i
T
i andP
i
b
i
T
i be two elements of Ahhr�1T ii. Then
X
i
a
i
T
i
! X
i
b
i
T
i
!=X
i
0
@iX
j=0
a
j
b
i�j
1
AT
i
and so
|| X
i
a
i
T
i
! X
i
b
i
T
i
!|| =
X
i
||0
@iX
j=0
a
j
b
i�j
1
A ||ri (3)
X
i
iX
j=0
(rj ||aj
||)(ri�j ||bi�j
||) (4)
=
X
i
||ai
||ri!
X
i
||bi
||ri!
(5)
= ||X
i
a
i
T
i|| · ||X
i
b
i
T
i||. (6)
This shows that || · || is submultiplicative, and additionally shows that Ahhr�1T ii is closed under multipli-
cation. The equalities ||0|| = 0 and ||1|| = 1 follow because the norm on Ahhr�1T ii extends the norm on A.
The fact that A is closed under negation is easy, as is the verification that ||↵|| > 0 for 0 6= ↵ 2 Ahhr�1T ii.
The proof of completeness is similar to the proof that L
p spaces are complete. We first show that if↵
j
=P
i
a
i,j
T
i is a sequence of elements of Ahhr�1T ii such that
Pj
||↵j
|| < 1, thenP
j
↵
j
converges. For
each i, riP
j
||ai,j
|| Pj
Pi
||ai,j
||ri < 1. ThereforeP
j
a
i,j
converges to some value b
i
. Then
X
i
||bi
||ri X
i
0
@X
j
||ai,j
||1
Ar
i
< 1,
so the formal power series � =P
i
b
i
T
i is an element of Ahhr�1T ii.
Then for j0 > 0
||� �j0X
j=0
↵
i
|| = ||X
i
0
@b
i
�j0X
j=0
a
i,j
1
AT
i||
=X
i
||bi
�j0X
j=0
a
i,j
||ri =X
i
||X
j>j0
a
i,j
||ri
X
j>j0
X
i
||ai,j
||ri =X
j>j0
||↵j
||
which goes to 0 as j0 ! 1 becauseP
j
||↵j
|| < 1. SoP
j
↵
j
= �.
Now suppose that {↵j
}1j=0 is an arbitrary Cauchy sequence in Ahhr�1
T ii. Choose n1 < n2 < n3 < . . . sothat ||↵
nj � ↵
nj+1 || 2�j for every j. ThenP
i
(↵ni � ↵
ni+1) converges by what we just showed. Thereforelim
j!1 ↵
nj exists. But since the original sequence {↵j
}1j=0 is a Cauchy sequence, it converges as long as
any subsequence converges.
3
With all this out of the way, we can prove our main result:
Theorem 1. (Berkovich) Let (A, || · ||) be a nonzero Banach ring. Then the Berkovich spectrum M(A) isnonempty.
Proof. Let M be a maximal ideal of A. By Lemma ??, M is closed. By Lemma ??, this makes the field A/M
a Banach ring with the residue seminorm. If we can show thatM(A/M) is empty, then there is some boundedmultiplicative seminorm | · | : A/M ! R�0. Composing with the projection ⇡ : A ⇣ A/M gives a boundedmultiplicative seminorm on A. Indeed, ||⇡(a)|| ||a|| for a 2 A/M , so |⇡(a)| ||⇡(a)|| ||a||, making thecomposition bounded. The composition is a multiplicative seminorm because ⇡ is a homomorphism and | · |is a multiplicative seminorm.
Replacing A with A/M , we can therefore assume that A is a field, without loss of generality.
Let S be the set of bounded (but not necessarily multiplicative) seminorms on A. The Banach norm || · || onA is an element of S, so S is nonempty. Assign a partial order on S by saying that | · |1 | · |2 if |f |1 |f |2for every f 2 A. It is straightforward to check that Zorn’s lemma applies to S, yielding a minimal element| · |. It remains to show that | · | is multiplicative.
Note that since A is a field, | · | must actually be a norm, since for nonzero a 2 A, 1 = |1| = |aa�1| |a||a�1|,so |a| 6= 0. Also, if {a
i
}1i=0 is any Cauchy sequence with respect to | · |, then {a
i
}1i=0 is also a Cauchy sequence
with respect to || · || (the Banach norm), since | · | is bounded by || · ||. If ↵ is limi!1 a
i
with respect to theBanach norm || · ||, then since |a� a
i
| ||a� a
i
|| ! 0, ↵ is also the limit with respect to the norm | · |. So A
is also a Banach ring with respect to | · |. Replacing || · || with | · |, we can assume without loss of generalitythat | · | = || · ||. (Note that S only gets smaller when we decrease the Banach norm.)
At this point, we have reduced to the case where A is a field, and the Banach norm | · | = || · || is the uniquebounded seminorm on A. We will show that | · | is multiplicative in two steps.
Lemma 6. The norm | · | is power-multiplicative: |fn| = |f |n for all n � 1.
Proof. Submultiplicativity already gives |fn| |f |n, so if power-multiplicativity fails, then |fn| < |f |n forsome nonzero f 2 A and some n > 1. Let r = n
p|fn|. Thus rn = |fn| < |f |n and r < |f |.Consider the Banach ring Ahhr�1
T ii provided by Lemma ??. We claim that f �T is not invertible. Since Ais a field, f�1 exists, and the inverse of f �T in A[[T ]] is f�1(1� f
�1T )�1 = f
�1(1+ f
�1T + f
�2T
2 + · · · ).If f � T had an inverse in Ahhr�1
T ii, it would need to agree with this inverse in A[[T ]], because Ahhr�1T ii
is a subring of A[[T ]] and multiplicative inverses in commutative rings are unique, when they exist. But iff
�1(1 + f
�1T + f
�2T
2 + · · · ) 2 Ahhr�1T ii, then certainly
1 + f
�1T + f
�2T
2 + · · · 2 Ahhr�1T ii (7)
since Ahhr�1T ii is closed under multiplication and contains f . But by definition of Ahhr�1
T ii, (??) impliesthat X
i
|f�i|ri < 1.
By Lemma ??, this makesP1
i=0 |f i|nrin < 1. However,
1 =1X
i=0
1 =1X
i=0
r
�in
r
in =1X
i=0
|fn|�i
r
in 1X
i=0
|f�i|nrin < 1,
where the penultimate inequality follows by Lemma ??. This is a contradiction, and so f�T is not invertible.
4
Since f�T is not invertible, Ahhr�1T ii/(f�T ) is nonzero, and has a residue seminorm | · |
res
by Lemma ??.Let � : A ! Ahhr�1
T ii/(f � T ) be the composite of the embedding of A into Ahhr�1T ii and the canonical
surjection. Define a seminorm | · |0 on A by pulling back the residue seminorm on Ahhr�1T ii/(f � T ):
|g|0 = |�(g)|res
.
Then | · |0 is a seminorm because � is a homomorphism and | · |res
is a seminorm. If a 2 A, then |a|0 =|�(a)|
res
|a|, since the norm on Ahhr�1T ii extends the norm on A. Thus | · |0 is a bounded seminorm.
However, it isn’t the same as | · | because
|f |0 = |�(f)|res
= |�(T )|res
||T || = r < |f |,
using the fact that f and T have the same image in the quotient Ahhr�1T ii/(f � T ). So | · |0 is a strictly
smaller seminorm than | · |, as an element of S, contradicting the choice of | · |.Lemma 7. For every nonzero f 2 A, |f�1| = |f |�1.
Proof. The proof is similar to the previous step. By submultiplicativity, 1 = |ff�1| |f�1||f |, so |f |�1 |f�1|. The only way that |f�1| = |f |�1 can fail to hold is thus if |f |�1
< |f�1|. In this case, let r = |f�1|�1,and consider the Banach ring Ahhr�1
T ii provided by Lemma ??. Again, we claim that f � T is not a unitin Ahhr�1
T ii. As before, invertibility of f � T would imply that
X
i
|f�i|ri < 1. (8)
However, by the previous lemma, |f�i| = |f�1|i = r
�i, so (??) says that
1 =X
i
1 =X
i
r
�i
r
i =X
i
|f�i|ri < 1,
a contradiction. Thus f � T is not invertible, as before.
Then consider the homomorphism � : A ! Ahhr�1T ii/(f�T ), and the pullback | · |0 of the residue seminorm
| · |res
on Ahhr�1T ii/(f � T ) as in the previous lemma. As before, | · |0 is a bounded seminorm on A, but
|f |0 = |�(f)|res
= |�(T )|res
||T || = r = |f�1|�1< |f |.
The last step follows by inverting |f�1| > |f |�1. Thus | · |0 is strictly smaller than | · | (as an element of S),a contradiction.
Multiplicativity of | · | then follows easily from the previous lemma. If f, g 2 A, then the lemma andsubmultiplicativity yield
|fg|�1 = |f�1g
�1| |f |�1|g|�1 = |f |�1|g|�1 |fg|�1
and so |f ||g| = |fg|, unless f or g vanishes, in which case multiplicativity is obvious.
This theorem has several corollaries. Here is one:
Corollary 1. If A is a nonzero Banach ring, then f 2 A is invertible if and only if f(x) 6= 0 for all x 2 M(A),
where f(x) is the image of f in H(x) = dk(x) under the canonical map A ! k(x) ! d
k(x) induced by x.
5
Proof. If f is invertible, then letting g be an inverse, we have f(x)g(x) = 1(x) = 1 for all x 2 M(A).Therefore, f(x) 6= 0 for all x 2 M(A). Conversely, suppose that f is not invertible. Then f 2 M forsome maximal ideal M . By Lemmas ?? and ??, the residue seminorm on A/M is a norm making A/M
into a Banach ring. Thus by the theorem, M(A/M) is nonempty. Taking any multiplicative seminorm| · | : A/M ! R�0, compose with the map A ! A/M and get a point x 2 M(A) such that |f |
x
= 0. Thusf 2 ker(A ! k(x)) = ker(A ! H(x)), by definition of k(x). So f(x) = 0.
In the next class, we will see additional corollaries of Theorem ??.
6
Lecture Notes: Math 274, Feb 8, 2012
Doosung Park
February 10, 2012
1 Spectral radii and Gel0fand transform
Let A be a nonzero Banach ring. Last time, we showed that M (A ) 6= ;.
Definition 1. For f 2 A , define the spectral radius of f to be
⇢(f) := infn
npkfk = lim
n!1npkfk.
Theorem 1. For all f 2 A,⇢(f) = max
x2M (A )|f(x)|.
Remarks 1. (i) Recall that f(x) 2 H (x), so it make sense to talk about its norm |f(x)|. Also, M (A ) iscompact, so |f(x)| achieves a maximum value on M (A ).
(ii) One can think of this result as a maximum modulus principle for Banach rings.
(iii) Consider the mapA !
Y
x2M (A )
H (x)
defined byf 7! f := (f(x))
x2M (A ).
This map is called the Gel
0fand transform of f . Here, the product is not the product in the category of
rings. Instead, if (Bi
, k · ki
)i2I
is a family of Banach ring, set
B =Y
i2I
Bi
to be a Banach rings whose elements are of the form f = (fi
) such that there exists c > 0 with kfi
ki
c forall i. The norm is defined by
kfk := supi
kfi
ki
.
By definition, the Gelfand transform is continuous. Moreover, the theorem says that
⇢(f) = kfk.
In particular, the kernel of the Gelfand transform is the kernel of ⇢. Note also that if k · k is powermultiplicative, then ⇢(f) = kfk = kfk.
Exercise 1. Prove that ⇢ is a bounded power-multiplicative seminorm.
1
Proof of Theorem 1. Write|f |sup = max
x2M (A )|f(x)|.
Then|f |nsup = |fn|sup kfnk,
so |f |sup ⇢(f). Now, suppose that there exists f 2 A such that |f |sup < ⇢(f). Choose a real number r
such that|f |sup < r < ⇢(f).
Consider the Banach ring B = A hhrT ii. We have kTk = r
�1, so for all x 2 M (B), |T (x)| r
�1. Therefore,
|(fT )(x)| < r · r�1 = 1
so (1 � fT )(x) 6= 0 for all x 2 M (B). Then by a result in the last lecture, 1 � fT is invertible in B withinverse
1X
i=0
f
i
T
i
.
Then1X
i=0
kf ikr�i
< 1,
so by the root test, ⇢(f) r, which is a contradiction.
2 Common extensions of two fields
Consider a diagramK1??y
K2 ����! k
of fields. Does there exist a field K extending this diagram to a diagram
K ����! K1??y
??y
K2 ����! k
which is commutative? One cannot answer the question using the compositum operation because the com-positum of K1 and K2 is only defined once we have constructed a common overfield K.”
Step 1. A := K1 ⌦k
K2 is a nonzero k-algebra.
This is true because if V1, V2 are nonzero k-vector spaces, then V1 ⌦k
V2 is also nonzero.
Step 2. By step 1, we have Spec (A) 6= ;, so there exists a prime ideal p in A, and we have natural maps
K
i
,! Frac (A/p).
Set K = Frac (A/p).
Fact. (This will not be used in this course.) Every compositum of K1, K2 over k can be obtained in thisway, and di↵erent prime ideals p yield non isomorphic K’s, i.e., Spec (K1 ⌦k
K2) classifies the classes ofcompositum.
2
Theorem 2. Given complete valued fields K1 and K2 over a field k, there exists a complete valued field K
extending K1 and K2.
Lemma 1. Any bounded map of valued fields is an isometric embedding.
Proof. Let ' : k ! K be the map. If |a| < |'(a)| for some a 2 k, then
|'(an)||an| !1
as n ! 1, contradicting boundedness. We get a similar contradiction if |a�1| < '(a�1), i.e., if |a| >
|'(a)|.
We used the tensor product K1 ⌦k
K2 to construct the common extension K. However, this is in generalnot a valued field. We need instead the completed tensor product. We have a norm on K1 ⌦k
K2 defined by
|z| = inf
(rX
i=1
kxi
kkyi
k : z =rX
i=1
x
i
⌦ y
i
, all writings
).
If K1, K2, k are non-archimedean fields, we use a di↵erent norm defined by
|z| = inf
(inf
1ir
kxi
kkyi
k : z =rX
i=1
x
i
⌦ y
i
, all writings
).
Set K1b⌦k
K2 to be the completion of K1 ⌦k
K2 respect to one of these norm. This K1b⌦k
K2 is called thecompleted tensor product.
Lemma 2. If K1, K2 are complete valued field extensions of k, their completed tensor product K1b⌦k
K2 isnonzero.
Proof. We can prove that k1⌦ 1k = 1. This is an exercise, or one can see ([BGR],4.19). More generally, wehave a canonical injection [Gruson]
K1 ⌦k
K2 ,! K1b⌦k
K2,
but we will not use this fact.
Proof of Theorem 2. Choose a point x 2 M (K1b⌦k
K2). Recall that H (x) is the complete residue field of x.The natural injections K
i
,! H (x) are bounded. By Lemma 1, they are isometric embeddings. Now, setK = H (x).
