ponti cia universidad javeriana facultad de ciencias

Pontificia Universidad JaverianaFacultad de CienciasDepartamento de Matematicas

Commutative Algebra and some results in AlgebraicGeometry

Daniel Felipe Luque DuqueAdvisor: Beatriz Grana Otero

Bogota - ColombiaNoviembre, 2016

Contents

Acknowledgements iii

Introduction iv

1 Commutative Algebra 11.1 Rings and Homomorphisms . . . . . . . . . . . . . . . . . . . 11.2 Ideals and Radicals . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Modules and Sequences . . . . . . . . . . . . . . . . . . . . . 91.4 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Valuation Rings . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Varieties 172.1 Noetherian Rings . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Affine Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3 Grobner Bases . . . . . . . . . . . . . . . . . . . . . . . . . . 302.4 Primary Decomposition . . . . . . . . . . . . . . . . . . . . . 35

3 Dimension 413.1 Krull Dimension . . . . . . . . . . . . . . . . . . . . . . . . . 413.2 Infinite Dimensional Noetherian Rings . . . . . . . . . . . . . 47

Bibliography 50

ii

My most heartfelt gratitude goes to every person that helped me through theprocess of writing this thesis.

To Miller Martınez, for teaching me at a young age the joy of numbersand mathematics.

To Beatriz Grana, for guiding me throughout the process of writing thisthesis.

And to my family, for always supporting my decisions and being thereall the way.

To all of you, thank you.

iii

Introduction

The goal of this thesis is to provide an introduction to the study of affinealgebraic varieties in their natural position within algebraic geometry. Thisbranch of mathematics is relatively modern, and it was developed throughseveral steps. Although, its starting point can be traced back to Riemann’swork on complex analysis, the study of one-dimensional complex manifolds,which are now called Riemann surfaces. In the early 20th century, Germanmathematician Emmy Noether made a historical contribution to abstractalgebra by defining the class of rings that now bears her name, which alsoprovided a crucial way to connect algebra and geometry. Another greatcontribution to this branch was made by David Hilbert, another Germanmathematician with interests in several branches of mathematics. He left,amongst his extraordinary legacy to the field, the Nullstellensatz, whichestablishes a correspondence between algebraic sets and ideals in a ring ofpolynomials over an algebraically closed field.

Naıvely speaking, algebraic geometry is the study of geometry usingalgebraic techniques, but being more accurate algebraic geometry is thestudy of zero loci of systems of polynomials in affine or projective spaces.These zero loci are what we refer to as algebraic sets. When algebraic setsare irreducible in a topological sense, they are called algebraic varieties andthey are the main object of study in algebraic geometry. Thus, algebraicgeometry developed mostly within an algebraic framework. This allowed thetheory to develop independent of the choice of a particular ambient space.

One of the latest breakthroughs in algebraic geometry was made byFrench mathematician Alexander Grothendieck, who passed away two yearsago. He introduced the concept of schemes as a generalization of algebraicvarieties to a far more abstract concept, allowing the use of sheaf theory asin differential and analytic geometry.

One would think that the fact that the equations in this context arepolynomials implies that the class of geometric objects described is too re-strictive. Nonetheless, a theorem from Chinese mathematician W.L. Chow

iv

INTRODUCTION v

states that any analytic subspace of a complex projective space which isclosed in the strong topology is also closed in the Zariski topology, thusevery closed analytic subspace of a complex projective space is an algebraicvariety.

The document is divided in three chapters. The first provides the alge-braic tools from commutative algebra used throughout the document, suchas rings, ideals, and modules amongst others. Which are the basis for un-derstanding the following chapters. This chapter also introduces the readerto the concept of polynomial rings, and the prime spectrum of a ring as theyare vital to Algebraic Geometry.

The second chapter introduces the concept of Noetherian rings and someimportant results on them, such as Hilbert’s Basis Theorem. This chapteralso introduces the first concept from Algebraic Geometry, namely affinealgebraic varieties, and the aforementioned Nullstellesatz.

The third and final chapter introduces two concepts of dimension, oneis the dimension of a topological space and the other is the Krull dimensionof a ring, and the relation between both in the case of affine varieties andalgebraic sets is noted. A part of the Dimension theorem is stated, althoughthe proof is omitted as it needs other concepts of Dimension theory andthe definition of Hilbert’s polynomial, which were beyond the scope of thisthesis. Lastly, we construct Nagata’s example of an infinite dimensionalNoetherian ring, which a priori seems like a contradicting idea, but theconstruction shows us that it is in fact possible for this rings to exist.

Chapter