3 One more example of M (A )
ConsiderQ
i2I
K
i
where K
i
are complete valued fields. Then
M (Ki
) = {pt}
for all i. This is an exercise. Also,
M
Y
i2I
K
i
!
is homeomorphic to the Stone-Cech compactification SC(I) := {ultrafilters on I} of I. In SC(I), subsets{F : J 2 F}, J ⇢ I are open and generate the topology. For the detail, see ([Berk],1.2.3).
3
Math 274: Non–archimedean geometry
Qingchun Ren
February 17, 2012
This lecture discusses the classification of points and the structure sheaf on M(Z). Also, this lecture brieflyintroduces rigid analytic geometry.
1 M(Z)
Recall that the Berkovich spectrum M(A) is the set of all nonzero bounded multiplicative seminorms onA, with the weakest topology such that | · | 7! |f | is continuous for all f 2 A. First we recall Ostrowski’stheorem.
Theorem 1. (Ostrowski) Every nonzero absolute value on Q is equivalent to the usual absolute value | · |1or the absolute value | · |
p
, where p is a prime number, defined as follows:
|0| = 0,
|pk ab
|p
= p
�k
,where k 2 Z, a, b 2 Z\{0}, gcd(a, p) =gcd(b, p) = 1.
The following theorem classifies the points in M(Z).
Theorem 2. Every point in M(Z) is of one of the following 4 types:
(Z1) The p-trivial seminorms | · |p,1, where p is a prime number:
|n|p,1 =
⇢0, if p|n,1, otherwise.
(Z2) The trivial seminorm | · |0:|n|0 =
⇢0, if n = 0,1, otherwise.
(Z3) The seminorm | · |p,✏
, where p is a prime number and ✏ 2 (0,+1):
|n|p,✏
= |n|✏p
.
(Z4) The seminorm | · |1,✏
, where p is a prime number and ✏ 2 (0, 1]:
|n|1,✏
= |n|✏1.
Proof. Exercise.
1
Figure 1: A picture of M(Z). Taken from Poineau, An introduction to the Berkovich line over Z.
Figure 1 illustrates the topology on M(Z). Each branch is homeomorphic to a real interval. The openneighborhoods U of the root | · |0 are such that
• The intersection of U with each branch is open.
• U contains all but finitely many branches.
It can be shown (exercise) that this space is compact and Hausdor↵.
Remark 1. The topology of A1Berk,k with k trivially valued is similar to that of M(Z), except that the top
point is missing.
Definition 1. For a 2 Z and x 2 M(Z), let a(x) denote the image of a under the canonical map Z ! H(x).A rational function without poles on an open set U ⇢ M(Z) is a map
f : U ! qx2U
H(x)
of the form f(x) = a(x)/b(x), where a, b 2 Z and b(x) is nonzero on U .
Definition 2. Let U be an open subset of M(Z). An analytic function on U is a map
f : U ! qx2U
H(x)
2
which is a locally uniform limit of elements of Q without poles, i.e. for any x 2 U , there exists an openneighborhood U
0 of x in U such that for any ✏ > 0, there exists a rational function g without poles on U
such that |f(x0)� g(x0)| < ✏ for all x0 2 U
0.
The analytic functions form the structure sheaf on M(Z).
Remark 2. We can get the definition of the structure sheaf on A1Berk,k by replacing Z by k[T ] and replacing
Q by k(T ). This works more generally for An
Berk,k. It follows from the Gelfand–Mazur theorem that An
Berk,Cis homeomorphic to Cn with the usual topology. Runge proved that up to this homeomorphism, the analyticfunctions on an open subset U of An
Berk,C is the same as the usual analytic functions on U . Thus, An
Berk,Cand Cn are isomorphic as locally ringed spaces.
Figure 2: A picture of the structure sheaf on M(Z). Taken from Poineau, An introduction to the Berkovichline over Z.
Figure 2 illustrates the structure sheaf on M(Z).
Let U = M(Z)� {| · |0}. Let j : U ! M(Z) be the canonical embedding. Berkovich observes that (j⇤OU
)|·|0is the ring of adeles of Q, and (j⇤O⇤
U
)|·|0 is the group of ideles of Q. This also works if Z is replaced by thering of integers of a number field.
2 A crash course in rigid analytic geometry
Let k be a complete non–trivially valued non–archimedean field. An a�noid algebra over k is A =khT1, . . . , Tn
i/I for some (closed) ideal I of khT1, . . . , Tn
i, with the residue norm. It follows that A is a
3
Banach algebra over k. If A is isomorphic to khT1, . . . , Tm
i/J for some ideal J as well, the resulting residuenorms on A may be di↵erent. However, it can be shown that they are always equivalent.
Fact: for every maximal ideal m of A, A/m is a finite extension of k. In particular, there is a canonicalextension of the absolute value on k to an absolute value on H(x) = k(x) := A/m
x
, where mx
is a maximalideal of A and x is the corresponding point in Sp(A) := {maximal ideals of A}. The canonical topology onSp(A) is the weakest topology such that x 7! |f(x)| is continuous for all f 2 A, where f(x) is the image off in k(x). Sp(A) is a totally disconnected, non–locally compact topological space.
Example 1. If k is algebraically closed and A = khT i, then Sp(A) with its canonical topology is homeo-morphic to k
0 with its usual topology.
Now we define the special G–topology on Sp(A). A rational subdomain of Sp(A) is V = {x 2 Sp(A) : |fi
(x)| |g(x)|} for some f1, . . . , fr, g 2 A having no common zeros. A special subdomain is a finite union of rationalsubdomains. The special G–topology on Sp(A) is the following data:
• Admissible opens: special subdomains.
• Admissible coverings: finite coverings by special subdomains.
Let V be a rational domain. Define
AV
:= Ahx1, . . . , xr
i/(gx1 � f1, . . . , gxr
� f
r
).
In fact, AV
is independent of the presentation. Moreover, Sp(AV
) is homeomorphic to V .
Theorem 3. (Tate) The association V 7! AV
extends to a sheaf on the special G–topology. In particular,If {V
i
} is a finite covering of V by special subdomains, then the sequence
0 ! AV
!Y
AVi !
Y
i,j
AVi\Vj
is exact.
4
MATH 274: NON-ARCHIMEDEAN GEOMETRY
SARAH BRODSKY AND MELODY CHAN
Let us begin by recalling the following two definitions:
Definition 0.1. A lattice is a partially ordered set (L,) which has pairwise greatest lowerbounds and least upper bounds, denoted ^ and _ respectively. If ^ and _ distribute over eachother, we say that L is distributive. If there is some 0 2 L with 0 x for all x 2 L, thenwe say that L is lower-bounded.
Definition 0.2. We write x C x
0—said ”x is inner in x
0”—if, for all z � x
0, there existsw with x ^ w = 0, x0 _ w = z. If L is a lower-bounded distributive lattice with the propertythat, for all x, y 2 L with x ^ y = 0, there exists xC x
0 with x
0 ^ y = 0, then we say that Lis an overconvergent lattice.
Let k be a complete, nontrivially valued non-archimedean field, let A be a (strict) a�noidalgebra over k, and let X = Sp(A). Let L be the lattice of special subdomains of X, so L
is generated by rational domains R(f0; f1, ..., fs) = {x 2 X : |fi
(x)| |f0(x)|, i = 1, ..., s},where f0, ..., fs generate the unit ideal in A.
Note that the rational domains form a basis for “clopen” sets for the canonical topologyon X. If Xan = M(A) is the Berkovich spectrum of X, then rational domains form a basisof compact neighborhoods of Xan (this follow from the definition of the Berkovich topology).We’ll recall the construction of Xan from the data of X and the lattice L.
Theorem 0.3. The lattice L is overconvergent.
For a thorough discussion when X is an a�noid, see Chapter 7 of Rigid Analytic Geometryand Its Applications by Fresnel and van der Put. The following lemma is the key point inthe proof of the theorem above:
Lemma 0.4. Suppose R = R(f0; f1, ..., fs) and R
0 = R(f 00; f
01, ..., f
0s
) are disjoint rational
domains in X. Then there exists ⇢ 2p|k⇤| with ⇢ > 1 such that R
⇢
= R(⇢f0; f1, ..., fs) andR
0 are still disjoint in X.
Date: February 17, 2012.
1
2 SARAH BRODSKY AND MELODY CHAN
Note that since k is nontrivially valued,p|k⇤| is dense in R. The proof of this lemma is
left as an exercise, noting the maximum modulus principle (see BGR) and that R is innerin R
⇢
.
Theorem 0.5. (Dudzik) Suppose X is a Hausdor↵ topological space and L is an overcon-vergent lattice of subsets of X that form a neighborhood base. Then:
(1) There is a canonical surjective map P(L) ! M(L) sending a prime filter P to theunique maximal filter containing it. Recall that the topology on the set P(L) of primefilters is generated by the open sets {P : V 2 P} for V 2 L. Hence we can equipM(L) with the quotient topology.
(2) With the quotiemt topology, M(L) is the unique locally compact Hausdor↵ spacecontaining X as a dense subspace such that {V : V 2 L} forms a basis of compactneighborhoods in M(L).
Corollary 0.6. Let X = Sp(A) as above. Then we have a homeomorphism M(L) ' X
an.
The corollary follows from the fact that Sp(A) is dense in M(A). We will now sketch theproof of this fact:
Lemma 0.7. Let k and A be as above. Then the natural map Sp(A) ! M(A) has a denseimage.
Proof. (Sketch) WLOG, let us assume that |k⇤| is dense. Then any point x0 2 M(A) has abasis of open neighborhoods of the form
U = {x 2 M(A) : |fi
(x)| < 1, |gj
(x)| > 1}
MATH 274: NON-ARCHIMEDEAN GEOMETRY 3
for i = 1, ...,m, j = 1, ..., n, and f
i
, g
j
2 A. Choose p
i
, q
j
2 k such that |fi
(x0)| < |pi
| < 1and |g
j
(x0)| > |qj
| > 1. Let B = AhTi
, S
j
i/(pi
T
i
� f
i
, g
j
S
j
� q
j
). Then we have
M(B) ⇠= {x 2 M(A) : |fi
(x)| |pi
| and |gj
(x)| � |qj
|}.In particular, we have M(B) ✓ U . Now B 6= 0 implies that there exists M 2 Sp(B) (it is
an exercise to show this). The image of the canonical map
Sp(B) ! Sp(A) ! M(A)#
M(B)
lies in U and contains the point corresponding to M . ⇤
In the next exercise, we will make the homeomorphism M(L) ' X
an more explicit bygiving a correspondence between between maximal filters and points of Xan.
Exercise: Let k and A be as above. Then there is a bijection M(A) $ M(L) sendingF 2 M(L) to the element | · |
F
, where for f 2 A, we let |f |F
= infV 2FkfkV . Note that if V is
a rational subdomain of X, then kfkV
= supx2V |f(x)|, or if ⇡ : A ! A
V
= B is the naturalmap then kfk
V
= k⇡(f)k. In the other direction, we send | · |x
2 M(A) to the maximal filter
F
x
= {V 2 L : V contains a rational domain R = R(f0; f1, ..., fs) such that |fi
|x
|f0|x8i}.Let’s illustrate this correspondence in the case of the unit disc.
Example 0.8. Let k be algebraically closed, and let X = Sp(khT i) ' B
1 = k
0. LetX
an = M(khT i).
Type 1 points: x $ a 2 k corresponds to F = {V : a 2 V }.
Type 2 and 3 points: x $ D = D(a, r) corresponds to F = {V : V ◆ D
0 for some D
0 =D\ [s
i=1 D(ai
, r
i
)�} where D(ai
, r
i
)� denotes the open disc. Note that in type 3, D0 won’titself be a rational domain.
Type 4 points: x $ D1 ) D2 ) · · · with \Dn
6= ;. Then x corresponds to F = {V : V ◆D
n
for some n}.
Math 274: Non-Archimedean Geometry
Sarah Brodsky and Melody Chan
February 22, 2012
Today, we will address the question “where are the type V points?”. Throughout, k = k is complete withrespect to a nonarchimedean valuation. Our running example shall be X = B1 = Sp(khT i), equipped withthe special G-topology: admissible opens are special subsets and admissible covers are covers with a finiterefinement by special sets. A prime G-filter is a collection F of special subsets such that:
• ; 62 F
• If U, V 2 F , then U \ V 2 F
• If U 2 F with U 0 � U , then U 0 2 F
• If V1 [ · · · [ Vn 2 F , then some Vi 2 F
In our example X = B1, the special sets are just finite unions of a�noid domains of the form D\ q D�i ,
where D ⇢ D(0, 1) is a closed disc and D�i ⇢ D are open discs.
It is an exercise to show that all rational domains have the above form. Then by the Gerritzen-Grauerttheorem, all a�noid domains have this form as well.
Here is an example of a type V point, i.e. a prime filter that is not maximal. Let a 2 k0, i.e. |a| 1. Thenlet us define
F := {V : V ◆ {z 2 B1 : r0 < |z � a| < 1} for some 0 < r0 < 1}
1
We have that F is a prime filter (it is an exercise to prove this). Furthermore, F is properly contained inthe type II maximal filter associated to the Gauss point, namely
FGauss := {V : V ◆ D(0, 1)\qD(ai, ri)� for some ai, ri}.
For example, V = D(0, 1)\D(0, 1)� 2 FGauss but V is not in the prime filter F above. Therefore, FGaussis the unique maximal filter containing F .
Similarly, for each open disc D(a, 1)� ⇢ D(0, 1), we get an associated “type V” point of the Huber adicspace P(X); each such point is associated to the Gauss point.
We remark that the tangent directions at ⇣Gauss are in 1 to 1 correspondence with closed points of A1k,
which are in 1 � 1 correspondence with open discs D(a, 1)� ⇢ D(0, 1). For more general points ⇣ = ⇣a,r oftype II corresponding to D(a, r), we have
{tangent directions at ⇣} ! {closed points of P1k}
! {open discs D(a0, r)� ⇢ D(a, r)} [ {D(0, 1)\D(a, r)}.
For each such ⇣ of type II and each tangent direction at ⇣, we obtain an associated type V point of Huberadic space.
What is the topology on the space P(X) of prime filters? Again, let X = B1, k = k.
Figure 1: Topology on P(X)
Figure 2: Topology on M(X)
The point ⇣Gauss is closed inM(X), whereas inXad = P(X), its closure is ⇣Gauss[{associated type V points}.Recall that P(X) is quasicompact but not Hausdor↵. The map ⇢ : P(X) ! M(X) is continuous, with a
2
natural section M(X) ! P(X). In fact, M(X) is the maximal Hausdor↵ quotient of P(X). Furthermore,we have a homeomorphism
⇢�1(⇣Gauss in (B1)an) ' A1k
considered as schemes.
3
Prime filters and semivaluations
Michael Daub
March 8, 2012
Let X = Sp(A), where A is an a�noid algebra over k, a complete non-trivally valued non-archimedianfield. In this lecture, we aim to show the equivalence between the set of prime filters P(X) on the specialG-topology of X and Huber’s adic space of continuous semivaluations on A. We begin with a few definitions.
Definition 1. Let K be a field. A valuation ring of K is an integral domain R with field of fractions K
such that for all x 2 K, either x 2 R or x
�1 2 R. The height or rank of a valuation ring R is its Krulldimension, and is denoted by ht(R).
Remark. The ring of integral elements of k is a valuation ring of height 1.
Definition 2. Let (A, || · ||) be a Banach algebra over k with || · || extending the norm on k. Choose anelement ⇡ 2 k
0 with 0 < ||⇡|| < 1. A semivaluation on A is a pair (p, B), where p 2 Spec(A) and B is avaluation ring of Frac(A/p). A semivaluation (p, B) is continuous if the following conditions hold:
1. The image of A0 in Frac(A/p) is contained in B.
2.T1
n=1 ⇡nB = 0.
It is easily checked that the notion of continuity is independent of the choice of ⇡. Additionally, a continuoussemivaluation (p, B) with B a height 1 valuation ring is equivalent to giving a bounded multiplicativeseminorm on A. Indeed, the valuation ring B gives rise to an equivalence class of norms on Frac(A/p). ForB with ht(B) = 1, condition (2) above is equivalent to ⇡ lying in the maximal ideal of B, so there is a uniquenorm | · |B in this equivalence class satisfying |⇡|B = ||⇡||. Then we define | · |(p,B) by setting |a|(p,B) = |a|B .Clearly, this is a multiplicative seminorm on A, so it remains to show that | · |(p,B) is bounded.
It is an easy exercise to show that the boundedness of | · |(p,B) is equivalent to showing that the projectionmap (A, || · ||) ! (A/p, | · |B) is continuous. Let B(a, r) be an open ball of radius r > 0 about a 2 A/p. Wemay assume r < 1. Additionally, since the map A/p ! A/p given by x 7! x � a is a homeomorphism, wemay assume a = 0. Thus, we are reduced to showing that there is some r
0> 0 such that ||a|| < r
0 implies|a|B < r. Choose n such that |⇡n|B < r. Then we can take r
0 = |⇡n|B . Indeed, if ||a|| < r
0, then
||⇡�na|| ||⇡�n|| · ||a|| = |⇡�n|B · ||a|| = |⇡n|�1
B · ||a|| = (r0)�1||a|| < 1.
This shows that ⇡�na 2 A
0, and so by (1) we have ⇡�na 2 B. It follows that |⇡�n
a|B 1, in other words|a|B |⇡n| = r
0< r.
On the other hand, if | · | is a bounded multiplicative seminorm on A, we can construct a height 1 continuoussemivaluation by setting p := ker(| · |) and B to be the valuation ring of Frac(A/p) determined by | · |. Since| · | is bounded, for a 2 A
0 we have |a| ||a||, and so a is power bounded under | · |. But the set of powerbounded elements of A/p under | · | is precisely B, so (1) holds. Also, |⇡| ||⇡|| < 1, which is equivalent to(2) as stated above. Therefore, (p, B) is continuous. It is clear that these two constructions are inverses ofeach other, and so we have established a bijection
{height 1 continuous semivaluations} $ {bounded multiplicative seminorms}.
1
We wish to generalize this bijection to the set of all continuous semivaluations on A, which we will denote byVal(A). Previously, we found a bijection between bounded multiplicative seminorms and the set of maximalfilters M(X) on X when A is an a�noid k-algebra, so a natural extension to consider is a bijection betweenVal(A) and the set of prime filters P(X), again when A is a�noid. First, given P 2 P(X), we can constructa continuous semivaluation as follows. Define OP := lim�!V
AV , where the limit is taken over all V 2 P
corresponding to rational subdomains of A. We can put a seminorm on OP by setting ||f ||P := infV 2P ||f ||V .Finally, set O0
P := lim�!VA
0V .
Lemma 1. 1. OP is a local ring with unique maximal ideal
MP = {f 2 OP : ||f ||P = 0} ✓ O0P .
2. k
0P := O0
P /MP is a valuation ring of kP := OP /MP .
3. Let p = ker(A ! kP ), : Frac(A/p) ! kP , and B :=
�1(k0P ). Then (p, B) is a continuoussemivaluation on A.
Proof. We provide a brief sketch. For complete details, see [F-vdP]. For (1), let f 2 OP , and chooseV 2 P such that f 2 AV . If f 62 MP , then ||f ||P > 0. Thus, there exists ✏ > 0 in
p|k⇥| such thatV✏ := {x 2 V : |f(x)| � ✏} 2 P . It follows that f is invertible in V✏, and hence f is invertible in OP .
To prove (2), consider f 2 kP with f 62 k
0P . Choose V 2 P such that f 2 AV . Then {x 2 V : |f(x)| 1} 62 P
since f 62 k
0P . If we write
V = {x 2 V : |f(x)| 1} [ {x 2 V : |f(x)| � 1},
then we conclude that {x 2 V : |f(x)| � 1} 2 V since P is a prime filter. It follows that f�1 2 k
0P .
Finally, for (3) note that if : F ! K is an extension of fields, and R is a valuation ring of K, then �1(R)is a valuation ring of F . Therefore B is a valuation ring of Frac(A/p). The image of A0 in OP is containedin O0
P , and thus in B. Finally, if f 2 O0P is contained in \1
n=1⇡nO0
P , then ⇡�n
f 2 O0P for all n and thus
||⇡�nf ||P is bounded. So ||f ||P = 0, showing that \1
n=1⇡nOP is contained in MP . Thus \1
n=1⇡nk
0P = 0,
implying \n=1⇡nB = 0.
Given a prime filter P 2 P(X), we can construct a continuous semivaluation, as in (3) of the lemma. Next,given a continuous semivaluation (p, B), we can associate to it the prime filter
P =
⇢special V ✓ X : V ◆ R(f0; f1, . . . , fr) with fi/f0 2 B
for some choice of f0, . . . , fr generating the unit ideal
�.
Theorem 1. 1. The above association gives a bijection between P(X) and Val(X).
2. Let M(X) denote the maximal filters on X. Under this bijection, M(X) corresponds to the set ofcontinuous height 1 semivaluations on A.
3. For any continuous semivaluation (p, B) on A, ht(B) dim(A) + 1.
4. The retraction map P(X) ! M(X) is given by (p, B) 7! (p, B0), where B
0 = Bq and q =p⇡B.
Proof. See [F-vdP].
2
We conclude with a concrete example of type V points on the unit disk B1k. For any r 2 p|k⇥| with
0 < r 1, define � := R>0 ⇥ �
Zr with the unique total order satisfying r
0< �r < r for any r
0< r. For any
a 2 k, set D� := D(a, r)�. Then we can define a continuous height 2 valuation on khT i by�����
1X
n=0
an(T � a)n
�����D�
:= maxn
|an|�nr .
Let B := {f 2 Frac(khT i) : |f |D� 1}. Then it is easy to check that (0, B) defines a continuoussemivaluation on khT i, giving rise to a type V point on B1
k.
References
[F-vdP] J. Fresnel, M. van der Put, Rigid Analytic Geometry and Its Applications, Progress in Mathematics,218, Birkhuser Boston, Inc., Boston, MA, 2004.
3
Math 274: Non-Archimedean Geometry
Andrew Niles
February 27, 2012
The Reduction Map
Let A be a commutative, non-Archimedean Banach algebra. Recall that we
have the spectral seminorm ⇢ on A, defined by
⇢(f) = lim
n!1np||fn|| = max
x2M(A)|f(x)|
for any f 2 A. We also write ⇢(f) as ||f ||sup.
Recall that A� denotes the subring of A consisting of its power-bounded ele-
ments:
A� := {f 2 A : ||f ||sup 1} = {f 2 A : 9 M s.t. ||fn|| M 8 n > 0}.
A� contains an ideal A�� consisting of the topologically nilpotent elements of A:
A�� := {f 2 A : ||f ||sup < 1} = {f 2 A : ||fn||! 0 as n !1}.
Note that A�� is not necessarily a maximal ideal in A�. We define the reduction
of A to be the quotient ring
eA := A�/A��.
The construction of
eA is functorial: if � : A! B is a bounded homomorphism
of commutative non-Archimedean Banach algebras, then � sends A� to B� and
A�� to B��, so we get an induced ring homomorphism
e� :
eA! eB.
We will now construct a canonical map
M(A) ! Spec(
eA).
Given an element x 2 M(A) with completed residue field H(x), we have a
corresponding character �
x
: A ! H(x). With respect to the norm on H(x)
1
inherited from A, �
x
is a bounded homomorphism of Banach rings. So as above,
we get an induced homomorphism
e�x
:
eA! ]H(x),
where
]H(x) denotes the residue field of H(x). In particular, the kernel of e�x
is
a prime ideal in
eA, so this defines a canonical reduction map
⇡ : M(A) ! Spec(
eA)
x 7! ker(e�x
).
Example 1. LetA = khT i (k complete, non-Archimedean, algebraically closed),
so M(A) is the Berkovich analytification of the unit ball. More precisely,
A =
� 1X
n=0
a
n
T
n
: a
n
2 k, |an
|! 0 as n !1 .
The norm on A is the sup norm: ||P a
n
T
n|| = max
n�0{|an
|}.In this case, A� = {P a
n
T
n
: a
n
2 k
�, |a
n
| ! 0}. Thus for
Pa
n
T
n 2 A�, we
have a
n
2 k
��for n >> 0, so we conclude that
eA is the polynomial ring
ek[T ],
where
ek is the residue field of k.
Thus Spec(
eA) is the a�ne line over
ek, and we have the reduction map ⇡ :
M(A) ! A1ek
. Working through the definitions, we can see that this map sends
the Gauss point ⇣Gauss to the generic point of A1ek
. If x 6= ⇣Gauss, then there
exists some a 2 k
�such that |T � a|
x
< 1, and it turns out that ⇡(x) = ea 2 ek,
viewing
ek as the set of closed points in A1
ek
.
Now we recall some facts about a�noid algebras. These may be found in [BGR].
Throughout, let k be a complete, non-Archimedean field (possibly trivially val-
ued).
Recall that a strictly k-a�noid algebra is a quotient of some Tate algebra T
n
=
khT1, ..., Tn
i by some ideal (which is automatically closed), equipped with the
residue norm induced by the sup norm on T
n
. The residue norm is not canonical,
but any two presentations of an algebra as a quotient of a Tate algebra yield
equivalent residue norms. For the rest of today’s lecture, A will denote a strictly
k-a�noid algebra.
Proposition 1. (1)
eA is a reduced, finitely generated
ek-algebra.
(2) If A is reduced, then the spectral seminorm ⇢ is a norm, and is equivalent
to the given Banach norm ||·|| on A.
(3) A is Noetherian.
(4) (“Maximum modulus principle for strictly k-a�noid algebras”) For any f 2A, ⇢(f) 2 |kalg| = p|k|.
2
Proof. For (1), cf. [BGR] §1.2, Prop. 7 and §6.3, Cor. 3. For (2) and (4), cf.
[BGR] §6.2, Prop. 4. For (3), cf. [BGR] §6.1, Prop. 3.
Proposition 2. (1) If f1, ..., fr
2 A� and
eI = h ef1, ...,
ef
r
i ⇢ eA, then for the
reduction map ⇡ : M(A) ! Spec(
eA) we have
⇡
�1(V (
eI)) = {x 2M(A) : |f
i
(x)| < 1 8 i}.
(2) If f 2 A� \ A��, then
⇡
�1(D(
ef)) = {x 2M(A) : |f(x)| = 1}.
(3) If f 2 A� \ A��, then ⇡
�1(D(
ef)) is a nonempty, compact subset of M(A).
Proof. The first two statements are left as easy exercises. To see that ⇡
�1(D(
ef))
is nonempty for f 2 A� \A��, note that for such an f we have |f |sup = 1, so by
the maximum modulus principle there is some x 2M(A) with |f(x)| = 1.
Corllary 1. The reduction map ⇡ : M(A) ! Spec(
eA) is anti-continuous: the
inverse image of an open set is closed.
Proof.
eA is Noetherian, so any open subset of Spec(
eA) is a finite union of sets
of the form D(
ef). In particular the inverse image of any open set is compact
by the previous lemma, hence closed.
Lemma 1. If p 2 Spec(
eA) is a minimal prime ideal, then ⇡
�1(p) is closed and
nonempty.
Proof. Since p is minimal, we have
\
ef /2p
D(
ef) = {p},
hence
⇡
�1(p) =
\
ef /2p
⇡
�1(D(
ef)).
This is an intersection of closed sets, hence closed.
For any finite subset I ⇢ eA \ p,
\
ef2I
⇡
�1(D(
ef)) = ⇡
�1(D(
Y
ef2I
ef)),
which is closed and nonempty. Since M(A) is compact, it follows that the entire
intersection \
ef /2p
D(
ef)
is nonempty.
3
Next time we will see that ⇡ : M(A) ! Spec(
eA) is surjective, and that the
inverse image of a minimal prime p is a single point. The preimage �(A) ⇢M(A) of all the minimal prime ideals in Spec(
eA) is called the Shilov boundary
of M(A).
4
Math 274: Non-archimedean geometry
Ralph Morrison
February 29, 2012
1 Reminders from Last Time
Let k be as usual, and let A be a strictly k-a�noid algebra. We define eA := A0/A00; this is a finitely
generated k algebra. For x 2M(A), we have a map
�
x
: A! H(x),
inducing a mape�
x
: eA! ]H(x).
This allows us to define the map⇡ : M(A) ! Spec(A)
by⇡(x) = ker(e�
x
) 2 SpecA.
Example 1. If A = k hT i, then⇡ : (B
k
)an ! A1ek
sends the Gauss point to the generic point, and sends x 2 (D(a, 1)�)an to a, a closed point.
2 Properties of ⇡
Theorem 1 (Berkovich). Let A be a strictly k-a�noid algebra.
(1) ⇡ : M(A) ! Spec⇣
eA⌘
is surjective.
(2) For each generic point x 2 Spec⇣
eA⌘
(that is, each point corresponding to a minimal prime), ⇡
�1(x)is a singleton {x}. We call x the Shilov point corresponding to x.
To prove (1), we will use the following result from [BGR].
Proposition 1. The mapMax(A) ! Max
⇣eA⌘
is surjective. (This map is defined since if x 2 Max(A), then H(x) is a finite extension of k; hence ker(e�x
)is a maximal ideal.)
1
Proof of Part (1) of the Theorem. Note that if K/k is any valued field extension, then we have the commu-tative diagram
M(AK
) //
✏✏
Spec⇣
eAK
⌘
✏✏
M(A) // Spec⇣
eA⌘
.
We claim that for any point x 2 eX := Spec( eA), there exists a complete non-Archimedean field extension
K/K and a closed point y of Spec⇣
eAK
⌘lying over x. Once we’ve shown this, the claim follows from the
surjectivity of Max(A) ! Max⇣
eA⌘.
In order to prove the claim, it su�ces to prove the following two statements:
(a) Given x 2 eX, there exists K/k such that the natural map Spec( e
A⌦k
eK) ! Spec(A) has a closed point
in the fiber over x.
(b) For every K/k, the natural map Spec( eA
K
) ! Spec( eA⌦
k
eK) is surjective.
Statement (a) can be proved by induction as follows. It is easy to find an extension F/k such that
Spec( eA⌦
k
F ) ! Spec( eA)
has a closed point in the fiber over x; in particular, take F = k(x) to be the residue field of x. Since F
is finitely generated over k, we may invoke the following general fact: If F/k is any finitely generated fieldextension, then there exists a complete valued field extension K/k such that e
K
⇠= F as extensions of k.(Proof: It su�ces to treat separately the cases (i) F/k finite and (ii) F = k(T )/k purely transcendental oftranscendence degree 1. Case (i) is an easy exercise, and case (ii) follows by letting K be the completedresidue field of the Gauss point of M(khT i).)To establish (b), we claim that the natural map e
A ⌦k
eK ! e
A
K
is finite and injective (and thereforeSpec( e
A
K
) ! Spec( eA ⌦
k
eK) is surjective by the going-up theorem). Since e
A
K
is of finite type over eK
and a homomorphism of K-algebras is finite if and only if it is integral and finite type, it su�ces to provethat the natural injective map B := A
� ⌦k
�K
� ! A
�K
is integral (since tensoring with eK then gives an
integral map eA ⌦
k
eK ! e
A
K
). By ([BGR], 6.1.2/1(ii)), there is a finite injective K-algebra homomorphismKhX1, . . . ,Xd
i ! A
K
such that K
�hX1, . . . ,Xd
i is contained in B, and by ([BGR], 6.3.4/1) the inducedmap K
�hX1, . . . ,Xd
i ! A
�K
is integral. It follows that A
�K
is integral over B. This completes the proof.
Proof of Part (2) of the Theorem. Let x 2 eXgen, where e
X = Spec( eA). Last time we proved that ⇡
�1(x) isnonempty. (This is also implied by part (1).)
Let x 2 ⇡
�1(x). Without loss of generality, A is reduced, as this doesn’t change the spectrum topologically.
For a first case, we’ll assume that eA is an integral domain (i.e. there exists a unique generic point of eX).
We claim that for all f 2 A, |f(x)| = ||f ||sup. (This uniquely determines x.)
We always have |f(x)| ||f ||sup, so without loss of generality we may take ||f ||sup > 0. Then there existsa 2 k
⇥ and n 2 N such that ||f ||nsup = |a|, since ||f ||sup 2p|k⇥|. Let g = a
�1f
n, so that ||g||sup = 1. Hence
2
g 2 A0 \A00, so g 6= 0. As eA is an integral domain, g(x) 6= 0. So x 2 ⇡
�1(D(g)), implying |g(x)| = 1, givingus ||g||sup = |g(x)|. Since eA is a domain, || · ||sup is multiplicative, so we may “unscale” this equality to get||f ||sup = |f(x)|.Now for the general case (meaning eA might not be an integral domain). Choose f 2 A0 such that f 2 eAvanishes at all poitns of e
Xgen except for x.
Let B = A hT i /(1� Tf). Then
• B is a (strict) k-a�noid algebra,
• eB is eAf
, an integral domain, and
• M(B) is canonically homeomorphic to ⇡
�1(D(f)) ⇢M(A) and ⇡|M(B) = ⇡M(B).
Hence we have reduced to the previous case, completing the proof.
Remark 1. A reasonable question that arises in the above proof is whether or not A is an integral domainif and only if eA is an integral domain. The answer is no. For an example, we may consider the a�noidsubdomain {1 |z| r}, where r > 1 is in the value group. In this case we have that A is an integraldomain (as an a�noid algebra), but eA yields two irreducible components (essentially looking like two linesintersecting at a point). This example will be worked out in more detail in the next set of notes.
3
Math 274: Non-archimedean geometry
Ralph Morrison
March 2, 2012
1 Surjectivity in the Diagram From Last Time
Let k be as usual, and let A be a strictly k-a�noid algebra. Let K/k be a valued field extension. Last timewe considered the commutative diagram
M(AK
) //
✏✏
Spec⇣
eAK
⌘
✏✏
M(A) // Spec⇣
eA⌘
.
and proved that the horizontal arrow ⇡ : M(A) ! Spec⇣
eA⌘
is surjective.
Remark 1. The map M(AK
) !M(A) is also surjective.
Proof. Let L be a valued field extension extension of k, and let A! L be given. Choose L
0 extending bothL and K (for the existence of L
0, see the notes from February 8). This gives a map from A to L
0 and fromK to L
0, inducing a map AK
! L
0 since AK
= Ab⌦k
K. This gives us the desired surjectivity.
Ab⌦k
K = AK
))L
//L
0
A
OO 99ssssssssssssk
OO
//K
OO
2 Adding Adics to the Picture
Let A be as before. Let X = Max(A), and X = Max( eA). We have a map red : X ! X, which is surjective(due to Tate). Now, X = Max(A) ⇢M(A) = M(X) ⇢ P (X), where M(X) denotes the maximal filters onL, the lattice of special subsets in X, and where P (X) is the prime filters on L. Now, we have a projectionmap P (X) ⇡�! P (X) which sends a prime filter F on X to F , where U 2 F if and only if red�1(U) 2 F .By a well-known result (cf. [FvdP]), P (X) ⇠= e
X := Spec( eA). This is summarized in the following diagram.
1
X = Max(A)� _
✏✏
red // // Max( eA) = X
M(A)
id
✏✏M(X) � � //
P (X) ⇡ //P (X) ⇠= e
X = Spec( eA)
By an argument in [FvdP], the map X
ad = P (X) ! P (X) = Spec( e
A) is continuous and surjective. Byrestriction of ⇡ to M(X), we also get a map X
an = M(X) ! Spec( eA):
X
ad // // Spec( eA)
X
an?�
OO ::uuuuuuuuu
However, we come across the curiosity that while X
ad ! Spec( eA) is continuous, X
an ! X
an is anticontin-uous. This occurs because M(X) has the quotient topology (induced by the surjection P (X) ⇣ M(X)),rather than the subspace topology. Also interesting is the fact that X
an (not just X
ad) surjects onto Spec(A).
3 Shilov Boundaries
Let ⇡ : M(A) ! Spec( eA) be the restriction of ⇡ to M(X) = M(A), and let � := ⇡
�1({generic points}).From last time, each generic point has one point in its preimage, so � is a finite set. We call � the Shilov
boundary of M(A).
Theorem 1. For every f 2 A,sup
x2M(A)|f(x)| = max
x2�|f(x)|.
Moreover, � is the unique minimal such subset of M(A) with this property (meaning that � is the Shilovboundary).
Proof of the first claim. If ||f ||sup = 1, then there exists x 2 e
Xgen such that f(x) 6= 0. Let x = ⇡
�1(x). Then|f(x)| = 1 = ||f ||sup, establishing the claim in this case. By scaling f to have norm 1 and then unscaling (asin the previous set of notes), this implies the general case.
Example 1. Let X = (B1)an = M(k hT i), so that e
X = Spec(A1k
) and � = {the Gauss point}. By thetheorem, each function attains its maximum at the Gauss point.
Example 2. Let X be the closed annulus r |x| 1, where r 2p
|k⇥| and 0 < r < 1. Explicitly, X is
{x 2M(k hT i)�
� |T |x
1, |T |x
� r} = M(k hT, Si /(TS � r)).
We have A = k hT, Si /(TS � r), and eA = k[T, S]/(TS). eA has two irreducible components, giving us that� has two points. These correspond to the inner and outer boundaries of the annulus.
In general, the set{x 2M(A)
�
� |fi
(x)| r
i
, |gj
(x)| � s
j
}i,j
2
(with i and j taking on finitely many values, and the r
i
’s and s
j
’s are in the value group) corresponds to
M�
A⌦
r
�1X, sY
↵
/(Xi
� f
i
, g
j
Y
j
� 1)�
,
whereA
⌦
r
�11 T1, ..., r
�1n
T
n
↵
=n
f =X
a
⌫
T
⌫
�
� |a⌫
r
⌫ |! 0 as ||⌫||! 0o
.
3
Math 274: Non-Archimedean Geometry
Andrew Niles
March 5, 2012
A�noids in the sense of Berkovich
In the classical setting of [BGR], [Tate], etc., one works solely with (strictly)k-a�noid algebras of the form khT1, ..., Tni/a, or equivalently of the form
khr�11 T1, ..., r
�1n Tni/a
with r1, ..., rn 2 p|k⇤|, where k is a complete non-Archimedean field whichis non-trivially valued. (Recall that khr�1
1 T1, ..., r�1n Tni denotes the algebra of
power seriesP
aITI with |aI |rI ! 0 as |I|!1.)
Berkovich generalizes this in two ways. First, one allows the possibility that k istrivially valued; second, one allows quotients as above for arbitrary r1, ..., rn 2R>0. For the rest of today, k will denote an arbitrary complete non-Archimedeanfield.
Definition 1. A Banach k-algebra A is a�noid if there exists an admissiblesurjection khr�1
1 T1, ..., r�1n Tni ⇣ A for some r1, ..., rn 2 R>0. A is strictly
k-a�noid if we can choose r1, ..., rn 2p|k⇤|.
Here a surjective homomorphism from a Banach k-algebra onto A is consideredadmissible if the Banach norm on A is equivalent to the residue norm comingfrom the given surjection.
Example 1. Let r 2 R>0. We define
Kr :=n
1X
i=�1aiT
i : ai 2 k, |ai|ri ! 0 as |i|!1o
.
Kr is a (commutative) Banach k-algebra with a multiplicative norm || · || definedby
�
�
�
�
�
�
X
aiTi�
�
�
�
�
�
:= maxi2Z
{|ai|ri}.
1
Fact: Kr is a (usually non-strict) k-a�noid algebra, and if r /2 p|k⇤| then Kr
is a field.
In fact, let x be the type III point of A1Berk,k corresponding to the closed disk
D(0, r).
Exercise: Kr⇠= H(x), the completed residue field of x.
To verify that Kr is k-a�noid, one defines a homomorphism
khr�1T1, rT2i/(T1T2 � 1) ! Kr
by sending T1 7! T and T2 7! T
�1, and then one checks that this is an iso-morphism. (Note that if instead we had r 2 p|k⇤|, then the algebra defined onthe left-hand side above would correspond to the circle of radius r centered atthe origin. For r /2 p|k⇤| the rigid space corresponding to this algebra has nopoints in the classical sense.)
Why is Kr a field? One can show that the norm || · || defined above is theunique bounded multiplicative norm on Kr, and then use this fact to deducethat Kr
⇠= H(x).
More generally, given r1, ..., rn 2 R>0 whose images in R>0/p|k⇤| are linearly
independent over Q, we have a k-a�noid algebra
Kr1,...,rn := Kr1b⌦k · · · b⌦kKrn .
This is a complete non-Archimedean field with value group
|K⇤r1,...,rn
| = h|k⇤|, r1, ..., rni.
This immediately yields:
Lemma 1. For every k-a�noid algebra A, there exists a (non-trivially valued)complete non-Archimedean field extension K = Kr1,...,rn of k such that Ab⌦kK
is strictly K-a�noid.
Proof. Choose an admissible surjection khr�11 T1, ..., r
�1n Tni ⇣ A and let K =
Kr1,...,rn .
Proposition 1. Let A be a k-a�noid algebra. Then A is Noetherian and everyideal a ✓ A is closed.
Proof. By the above lemma and Tate’s corresponding theorems for strict a�-noids, it su�ces to show that if r /2 p|k⇤| and Ab⌦kKr is Noetherian with allof its ideals closed, then the same holds for A.
Identify A with its image in Ab⌦kKr, and let a ✓ A be an ideal. Then the ideala(Ab⌦kKr) of Ab⌦kKr is finitely generated, say by elements f1, ..., fm. Withoutloss of generality we may assume each fi lies in a.
2
Let f 2 a. Then there exist g1, ..., gm 2 Ab⌦kKr such that f =Pm
i=1 figi. Write
gi =1X
j=�1gijT
j
with each gij 2 A and ||gij ||rj ! 0 as |j| ! 1. Since ||T || = r /2 p|k⇤|, we infact have f =
Pmi=1 gi0fi. Therefore a is generated by f1, ..., fm, so we conclude
that A is Noetherian.
This also shows that a = A \ a(Ab⌦kKr). Both A and a(Ab⌦kKr) are closed inAb⌦kKr, so a is closed in A.
Next time, we will define k-a�noid subdomains of M(A) for arbitrary k-a�noidalgebrasA, and we will see that an analogue of Tate’s acyclicity theorem holds inthis more general setting: given a finite cover of M(A) by a�noid subdomains,the corresponding Cech complex is exact and admissible.
3
Lecture Notes: Math 274, Mar 7, 2012
Doosung Park
March 8, 2012
1 A�noid domains
Definition 1. Let k be a complete non-Archimedean field (possibly trivially valued), A a�noid k-algebra,
X = M(A). A subset U ⇢ M(A) is a k-a�noid subdomain if there exists a bounded homomorphism
A ! AU of k-a�noid algebras with
M(AU ) ! U ⇢ M(A)
such that whenever A ! B is a bounded homomorphism of k-a�noid algebras with
M(B) ! U ⇢ M(A),
there exists the unique bounded homomorphism AU ! B such that the diagram
A B
AU
commutes.
Remark 1. Any k-algebra homomorphism A ! B between strictly k-a�noid algebras is bounded ([BGR],
6.1.3/1). This is not necessarily true in the non-strict case (cf. [Berk] Remark. 2.1.13).
Remark 2. The definition implies that AU and A ! AU are unique up to isomorphism.
Remark 3. The definition does not immediately imply that M(AU )⇠! U , but this is true. The idea of the
proof is to reduce to the strict case, which is proved in [BGR]. In order to do this reduction, we need to use
the surjectivity of M(AU b⌦kK) ! M(AU ) (cf. [Berk], Prop. 2.2.4).
Exercise 1. Weierstrass (resp. Laurent, rational) subdomains are k-a�noid subdomains.
Definition 2. A strict k-a�noid subdomain of a strictly k-a�noid space is defined similarly but with A ! B
(bounded) homomorphism of strictly k-a�noid algebras and AU strict.
Remark 4. It is not clear a priori that a strictly k-a�noid subdomain is a k-a�noid subdomain. But, it
is true! To prove this, one needs the Gerritzen-Grauert theorem, which says that every (strictly) k-a�noid
subdomain of a (strict) k-a�noid space is a finite union of (strict) rational subdomains. (It is an exercise that
rational subdomains have a stronger universal property). For the moment, we assume the Gerritzen-Grauert
theorem.
Remark 5. Another consequence of the Gerritzen-Grauert theorem is that if U ⇢ M(A) is a (strictly)
k-a�noid subdomain and x 2 U = M(AU ) ⇢ M(A), then HM(A)(x) and HM(AU )(x) coincide. Temkin
found an independent proof of this fact not using the Gerritzen-Grauert theorem, and he uses this to give a
new proof of the Gerritzen-Grauert theorem.
1
2 Tate’s acyclicity theorem
Proposition 1. Let X be a Banach space over k, K = Kr as in the last lecture where r /2 p|K⇥|.1. We have an isomorphism of K-Banach spaces
X
b⌦kK⇠=
{f =
1X
i=�1xiT
i | xi ! X, kxikri ! 0 as i ! ±1}
where the norm of the second algebra is defined by kfk = supi kxikri.2. The canonical map X ! X
b⌦kK is an isometric embedding.
3. A sequence of bounded homomorphisms X
f! Y
g! Z of k-Banach spaces is exact and admissible, i.e.,
ker (g) = im(f), and the norm k · kY on ker (g) is equivalent to the quotient seminorm on im (f) induced by
k · kX , if and only if
X
b⌦kK ! Y
b⌦kK ! Z
b⌦kK
is exact and admissible.
Next time: Given a finite covering of a k-a�noid space X = M(A) by k-a�noid subdomains Vi, the
corresponding
ˇ
Cech complex
0 ! A !Y
i
AVi !Y
ij
AVi\Vj ! · · ·
is exact and admissible.
2
Math 274: Non-Archimedean Geometry
Andrew Niles
March 12, 2012
The structure sheaf of a Berkovich space
Recall that in algebraic geometry, to define the structure sheaf OX
of an a�ne scheme X = Spec(R), we set
OX
(D(f)) := R
f
for any f 2 R, where D(f) = {p 2 X : f /2 p}. The open sets D(f) form a basis for the Zariski topology onX, and one proves that this definition extends uniquely to give a sheaf O
X
on X.
In rigid and Berkovich geometry, this is more complicated. Let A be a k-a�noid algebra. Given f 2 A, thelocalization A
f
is not generally an a�noid algebra. We can replace it with the k-a�noid algebra
bAf
:= AhT i/(Tf � 1),
so M( bAf
) is the Laurent domain {x 2M(A) : |f(x)| � 1}. But it is insu�cient to only consider domains
of this form, because the canonical topology is too fine: the Laurent domains M( bAf
) for f 2 A do not givea basis for the canonical topology.
Instead, we need to consider more general Laurent domains of the form
V = {x 2M(A) : |fi
(x)| s
i
for i = 1, ...,m, |gj
(x)| � t
j
for j = 1, ..., n}
for f1, ..., fm, g1, ..., gn 2 A and a
i
, b
j
2 A with |ai
| = s
i
, |bj
| = t
j
. For such a Laurent domain V we set
AV
:= AhS1, ..., Sm
, T1, ..., Tn
i/(fi
� a
i
S
i
, g
j
T
j
� b
j
).
It is not immediately clear that AV
is independent of the choice of presentation of V . But one checks readilythat the natural map A! A
V
has the universal property in the definition of an a�noid domain, so indeedA
V
is canonically determined by V and the assignment V 7! AV
is functorial in the appropriate sense.
Remark 1. It will be important to work with rational domains, rather than only Laurent domains, for tworeasons:
1. Being a Laurent subdomain is not a transitive concept: cf. [BGR, §7.2.4] for an example where W isa Laurent subdomain of V and V is a Laurent subdomain of X, but W is not a Laurent subdomain ofX. In contrast, being a rational subdomain is a transitive concept.
2. The Gerritzen-Grauert theorem tells us that every strict k-a�noid domain is a finite union of strictrational domains. This statement is no longer true if we replace “rational” with “Laurent.”
1
Definition 1. Let A be a k-a�noid algebra and X = M(A). Let U ✓ X be open. We define
OX
(U) := lim �AV
,
where the inverse limit is taken over all special sets V ✓ U .
One checks easily that OX
is a presheaf on X. In fact it is a sheaf, which is proved using the following twofacts:
1. Tate’s acyclicity theorem (which implies OX
is a “G-sheaf”).
2. Exercise: If {Ui
}i2I
is an open cover of some open subset U ✓M(A), and V ✓ U is a special set, thenthere is a finite open covering {V1, ..., Vm
} of V with each V
i
contained in some U
ji for ji
2 I.
A straightforward formal argument, left as an exercise, shows that 1 and 2 imply OX
is a sheaf.
Remark 2. Last time we gave two examples of non-admissible covers. One was the cover
B1k
= Sp(A) = {z 2 k : |z| 1} =a
i
D(ai
, 1)�
for A = khT i, where the disjoint union is over a collection {ai
} of representatives of the di↵erent residueclasses. HereD(a
i
, 1)� = {z 2 k : |z�ai
| < 1}. We may extend this to the subset {x 2M(A) : |T�ai
|x
< 1}of A1
Berk,k, which is an open subset in the Berkovich topology. But this is no longer an open cover:
(B1k
)an = M(A) = {x 2M(A) : |T |x
1} 6=[
{x 2M(A) : |T � a
i
|x
< 1},
because none of the open subsets on the right contain the Gauss point.
2
Math 274: Nonarchimedean Geometry
Will Johnson
March 14, 2012
Remark 1. A k-a�noid space is defined to be a locally ringed space (X,OX
) where X = M(A), A is ak-a�noid algebra, and O
X
is the structure sheaf defined in the previous lecture. A morphism (X,OX
) !(Y,O
Y
) of k-a�noid spaces is defined to be a morphism of the locally ringed spaces which is induced by abounded homomorphism A ! B of the corresponding k-a�noid algebras. Thus the category of k-a�noidspaces is antiequivalent to the category of k-a�noid algebras. One can define the category of a�noid k-spacesanalogously. Finally, one can define a (global) k-analytic space to be a locally ringed space which is locallyisomorphic to a�noid spaces. This is the approach taken in Berkovich’s book, but it isn’t the “modern”definition, which we will discuss in a few weeks.
1 Relative Interior and Relative Boundary
Definition 1. Let � : Y ! X be a morphism of k-a�noid spaces Y = M(B) and X = M(A). The relative
interior Int(Y/X) is the set of all y 2 Y for which there is an admissible surjection Ahr�11 T1, . . . , r
�1n
T
n
i ⇣ B
sending T
i
7! f
i
2 B, such that |fi
(y)| < r
i
for all i.
The relative boundary @(Y/X) is defined to be Y \ Int(Y/X).
In other words, y 2 Y is in the relative interior of Y over X if Y can be admissibly embedded into X timesa polydisc, in such a way that y is sent to a point in the “naive interior” of the polydisc. (Here, the “naiveinterior” of a polydisc {~y : |y
i
| r
i
8i} is the set {~y : |yi
| < r
i
8i}. This isn’t the topological interior in anyreasonable sense.) The point is that giving an admissible surjection Ahr�1
1 T1, . . . , r�1n
T
n
i ⇣ B is equivalentto giving a closed admissible embedding M(B) ,! M(Ahr�1
1 T1, . . . , r�1n
T
n
i), and M(Ahr�11 T1, . . . , r
�1n
T
n
i)is M(A) times a polydisc.
Exercise 1. Int(Y/X) is an open subset of Y , and so @(Y/X) is a closed subset of Y .
A boundaryless morphism Y ! X is one for which @(Y/X) = ;. This notion will be important later - forexample the correct definition of a proper morphism of global analytic spaces will be one that is topologicallyproper, and boundaryless (once we extend the definition of relative interior and relative boundary to generalanalytic spaces). It will also make an appearance in Temkin’s Berkovich-theoretic proof of the Gerritzen-Grauert Theorem.
We now state a number of important properties and examples of relative interior and relative boundary,without proof:
1. If Y is an a�noid domain in X, and we consider the inclusion Y ,! X, then Int(Y/X) is just thetopological interior of Y as a subset of X. This is relatively straightforward to see in the case when Y
is a rational subdomain, and the general case follows by the Gerritzen-Grauert theorem.
1
We remark that the inclusion Y ,! X of Berkovich spaces shares features of both open and closedimmersions. Topologically, it is a map from one compact space to another, so it is a closed immersion.Algebraically, it acts more like a scheme-theoretic open immersion.
2. A morphism � : Y ! X of k-a�noid spaces is boundaryless if and only if � : Y ! X is finite, meaningthat A ! B makes B a finite A-module. This is interesting because being finite is a global condition,while the definition of relative boundary is more local in nature.
3. If X = M(A) and Y = M(B) are strictly k-a�noid spaces, then
Int(Y/X) = {y 2 Y |�y
(B) is finite over �x
(A) where x = �(y)}
where � : Y ! X is a morphism of k-a�noid spaces, �y
is the character corresponding to y, i.e., the
map �
y
: B ! H(y), and �
y
: b ! eH(y) is the reduction. The reduction map is functorial, giving acommutative diagram
A
// //
✏✏
�
x
(A)
✏✏
� � // eH(x)
B
// //�
y
(A) ��
// eH(y)
Saying that �y
(B) is finite over �x
(A) means that the dotted arrow is a finite morphism of k-algebras.
This gives a convenient way of determining the relative interior or boundary in the strict a�noid case.Using Definition 1 by itself, it is di�cult to prove that a point belongs to the relative boundary, sinceone must consider all possible embeddings of Y into X times a polydisc.
This fact also shows that finiteness of a map A ! B can somehow be checked pointwise.
Example 1. Let r < 1, let Y = B1(0, r) be the ball of radius r centered at 0, let X = B1(0, 1) be the unitball, and consider the inclusion Y ,! X. Then @(Y/X) is the topological boundary of Y in X. By viewingthe Berkovich spaces as trees, one can see that this topological boundary consists of a single point, the Gausspoint of Y , or equivalently, the type II or type III point associated to the disk of radius r centered at 0.
Definition 2. If Y = M(B) is a k-a�noid space, the absolute interior Int(Y ) and absolute boundary @(Y )are defined to be Int(Y/M(k)) and @(Y/M(k)), where Y ! M(k) is the map induced by k ! B.
4. If Y is strictly k-a�noid, then Int(Y ) = ⇡
�1(Max(B)) where ⇡ : M(B) ! Spec B is the reduction mapand Max(B) is the set of closed (maximal) points of Spec B. By contrast, the Shilov boundary �(Y )is the preimage of the generic (minimal) points of Spec B. So in general, �(Y ) ✓ @(Y ), with equalitywhen Y has dimension 1. Note that �(Y ) is always finite, but @(Y ) usually isn’t.
As an example, if Y is a strict closed annulus {r |z| R}, then @(Y ) = �(Y ) consists of 2 points:the Gauss point of the outer boundary and the Gauss point of the inner boundary. The skeleton ofthe annulus is the unique path between these two points.
Unlike much of what we’ve been doing, the concepts of relative interior and relative boundary have no clearanalogs in either scheme theory or classical rigid analysis.
2 Inner Homomorphisms
Inner homomorphisms are a generalization of the notion of relative interior and relative boundary. Let A bea k-a�noid algebra, B and C be Banach A-algebras with B an A-a�noid algebra, and let � : B ! C be a
2
bounded homomorphism of A-algebras. This yields diagrams
A
����B
� //C
and M(C)� //
$$
M(B)
zzM(A)
Definition 3. The morphism � is inner with respect to A if there exists an admissible surjection of BanachA-algebras
Ahr�11 T1, . . . , r
�1n
T
n
i ⇣ B
sending T
i
7! f
i
, such that ||�(fi
)||sup < r
i
for all i, where || · ||sup denotes the spectral norm on C, orequivalently, the supremum-norm on M(C).
The case where M(C) is a point corresponds to the previous notion of relative interior. Specifically, ifX = M(A) ! Y = M(B) is a morphism of k-a�noid spaces, then
Int(Y/X) = {y 2 Y |�y
: B ! H(y) is inner with respect to A}
We close by noting that if M(C) and M(B) are strict k-a�noids, then � : B ! C is inner with respect toM(k) if and only if Spec C ! Spec B is a constant map. We will observe this in the next lecture.
3
Math 274: Nonarchimedean Geometry
Will Johnson
March 16, 2012
Suppose A and B are k-a�noid algebras. Recall that a bounded homomorphism � : B ! C of BanachA-algebras is inner with respect to A if there is an admissible surjection1
Ahr�11 T1, . . . , r
�1n Tni ⇣ B
sending Ti 7! fi, such that ||�(fi)||sup < ri for all i, where || · ||sup is the spectral norm (supremum-norm)on C.
One case of interest is when � : B ! C = B is the identity map. Then � is inner with respect to A if andonly if A ! B is boundaryless, i.e., @(M(B)/M(A)) = ;. This is a consequence of Proposition 2.5.9 in[Berk], which says that a map M(C) ! M(B) is inner with respect to M(A) if and only if the image is inthe relative interior of M(B) over M(A).
Another special case of the notion of inner homomorphism is when C is H(y) for some y 2 M(B) and � is �y,the characterB ! H(y). Then �y is inner with respect toA if and only if y 2 Int(M(A)/M(B)). This followsfrom the definitions, since �(fi) = fi(y) and || · ||sup on H(y) is the absolute value, so ||�(fi)||sup = |fi(y)|.As a convention, “strict” will henceforth mean “strict and non-trivially valued.”
Theorem 1. Let A and B be strictly k-a�noid algebras. Let ' : A ! B be a bounded homomorphism ofa�noid algebras. Then the following are equivalent:
(1) ' is finite
(2) ' is finite
(3) idB is inner with respect to A.
As noted above, the third condition is also equivalent to ' being boundaryless. The equivalence of (1) and(2) is nontrivial, but proved in Theorem 6.3.5/1 of [BGR].
Berkovich generalized Theorem 1 to the following relative form, which is a component of Proposition 2.5.2in [Berk]:
Theorem 2. (Berkovich) Let A,B be strictly k-a�noid algebras, and � : B ! C be a bounded homomor-phism of A-algebras. Then � is inner with respect to A if and only if �(B) is finite over �(A) (as subringsof C).
Note that when � is the identity map, this yields the equivalence of (2) and (3) in Theorem 1. Also notethat �(B) is finite over �(A) if and only if it is integral, because the map �(A) ! �(B) is finite type.
1It can be shown using the Banach open mapping theorem, that when |k| is nontrivial, admissibility follows from surjectivity
and boundedness.
1
We will prove Theorem 1 (assuming [BGR] Theorem 6.3.5/1), since the proof of Theorem 2 seems to bemore di�cult. We first state a lemma, which is [BGR 6.1.1/4] and [Berk 2.1.5] for the nonstrict case:
Lemma 1. If ' : A ! B is a bounded homomorphism of k-a�noid algebras and r1, . . . , rn > 0, then givenf1, . . . , fn 2 B with ||fi||sup ri, there is a unique bounded homomorphism � : Ahr�1
1 T1, . . . , r�1n Tni ! B
such that �(Ti) = fi and �|A = '. Conversely, if � : Ahr�11 T1, . . . , r�1
n Tni ! B is given and fi = �(Ti),then ||fi||sup ri.
Given this, we can prove Theorem 1:
Proof (of Theorem 1). As noted above, the equivalence of (1) and (2) is Theorem 6.3.5/1 of [BGR]. We prove(3) =) (2) and (1) =) (3). For (3) =) (2), suppose that idB is inner with respect to A. Then there is anadmissible surjection ⇡ : Ahr�1
1 T1, . . . , r�1n Tni ⇣ B with ||⇡(Ti)||sup < ri for each i. Since
p|k⇥| is dense, wecan find ⇢i 2
p|k⇥| with ||⇡(Ti)||sup < ⇢i ri. By Lemma 1, there is a map ⇡0 : Ah⇢�11 T1, . . . , ⇢�1
n Tni ! Bsending Ti to ⇡(Ti). Since ⇡ factors through ⇡0, ⇡0 is also surjective, and therefore admissible. Replacing ⇡with ⇡0 and ri with ⇢i, we can assume without loss of generality that all the ri live in
p|k⇥|.Write rsii = |c�1
i | for some ci 2 k⇥. Define ⌫ : A0 = AhT1, . . . , Tni ! Ahr�11 T1, . . . , r�1
n Tni by ⌫(Ti) = ciTsii .
This uniquely determines ⌫ by Lemma 1. Moreover, ⌫ is a finite map; this follows from the proof of Theorem6.1.5/4 in [BGR].
Let ⌫0 = ⇡ � ⌫ : A0 ! B. The map ⇡ is finite because it is a surjection, so therefore ⌫0 is finite. By [BGR]6.3.5/1, the map ⌫0 : A0 = A[T1, . . . , Tn] ! B is finite. This implies that B is finite over ⌫0(A0). But⌫0(A0) = ⌫0(A) because ⌫0(Ti) = 0 because ||⌫0(Ti)||sup < 1 because ||⇡(Ti)||sup < ri. Thus A ! B is finite.
For the converse, we show that (1) implies (3), i.e., that if ' : A ! B is finite, then idB is inner with respectto A. Choose generators b1, . . . , bn for B as an A-module. Rescaling them by scalars, we can assume thatevery bi has spectral norm at most 1, so that bi 2 B�. Now let ci 2 k be any scalar with |ci| < 1. ByLemma 1 we can construct a map AhT1, . . . , Tni ! B by sending Ti 7! cbi. This map is clearly surjective,since the cbi generate B as an A-module. Since ||cbi||sup < 1, we see that idB is inner with respect to A.
This argument is hard to generalize to a proof of Theorem 2. If � : B ! C is a bounded morphism ofA-algebras, with C not necessarily k-a�noid, then we don’t know that � is finite if and only if � is. Togeneralize the implication (2) =) (3), we therefore can’t use the intermediate step (2) =) (1) =) (3), sincethe generalization of (2) =) (1) is probably false.
The rough idea of the proof is as follows. We need to show that if � : A ! B is finite, then there existsan admissible surjection ⇡ : AhT1, . . . , Tni ⇣ B with ||⇡(Ti)||sup < 1. The idea is to choose a set of integral
generators of B over A, then lift them to B�. Also, lift their defining polynomial relators to polynomialsover A�. This yields inequalities of the form ||pi(bi)||sup < 1, where the bi’s are the lifts of the generatorsand the pi(X)’s are the polynomials over A�. For each such bi, we use 1+deg p of the T ’s, sending T0 7! cb,T1 7! p(b), T2 7! bp(b) and so on up to Tn 7! bn�1p(b), where n = deg p. This map turns out to be surjective.Details can be found in Lemma 2.5.4 of [Berk].
2
Math 274: Non-Archimedean Geometry
Sarah Brodsky and Melody Chan
March 19, 2012
Let k be a complete nonarchimedean valued field. Last time, we discussed inner homomorphisms. Recall(see Proposition 2.5.9 in Berkovich) that if A and B are a�noid k-algebras and C is a Banach A-algebra,a bounded A-homomorphism � : B ! C is inner with respect to A if and only if the image of M(C) liesin Int(M(B)/M(A)). Furthermore, a bounded homomorphism � of strictly k-a�noid algebras is finite ifand only if the map id
B
is A-inner, if and only if for all y 2 Y = M(B), we have that �y
(B) is finite
over �x
(A), where x = �(y). Also, we have @(Y/X) = ; if and only if � is a finite morphism. Recall thatInt(Y/X) = {y 2 Y : �
y
: B ! H(y) is A-inner}. So id
B
is A-inner if and only if for all y 2 M(B), we havethat �
y
: B ! H(y) is inner.
Let Y = M(B) be a strict k-a�noid space. We defined the absolute interior of Y to be Int(Y ) =Int(M(B)/M(k)). Let ⇡ : M(B) ! Spec(B) be the canonical reduction map.
Proposition 1. Int(Y ) = ⇡
�1(Max(B)).
Proof. Given y 2 Y , we have y 2 Int(Y ) if and only if �y
(B) is finite over �x
(A) = k; here A = k. This
happens if and only if �y
(B) is a field. Note that �y
(B) = B/ker(�y
), so �y
(B) is a field if and only if ⇡(y)is a maximal ideal.
Functorial properties of Int(Y/X) (see Proposition 2.5.8 in Berkovich).
1. Suppose we have a cartesian diagram of a�noids
Y
0 //
0
✏✏
X
0
✏✏
Y
//X
obtained via the completed tensor product. That is, if Y = M(B), X = M(A), and X
0 = M(A0),then Y
0 = M(B⌦A
A
0) is the fiber product. Then
( 0)�1(Int(Y/X)) ✓ Int(Y 0/X
0).
2. Let K be an extension of k, and let : Y ⌦k
K ! Y be the base change map. Then
�1(Int(Y/X)) ✓ Int(Y ⌦K/X⌦K).
3. Given a composition Z
! Y
�! X, we have
Int(Z/X) = Int(Z/Y ) \ �1(Int(Y/X)).
1
Next, we turn to Temkin’s proof of the Gerritzen-Grauert theorem. Let X = M(A) be a strictly k-a�noidspace.
Theorem 2 (Gerritzen-Grauert). Every strict a�noid subdomain in X is a finite union of strict rational
domains.
Proof sketch. Let Y = M(B) be an a�noid subdomain of X = M(A) (all a�noid and rational domains arestrict for the rest of today). Our basic strategy is to prove the following:
(⇤) For each y 2 Y , there exists a rational domain X
0 ⇢ X such that
(1) Y
0 = X
0 \ Y is a neighborhood of y in Y , and
(2) @(Y 0/X
0) = ;.
Assuming (⇤), we claim that Y
0 is a rational subdomain of X. Given the claim, for each y 2 Y , we geta rational domain Y
0y
⇢ Y satisfying conditions (1) and (2), and the collection {Y 0y
} covers Y . Since Y iscompact, {Y 0
y
} has a finite subcover Y = Y
0y1
[ · · ·Y 0ym
, and we are done.
We follow a suggestion of Ducros to prove the claim. Let Y
0 = M(B0) and X
0 = M(A0). Since the mapY
0 ! X
0 is a boundaryless map of a�noids by (2), it is a finite map (Berkovich 2.5.13). Also, since Y
0 is ana�noid subdomain of X 0 (for the intersection of an a�noid domain and rational domain is a�noid), the mapY
0 ! X
0 is flat by 7.3.2/6 in [BGR]. In other words, we have a ring map A
0 ! B
0 which is finite and flat.Then the scheme map Spec(B0) ! Spec(A0) is both open (by flatness) and closed (by finiteness), hence theimage of Spec(B0) in Spec(A0) is a union of connected components in the Zariski topology. Thus, there existsan idempotent e0 2 A
0 such that Y 0 = {p 2 X
0 : e0(p) = 0}. (Thus e0 = 0 on the components correspondingto Y
0, and e
0 = 1 on the other components). Rewriting, Y 0 = {p 2 X
0 : |e0(p)| 12} is a Weierstrass, and
hence rational, subdomain of X 0 ⇢ X. Since “rational subdomain” is transitive, we conclude that Y
0 is arational subdomain of X.
2
Harmonic functions on M(Z) and global k-analytic spaces
Michael Daub
April 17, 2012
1 Harmonic functions on M(Z)
We conclude the discussion on harmonic functions from last time by describing a class of harmonic functionson M(Z). First, let us set some notation. We shall denote by (1, ✏) the point of M(Z) corresponding tothe seminorm | · |✏1, where | · |1 is the usual Archimedean absolute value on Z and 0 ✏ 1. For any primep, write (p, ✏) for the point of M(Z) corresponding to the seminorm | · |✏
p
, where | · |p
is the p-adic absolutevalue on Z satisfying |p|
p
= 1/p and 0 ✏ 1. Here | · |1p
is the seminorm on Z such that |p|1p
= 0 and|n|1
p
= 1 when p - n. Note that when ✏ = 0, all of these norms coincide.
Let f 2 Q⇥. Then we can define a function log |f | : M(Z) ! R [ {±1} by (v, ✏) 7! log |n|✏v
. Writing f = a
b
with gcd(a, b) = 1, then log |f | defines a map from U to R, where U = M(Z) \ {(p,1) : p | ab}.
Proposition 1. log |f | is harmonic on U .
Proof. For any fixed place v of Z, on the branch (v, ✏) the function takes the form log |n|✏v
= ✏ log |n|v
. Hence,the function is linear on each branch of M(Z). So it su�ces to verify that log |n| is harmonic at the trivialseminorm. The slope on the branch (1, ✏) is equal to log |n|1, while on any branch (p, ✏) the slope is log |n|
p
.Thus, by the product formula, the sums of the slopes on all branches of | · |0 is
X
v
log |n|v
= log
Y
v
|n|v
!= log(1) = 0.
2 Global k-analytic spaces
In this section, X will be a locally Hausdor↵ topological space. We start o↵ with a few definitions.
Definition 1. A quasinet on X is a collection ⌧ of compact Hausdor↵ subsets V ✓ X such that each x 2 X
has a neighborhood that is a finite union of elements of ⌧ . A quasinet is called a net if for all V, V 0 2 ⌧ ,{W 2 ⌧ : W ✓ V \ V
0} is a quasinet on V \ V
0.
Definition 2. A k-a�noid atlas on (X, ⌧) is an assignment of a k-a�noid algebra A
V
to each V 2 ⌧ alongwith a homeomorphism V
⇠��! M(AV
), which is functorial in the sense that if V, V 0 2 ⌧ and V
0 ✓ V , thenthere is a bounded homomorphism of k-a�noid algebras A
V
! A
V
0 identifying (V 0, A
V
0) with an a�noidsubdomain of (V,A
V
). A k-analytic space is a triple (X,A, ⌧) consisting of a topological space X, a k-a�noidatlas A, and a net ⌧ .
1
Remark. If all AV
are strict k-a�noid algebras, then (X,A, ⌧) is called a strict k-analytic space.
We conclude with a few examples.
Example 1. Let X = (A1k
)an, the multiplicative seminorms on k[T ] extending the absolute value on k. Forevery r > 0, let X
r
= {x 2 X : |T |x
r}. Restriction to k[T ] induces a homeomorphism
i
r
: M(khr�1T i) ⇠��! X
r
.
Let ⌧ = {Xr
}r>0. This forms a net on X, and X
r
7! khr�1T i is a k-a�noid atlas.
Example 2. A k-a�noid atlas on X = (P1k
)an is given by V1 = B(0, 1), V2 = X \ B(0, 1)� ' B(0, 1), andV3 = V1 \ V2 = B(0, 1) \B(0, 1)�. Notice that V1 [ V2 is a neighborhood in X of the Gauss point ⇠Gauss, butno V
i
is itself. This justifies the inclusion of finite unions in the definition of quasinet instead of requiringeach point to have a neighborhood in ⌧ . Of course, there is a net ⌧ on (P1
k
)an such that every point has aneighborhood in ⌧ , but the definition we are using allows for the use of more convenient nets, such as theone in this example.
Example 3. Let A be a k-a�noid algebra, and set X = M(A). Then we can put many di↵erent nets onX. In particular, ⌧ = {X} is a net with atlas X 7! A, and ⌧
0 = {V ✓ A : V a�noid subdomain} is a netwith atlas V 7! A
V
. We would like to form a category of k-analytic spaces in which (X,A, ⌧) and (X,A0, ⌧
0)are isomorphic. There is an obvious definition of morphism of k-analytic spaces, and next time we will seethat by inverting a class of these morphisms, we achieve the desired category.
2
Morphisms of k-analytic spaces and localizations of categories
Michael Daub
April 17, 2012
1 Morphisms of k-analytic spaces
Recall from last time that a k-analytic space is a triple (X,A, ⌧) where X is a locally Hausdor↵ topologicalspace, ⌧ is a net on X, and A is a k-a�noid atlas. That is, ⌧ is a quasinet, i.e. a collection {Vi} of compactHausdor↵ subsets of X such that every x 2 X has a neighborhood of the form Vi1 [ · · · [ Vin . Furthermore,⌧ is a net, meaning that for all Vi, Vj 2 ⌧ , {W 2 ⌧ : W ✓ Vi \ Vj} is a quasinet on Vi \ Vj . Finally, A is ak-a�noid atlas, associating a k-a�noid algebra AV for each V 2 ⌧ , a homeomorphism V
⇠��!M(AV ), andfor each V
0 ✓ V a bounded homomorphism AV ! AV 0 .
Example 1. Let D ✓ R2 be the unit disk. Consider the collection ⌧ = {�j}1j6, where �j is the closedsector bounded by the line segments connecting the origin to e
2⇡ij/6. Then ⌧ is a quasinet, but not a net.Indeed, the collection {W 2 ⌧ : W ✓ �j\�k} is empty for any j 6= k. If we let ⇢1 = �1\�6 and ⇢j = �j\�j�1
for 2 j 6, then the collection {�j , ⇢k, 0}1j,k6 forms a net.
Definition 1. A strong morphism ' : (X,A, ⌧) ! (X 0,A0
, ⌧
0) of k-analytic spaces is a continuous map' : X ! X
0 such that for every V 2 ⌧ there is a V
0 2 ⌧
0 with '(V ) ✓ V
0, together with a compatible systemof morphisms of k-a�noid spaces
'V/V 0 : (V,AV )! (V 0, AV 0)
for all V 2 ⌧ and V
0 2 ⌧
0 with '(V ) ✓ V
0.
Define ]k-An as the category with triples (X,A, ⌧) as objects and strong morphisms of k-analytic spaces as
morphisms. A strong morphism is called a quasi-isomorphism if ' : X⇠��! X
0 is a homeomorphism andeach 'V/V 0 identifies V with an a�noid subdomain of V 0. The category k-An of k-analytic spaces is the
localization of ]k-An with respect to the quasi-isomorphisms.
2 Localizations of categories
In this brief digression from k-analytic spaces, set-theoretic issues will be ignored, as they will not causeproblems in the case we are interested in. Let C be a category, and S a class of morphisms. We want todefine a new category C[S�1] having the following universal property: there is a functor Q : C ! C[S�1]satisfying
1. Q(s) is an isomorphism for all s 2 S.
2. Any functor F : C ! D such that F (s) is an isomorphism for all s 2 S factors uniquely through Q.
1
Although we will not prove it here, it is true that localizations always exist. If S is particularly nice, thenC[S�1] admits a concrete description. Namely, if S admits a “calculus of right fractions” in the sense of[GZ], then
• Ob(C[S�1]) = Ob(C), and
• Mor(A,B) consists of equivalence classes of diagrams
A
s � A
0 f�! B
with s 2 S and f 2 Mor(A0, B).
Furthermore, if f, g 2 Mor(A,B), then Q(f) = Q(g) if and only if there exists s 2 S such that f � s = g � s.
It is not hard to verify that ]k-An with S the quasi-isomorphisms admits a calculus of right fractions. A
much more di�cult fact due to Temkin is that the category of strictly k-analytic spaces is a full subcategoryof k-An.
References
[GZ] P. Gabriel, M. Zisman. Calculus of fractions and homotopy theory. Ergebnisse der Mathematik undihrer Grenzgebiete, Band 35 Springer-Verlag New York, Inc., New York 1967.
2
Math 274: Non-Archimedean Geometry
Andrew Niles
April 11, 2012
“Good” spaces
Definition 1. A k-analytic space X is good if every point x 2 X has an a�noid neighborhood. (Recall ingeneral that x only has a neighborhood which is a finite union of a�noid subdomains.)
Remark 1. 1. In Berkovich’s red book, he defined a category of analytic spaces over k which he latergeneralized (in his IHES paper). The spaces from his book turn out to be exactly the good spaces.
2. There are natural analytification functors
Schemes of finite type over k
Quasi-separated, quasi-paracompact rigidk-analytic spaces
k-An
k-An.
If X is a scheme (of finite type over k), then X
an is good. But if X is a rigid space, X
an is notnecessarily good.
Example 1. Let B(0, 1) = M(khS, T i) ✓ (A2)an be the closed unit polydisk, and let
B(0, 1)� = {x 2 B(0, 1) : |S|x
< 1, |T |x
< 1}
be the open unit polydisk. Let
X = B(0, 1) \ B(0, 1)�
= {x 2 B(0, 1) : |S|x
= 1 or |T |x
= 1}= {x 2 B(0, 1) : |S|
x
= 1} [ {x 2 B(0, 1) : |T |x
= 1}.
It turns out that X is not good. (We will see this using Temkin’s criterion. The idea is that the “reduction”of X is A2
ek
\ {0}, which is not an a�ne scheme.)
Analytic domains
Definition 2. • A subset V ✓ X is a k-analytic domain if there is a covering {Vi
} of V with each V
i
k-a�noid, such that {Vi
} is a quasi-net on V . (This is the analogue of an “admissible open” for thestrong G-topology in rigid analysis.)
1
• A covering of an analytic domain V by analytic domains Vi
is admissible if {Vi
} is a quasi-net on V .(This is the analogue of an “admissible covering by admissible opens” in rigid analysis.)
Remark 2. Every open subset of X is an analytic domain in a natural way, as is any finite union of a�noiddomains.
Remark 3. The category k-An admits fiber products. These are constructed similarly to fiber products ofschemes. The category of k-a�noid spaces is a full subcategory of k-An, and forX = M(A), Y = M(B), Z =M(C), and morphisms Y ! X,Z ! X, the fiber product is
Y ⇥X
Z := M(Bb⌦A
C).
Morphisms
Definition 3. A morphism f : Y ! X of k-analytic spaces is a closed immersion (resp. finite) if there isan admissible covering {X
i
} of X by k-a�noid domains such that Yi
= f
�1(Xi
) = X
i
⇥X
Y is k-a�noid forall i, and the corresponding homomorphisms of k-Banach algebras A
Xi ! A
Yi are surjective (resp. finite)and admissible. (If |k⇤| 6= 1, then admissibility is automatic by the Banach open mapping theorem.)
Exercise 1. The class of closed immersions (resp. finite morphisms) is closed under composition, and anybase change of a closed immersion (resp. finite morphism) is a closed immersion (resp. finite).
Definition 4. Let f : Y ! X be a morphism of k-analytic spaces. The relative interior of f is the setInt(Y/X) of all y 2 Y such that, for any a�noid U ✓ X containing x = f(y), there is an a�noid V ✓ f
�1(U)which is a neighborhood of y in f
�1(U), and y 2 Int(V/U) as previously defined for a�noids.
The relative boundary of f is @(Y/X) := Y \ Int(Y/X).
We say f is boundaryless if @(Y/X) = ;. (Berkovich would say f is “closed”.)
Remark 4. If @(X) = ;, then X is good.
Definition 5. A morphism f : Y ! X of k-analytic spaces is proper if it is boundaryless and topologicallyproper (i.e. the inverse image of every compact subset of X is compact).
2
Math 274: Non-Archimedean Geometry
Doosung Park
April 13, 2012
1 Properties of morphisms
Definition 1. Recall that a morphism f : Y ! X of k-analytic spaces is a closed immersion (resp. finite)if there exists an admissible covering of X by a�noid domains such that Y
i
= X
i
⇥X
Y is a k-a�noid andA
Xi ! AYi are surjective (resp. finite) and admissible. Note that the admissible condition is automatic if k
is non-trivially valued.
Definition 2. A morphism f : Y ! X is separated if � : Y ! Y ⇥X
Y is a closed immersion.
Definition 3. A morphism f : Y ! X is boundaryless if @(Y/X) = 0, where @(Y/X) = Y \Int (Y/X),and Int (Y/X) is defined by the set of points y 2 Y such that for any a�noid domain U ⇢ X containingx = f(y), there exists an a�noid neighborhood V ⇢ f
�1(U) of y such that y 2 Int (V/U).
Remark 1. If Y has no absolute boundary (i.e., @(Y/k) = ;), then Y is good.
Definition 4. A morphism f : Y ! X is proper if @(Y/X) = ; and f is proper as a map of topologicalspaces.
Facts 1. 1) Finite morphisms are proper.2) More generally, f : Y ! X is finite if and only if it is proper and has finite fibers.3) Properness implies separatedness.
Lemma 1. (Transitivity property of the relative interior) If Z ! Y
�! X are morphisms of k-analyticspaces, then
Int (Z/X) = Int (Z/Y ) \ �1(Int (Y/X)).
2 Application of the relative interior
Let k be a non-trivially valued field, A a strict k-a�noid algebra, X = Sp (A), Z = Sp (A/I) where I =(f1, . . . , fn) is an ideal in A, which is automatically closed. Suppose that Z ⇢ U ⇢ X where U is a finiteunion of a�noid subdomains of X.
Theorem 1. There exists ✏ > 0 such that
{x 2 X : |fi
(x)| ✏, . . . , |fn
(x)| ✏} ⇢ U.
Proof. Since X
an = M(A) is compact and Hausdor↵, it su�ces to prove that Uan is a neighborhood of Zan
in X
an. Since U
an is a k-analytic domain in X
an, (Uan)� = Int (Uan/X
an) where (Uan)� is the topological
1
interior of Uan in X
an. Since Zan ! X
an is a closed immersion, it is finite and therefore boundaryless. Thus,Z
an = Int (Zan/X
an). Now, by the transitivity property of the relative interior,
Z
an = Int (Zan/X
an) = Int (Zan/U
an) \ Int (Uan/X
an) ⇢ Int (Uan/X
an) = (Uan)�.
3 Analytification
We want to define a functor X X
an from the category of schemes of finite type over k to k-analytic spaces.
A�ne Case. If X = SpecA, where A = k[x1, . . . , xn
]/I for an ideal I, let Xan be the set of multiplicativeseminorms on A extending | · | on k with the weakest topology such that x 7! |f |
x
is continuous for all f 2 A.Let ⌧ = {X
r
}r>0, where X
r
= M(khr�1x1, . . . , r
�1x
n
i/I), AXr = khr�1
x1, . . . , r�1
x
n
i/I.General Case. If X is a scheme of finite type over k, let Xan be the set of x = (⇣, | · |
⇣
) such that ⇣ 2 X
and | · |⇣
is an extension of | · | on k to k(⇣). Note that this definition coincides with the definition in thea�ne case. The topology is the weakest one for which U
an ⇢ X
an is open for all open a�ne subsets U ⇢ X
and x 7! |f(x)|, Uan ! R is continuous for all f 2 OX
(U).
Next time. We will discuss the universal property of analytification.
2
Math 274: Non-archimedean geometry
Ralph Morrison
April 16, 2012
1 Analytification of Schemes
Let k be a non-archimedean, complete field. We are interested in a functor
{schemes of finite type/k} �! {good strictly k-analytic spaces}
sending X to X
an. This represents the functor
{good strictly k-analytic spaces}! Sets
defined byY 7�! {morphisms of locally ringed spaces (Y,O
Y
)! (X,OX
)}.In particular, one gets a canonical morphism of locally ringed spaces
(Xan
,OX
an)! (X,OX
).
Let X be a k-analytic space.
Fact 1. The association V 7! AV
(where V is an a�noid domain in X) extends uniquely to a G-sheaf OXG
mapping V 7! AV
with respect to the strong G-topology
{k-analytic domains, admissible coverings}.
The restriction of OXG to the open subsets U ⇢ X is denoted O
X
. It is a sheaf in the usual (topological)sense.
In practice, one only makes use of OX
when X is a good space.
Fact 2. If X is good at x, then H(x) (computed with respect to any a�noid domain containing x) iscanonically isomorphic to the completion of K(x), where K(x) is the residue field of the local ring O
X,x
.
This can fail for non-good spaces.
There is also an analytification functor
{quasicompact, quasiseparated rigid spaces/k}! {compact, Hausdor↵, strictly k-analytic spaces}
which “locally” sends Sp(A) to M(A).
Berkovich shows in his IHES paper that this is an equivalence of categories.
1
Remark 1. In this setting, a rigid space is quasicompact if it is an admissible finite union of a�noids (whereadmissible is automatic if the space is quasiseparated). A rigid space is quasiseparated if the intersection ofa�noids is a finite admissible union of a�noids.
More generally, assuming a local finiteness condition on both sides (instead of quasicompactness), one getsan equivalence of categories
{quasiseparated rigid spaces of finite type} �! {paracompact Hausdor↵ strictly k-analytic spaces}
Remark 2. Recall that finite type means that there exists a locally finite admissible covering by a�noidsubdomains, and paracompact means that every open cover has a locally finite subcover.
We have functors from {separated schemes of finite type/k} to each of the above categories. Adding it tothe diagram gives a commutative diagram of functors:
{quasiseparated rigid spaces of finite type} // {paracompact Hausdor↵ strictly k-analytic spaces}
{separated schemes of finite type/k}
OO 22ddddddddddddddddddddddddddddd
2 Properties of Sch! k-An
(1) A morphism f : Y ! X between schemes of finite type over k is a closed immersion (respectively finite,proper, separated, an isomorphism, etc.) if and only if f
an : Y
an ! X
an is.
(2) (A GAGA-type fact.) If X is proper then analytification induces an equivalence of categories
Coh(X,OX
)! Coh(Xan
,OX
an),
where Coh denotes the category of coherent sheaves (on the given locally ringed space). The functorX ! X
an is fully faithful on proper schemes.
Remark 3. For a locally ringed space (Y,OY
), an OY
-module is coherent if it is locally isomorphicto the cokernel of a morphism of locally free O
Y
-modules of finite rank. (This definition is equivalentto the characterization found in such texts as Hartshorne’s Algebraic Geometry.)
(3) – X
an is Hausdor↵ if and only if X is separated.
– X
an is compact if and only if X is proper.
– X
an is (path) connected if and only if X is connected. (Xan connected implies that X
an path-connected.)
2
Math 274, Apr 18 & 20, 2012:
Reductions of Germs of Analytic Spaces
Andrew Dudzik
1 Germs of analytic spaces and Riemann-Zariski spaces
We let Germsk
denote the localization of the category of pointed strictly k-analytic spaces (X,x) with respectto the morphisms ' : (X,x) ! (Y, y) which induce an isomorphism from X to a neighborhood of y.
Henceforth, when we write (X,x), we mean its image in this localization, which we call the germ of X at x.
Definition 1. A germ (X,x) is good if x has an a�noid neighborhood.
Let k be a field (for the moment, it is not necessary to assume that k is the residue field of k). Let K/k bea field extension. We will define an associated Riemann-Zariski space associated to K/k as follows.
As a set, RZ
k
(K) = {valuation rings k ⇢ O ⇢ K with fraction field K}. We give RZ
k
(K) the weakesttopology for which {O
��f 2 O} is open for every f 2 K.
Fact 1. (Zariski) RZ
k
(K) is a quasicompact and quasiseparated topological space.
We also define a category bir
K
of birational spaces over k. The objects of this category are triples (X,K,�),where:
• X is a connected, quasicompact and quasiseparated topological space
• K is a field extension of k
• � : X ! RZ
k
(K) is a continuous map which is a local homeomorphism, i.e. for all x 2 X, � induces ahomeomorphism between an open neighborhood of x and an open subset of RZ
k
(K).
The morphisms (X,K,�) ! (Y, L, ) in bir
K
are pairs (h, i), where h : X ! Y is continuous, i : L ! K is
a k-homomorphism, and the following diagram commutes:
X Y
RZ
k
(K) RZ
k
(L)
h //
�
✏✏
✏✏i
#//
1
2 Classical Motivation
Let V ar
k
be the category of pointed k-varieties (here a variety is an integral scheme of finite type), whose
objects are maps Spec(K) ! Z, where K/k is a field extension, Z is a variety over k, and morphisms arecommutative diagrams:
Spec(K 0)Z
0
Spec(K)Z
//
f⌘
✏✏
fs
✏✏//
Let Bir
k
be the localization of this category with respect to the family of maps:
B := {(f⌘
, f
s
)��f
⌘
is an isomorphism and f
s
is proper}
Fact 2. B admits a calculus of right fractions.
For any f : Spec(K) ! X in V ar
k
, we can construct a birational space over k consisting of all pairs (O, g),with O 2 RZ
k
(K), and g : Spec(O) ! X a morphism such that the following diagram commutes:
Spec(K)X
Spec(O)
f //
✏✏
g
??
(note that when X is separated, g is unique)
Fact 3. The above map defines a functor V ar
k
! bir
k
inducing an equivalence of categories Bir
k
! bir
k
.
3 Reduction of an analytic germ
Now, assume that k is the residue field of k. We sketch the definition of the reduction functor Germsk
! bir
k
.
Let X be a paracompact, Hausdor↵, strictly k-analytic space, and X
ad the corresponding adic space. Interms of earlier lectures, we can interpret Xad (as a topological space) as the space of prime filters of specialsubdomains of X. Then there is a retraction map r : Xad ! X.
Take a point x 2 X. For any ⌫ 2 r
�1(x), which can be regarded as a Krull valuation on H(x) with valuation
ring O⌫
, we have k
o ⇢ O⌫
⇢ H(x)o, so it makes sense to consider the reduction fO⌫
(which is a valuation
ring over k with fraction field ]H(x)).
Definition 2. We define the reduction (X,x) of a germ (X,x) to be the birational space (X,K,�), where
X = r
�1(x), K = ]H(x), and � : X ! RZ
k
(K) maps ⌫ to fO⌫
.
4 Germ subdomains
Definition 3. A subdomain of a germ (X,x) is an equivalence class of pairs (Y, x) with Y ⇢ X an analyticsubdomain containing x, where (Y, x) ⇠ (Y 0
, x) if Y \ Y
0 is a neighborhood of x in both Y and Y
0.
2
Theorem 1. (Temkin) The reduction functor Germsk
! bir
k
establishes a bijection between germ subdo-
mains of (X,x) and quasicompact open subsets of the space X associated to (X,x). This bijection preservesinclusions, finite unions, and intersections.
Definition 4. A birational space (X,K,�) is a�ne if � is injective and its image is of the formRZ
k
(K)[f1, . . . , fn] :={O 2 RZ
k
(K)��f
i
2 O for all i} for some f
i
2 K.
Theorem 2. 2 The germ (X,x) is good if and only if (X,x) is a�ne.
Definition 5. A morphism Y ! X of analytic spaces is boundaryless (or inner) at y 2 Y if y 2 Int(Y/X).
A map f : Y ! X of birational spaces over k is separated (resp. proper) if Y ! X ⇥RZk(K) RZ
k
(K 0) isinjective (resp. bijective).
As an absolute version of the last definition, we note that (X,K,�) is separated (resp. proper) if � is injective(resp. bijective).
Theorem 3. A morphism Y ! X of k-analytic spaces is separated (resp. boundaryless) at y 2 Y i↵
f : (Y, y) ! (X,x) is separated (resp. proper).
(Recall that for an a�noid space X = M(A), x 2 Int(X) i↵ red(x) 2 Max(A) is a closed point.)
Corollary 1. f : Y ! X is proper i↵ f is topologically proper, and f
y
: (Y, y) ! ^(X, f(y)) is proper for ally 2 Y .
Example 1. Let B(0, 1) = {|z| 1} = M(khT i). Then red(B(0, 1)) ⇠= Spec(k[T]) ⇠= A1k.
If x is a type II point, x 6= ⇣
Gauss
, then x 2 Int(X), and (X,x) ⇠= P1k
, the entire Riemann-Zariski space. On
the other hand, if x = ⇣
Gauss
, then x /2 Int(X), and (X,x) ⇠= A1k
, which is not the entire Riemann-Zariskispace.
3
Math 274: Non-Archimedean Geometry
Doosung Park
April 23, 2012
1 Formal schemes
Let k be a nontrivially vauled complete non-archimedean field. All k-analytic spaces today are strict. LetR = k
�, and fix ⇡ 2 R with 0 < |⇡| < 1. We have an algebra
Rhx1, . . . , xn
i :=(
X
I2Nn
a
I
X
I | |ai
| 1 8I and |aI
| ! 0 as kIk ! 1).
A topoligically finitely presented R-algebra is an R-algebra A of the form
A = Rhx1, . . . , xn
i/I
with I = (f1, . . . , fn) a finitely generated ideal. An admissible R-algebra is a topologically finitely presentedalgebra which is flat over R (equivalently, ⇡-torsion free). If A is a topologically finitely presented R-algebra,we set X = Spec (A/⇡A) as a topological space. This is a finite type scheme over k. For f 2 A, we set
X
f
= {x 2 X : f(x) 6= 0},
and let A{f} = AhXi/(1� fx).
Exercise 1. The ring A{f} depends only on X
f
and not on f .
Theorem 1. The assignment Xf
A{f} uniquely extends to a sheaf OA of R-algebras on X. The stalksare local rings.
Definition 1. A (topologically finitely presented) a�ne formal scheme Spf(A) over R is a locally ringedspace of the form (X,OA). A (topologically finitely presented) formal scheme over R is a locally ringed space(X,OX) with OX a sheaf of R-algebras, which is locally isomorphic (over R) to an a�ne formal R-scheme.
2 Raynaud’s generic fiber functor
We have a functorSpf(A) Sp(A⌦
R
k) or M(A⌦R
k),
which carries Zariski open subsets to quasi-compact admissible opens in Sp(A) and compact special sets inM(A⌦
R
k). This construction extends canonically to a functor
{topologically finitely presented formal R-schemes} ! {rigid spaces over k}or ! {k-analytic spaces}
defined by X Xk
or Xan.
1
Proposition 1. This functor is faithful on the admissible formal R-schemes.
Given a formal scheme X, we have specialization (reduction) maps sp : Xk
! X where X is the scheme overek underlying X, and sp : Xan ! X. The second map is defined by the following: for x 2 Xan, there is a map
�
x
: M(H(x)) ! Xan. This gives rise to �
�x
: Spf(H(x)�) ! X, so we get fXx
: Spec(]H(x)) ! X.
Example 1. If Ak
is a reduced a�noid algebra, then A�k
is an admissible R-algebra. However, when X
is a gluing of Aan and Ban along Uan, it may be impossible to glue Spf(A�k
) and Spf(B�k
) along Spf(U�k
).Raynaud solved this problem by using admissible formal blowing-ups.
Theorem 2. (Raynaud) Every paracompact Hausdor↵ k-analytic space X admits an admissible formalmodel, i.e., an admissible formal scheme X such that Xan ⇠= X, and every morphism of such space is inducedby a morphism of admissible formal models.
3 Temkin’s reduction functor
Let X be a strict k-analytic space, x 2 X, X 0 a compact neighborhood of x, and X0 a formal model for X 0.
Define (X,x) to be the class of Spec(]H(x)) ! (X0, sp(x)) in Bire
k
.
Fact 1. This only depends on (X,x), not on X
0,X0, and gives a functor. This functor agrees with the
reduction functor which was defined previously using adic spaces.
Definition 2. A scheme model for an object (morphism) in birek
is an object (morphism) of Varek
in thecorresponding class.
Theorem 3. (Temkin) 1) A birational space is a�ne if and only if it admits an a�ne scheme model.
2) A morphism of birational spaces is separated (resp. proper) if and only if it admits a separated (resp.proper) scheme model if and only if every scheme model is separated (resp. proper).
Theorem 4. (Temkin, Conrad) Let X,X0, x 2 X as before. The following are equivalent:
1) x is a good point,
2) (gX,x) is a�ne (i.e., has an a�ne scheme model),
3) The normalization S of the Zariski closure of sp(x) in X0is proper over an a�ne e
k-scheme of finite type,
4) �(S,OS
) is finitely generated over ek and S ! Spec(�(S,OS
)) is proper.
Corllary 1. If S is quasi-a�ne but not a�ne, then x is not a good point.
Example 2. Let X = B2k
/(B2k
)�, ⇣ Gauss point of X ⇢ B2k
, X the complement of 0 in Spf(Rht, t0i). ThenX = A2
ek
� {0} is quasi-a�ne but not a�ne, and sp(⇣) is the generic point of X. By 4) ! 1), ⇣ is not good.
2
Math 274: Non-Archimedean Geometry
Andrew Niles
April 25, 2012
Structure of analytic curves (following [BPR, §5])
Assume for simplicity that k is an algebraically closed, non-trivially valued field. Let X be a smoothconnected algebraic curve over k, not necessarily proper, and let b
X be the (unique) smooth proper curve
birational to X. Let D = bX \X, the set of “punctures” or “points at infinity” (so D is a finite set of closed
points).
Theorem 1. There exists a finite set V of type 2 points in X
an such that bX
an \ V is a disjoint union ofopen balls and finitely many open annuli, such that each point of D is contained in a distinct such openball. Equivalently, Xan \ V is a disjoint union of open balls, finitely many open annuli, and finitely manypunctured open balls, with the punctures corresponding bijectively to the points of D.
Terminology: We call such a set V a semistable vertex set of X.
Idea of proof: By the semistable reduction of Bosch and Lutkebohmert, there exists a semistable (formal)
model X/R for bX, i.e. an (admissible) formal scheme X/R with Raynaud generic fiber b
X
an and special fiber
Xek a semistable curve over ek (a reduced separated one-dimensional scheme of finite type over ek whose only
singularities are ordinary double points). We have the reduction map red : bX
an ! Xek; recall it is surjectiveand anticontinuous. From Berkovich and Bosch/Lutkebohmert, we have:
• If ⇣ is a generic point of Xek, then red�1(⇣) = {pt}.
• If z 2 Xek is a smooth point, then red�1(z) is isomorphic as a k-analytic space to the open unit ballB(1)� (a similar fact holds in arbitrary dimensions).
• If z 2 Xek is an ordinary double point, then red�1(z) is isomorphic as a k-analytic space to an openannulus.
Remark 1. By suitable admissible formal blowups, one can arrange that all points of D reduce to distinctsmooth points.
Remark 2. One can also prove the above structure theorem for analytic curves without using the semistablereduction theorem, and then deduce the semistable reduction theorem as a result (cf. Temkin, Ducros,Thuiller).
Recall that any open annulus A is isomorphic to a standard open annulus
{z 2 A1Berk : r < |z| < 1}
for a unique r < 1. We call � log(r) the modulus of A.
1
Under such an isomorphism, the set{z = ⇣0,↵ : r < ↵ < 1}
is intrinsic to the annulus A; we call it the skeleton ⌃(A) of A. We can parametrize it by � log(↵), and thisallows us to identify ⌃(A) with the open interval (0,� log(r)).
For the punctured ball B = {z 2 A1Berk : 0 < |z| < 1}, we define the skeleton of B to be
⌃(B) = {⇣0,↵ : 0 < ↵ < 1}.
Via the parameter � log(↵), we may identify ⌃(B) with the open ray (0,1).
The skeleton ⌃ = ⌃(Xan, V ) of V is the metric graph obtained as
V t�a
A
⌃(A)�t�a
B
⌃(B)�⇢ X
an,
with the disjoint unions taken over the open annuli A and punctured open balls B of the structure theoremfor analytic curves.
Fact: ⌃ is connected.
Example 1. X = P1 \ {0, 1,1}, V = {⇣Gauss}.
⌃ =
1
01
Example 2. X = E an elliptic curve.
Case 1: E has good reduction.
⌃ = (point)
Case 2: E has bad reduction.
⌃ = (circle with length = log |jE |)
Example 3. E an elliptic curve with bad reduction, X = E \ E[2].
Case 1: k has residue characteristic 6= 2.
⌃ =
2
Case 2: k has residue characteristic 2.
⌃ =
Proposition 1. Let X,V be as above. Then
⌃(X,V ) = {x 2 X
an : x does not have an a�noid neighborhood isomorphic to B1k and disjoint from V }.
3
Math 274: Non-archimedean geometry
Ralph Morrison
April 27, 2012
1 Retraction to the Skeleton
Let X/k be a smooth algebraic curve as in the previous lecture. Let V be a semistable vertex set of Xan.Let ⌃ denote the skeleton ⌃(Xan, V ). We can define a retraction
⌧V
: Xan ⇣ ⌃
as follows. For x 2 Xan \ ⌃, there exists a unique connected component Bx
of Xan \ ⌃ containing x, withB
x
isomorphic to an open ball such that @Bx
2 ⌃. Then we define
⌧V
(x) := @Bx
.
Theorem 1 (Berkovich). The map ⌧V
is continuous and induces a homotopy equivalence.
Theorem 2. The natural mapXan �! lim
�V
⌃(Xan, V )
is a homeomorphism, where the limit is over semistable vertex sets V .
Definition 1. For x 2 Xan, set g(x) = 0 if x is not of Type II, and if x is of Type II, set g(x) to be thegenus of the unique nonsingular projective curve over k with function field ]H(x).
Theorem 3 (Genus formula).g(X) = g(⌃(Xan, V )) +
X
x2X
an
g(x),
where g(⌃(Xan, V )) is just dimR H1
(⌃, R).
Example 1. If g(X) = 1, there are two possibilities: either the genus comes from a Type II point, or itcomes from genus in the skeleton. These possibilities are pictured in Figure 1.
Definition 2. The Euler characteristic of X is
�(X) = 2� 2g(X)�#D,
where D = X \ X.
Definition 3. A semistable vertex set V is stable if there’s no x 2 V with g(x) = 0 and deg⌃
(x) < 3.
Theorem 4 (Stable Reduction Theorem). If �(x) < 0, then there exists a unique stable vertex set for X.If �(x) 0, then there exists a unique set-theoretically minimal skeleton of X.
1
g(⌃) = 1g(x) = 1
x
Figure 1: The two possibilities when g(X) = 1, from Example 1.
Example 2. First we’ll consider the case of g = 0. If D = X \ X has only two points, then �(X) = 0 andwe’re in the weaker case of having a unique set-theoretically minimal skeleton of X, rather than a uniquestable vertex set. The minimal skeleton in this case is the line segment connecting the two points of D, say0 and 1, which is not a stable vertex set. However, if D has at least three points, then the stable reductiontheorem says there will be a stable vertex set.
Now consider g = 1, so we’re considering X an elliptic curve. As argued in the previous lecture, in the caseof good reduction, the skeleton is simply a point. However, for bad reduction, the skeleton ⌃ is a loop, andthe minimal semistable vertex set is any Type II point on ⌃.
Once g � 2, there’s no need to mark any points whatsoever, since we’re guaranteed to have �(X) < 0regardless of how many points are in D.
2 Non-Archimedean Picard Theorems
Back in the Archimedean world, we have the following theorems.
Theorem 5 (Picard’s Little Theorem). A nonconstant entire function on C omits at most one complexvalue.
Theorem 6 (Picard’s Moderately-Sized Theorem). Let X be a Riemann surface with �(X) < 0. Then anyanalytic map C ! X is constant.
Proof. Since �(X) < 0, the analytic universal cover ⌦ of X is the open unit disk. By the standard propertiesof universal covers, our map C ! X lifts as pictured:
⌦
✏✏C //
>>~
~
~
~
X
The map C ! ⌦ is a bounded analytic function since ⌦ is the unit disk, and is thus constant, implying thatthe map C ! X is constant as well.
2
Berkovich proved a non-Archimedean analogue of Theorem 6.
Theorem 7 (Berkovich). Let k be a complete, nontrivially valued non-Archimedean field. Any analyticmap f : A1 ! X where X is a (smooth) analytic curve with �(X) 0 is constant. (Here the �(X) 0 isnot strict, unlike in the Archimedean case.)
To prove this, we’ll need a definition and some facts.
Definition 4. We say that X is totally degenerate if g(X) = g(⌃), that is, g(x) = 0 for all x 2 X.
Facts:
(1) X has a topological (and analytic) universal cover ⌦. (In particular, ⌦ is a topological cover endowedwith a unique analytic structure such that any map from a simply connected space to X factors through⌦.)
(2) (Due to Tate.) If g(X) = 1 and X is totally degenerate, then
⌦ ⇠= Gan
m
= (P1)an \ {0,1}.(3) (Due to Mumford.) If g(X) � 2 and X is totally degenerate, then
⌦ ⇠= (P1)an \ L,
where L ⇢ P1(k) is a nonempty perfect set.
Proof of Berkovich’s Theorem. Without loss of generality, we’ll assume k = k.
We now split into three cases.
• First o↵, if g(X) = 0, then all we need to show is that an entire nonconstant function A1 ! A1 mustbe surjective. This may be shown using Newton polygons.
• If Xan is totally degenerate, then ⌦ ⇢ (P1)an \ L where L ⇢ P1(k), with |L| � 2. This means we mayview ⌦ ⇢ (A1)an \ {0}. By the universal property of ⌦, we obtain the following diagram:
⌦
✏✏(A1)an
f
//
<<x
x
x
x
x
X
The map (A1)an ! ⌦ ,! (A1)an \ {0} ,! (A1)an is not surjective and so must be constant (by the firstcase). Hence the map (A1)an ! Xan is constant.
• If we’re not in one of the above cases, then there exists some x 2 Xan with g(x) � 1. This means thatx is not in the image of f , since if f(y) = x, we would have an injection eH(x) ,! eH(y) of a functionfield of genus � 1 into a function field of genus 0, which is impossible.Let U
1
, U2
be distinct open balls in Xan \ {x}. Now, f((A1)an) is path connected, but any path froma point of U
1
to a point of U2
must go through x. Hence some Ui
= U is disjoint from im(f).Choose y 2 U in X(k). By Riemann-Roch, we may pick a rational function h with poles only at y.Consider the bounded map h � f : A1 ! Xan \ U
h! P1 \ {1} ⇠= A1. It follows that h � f is constant,implying in turn that f is constant.
3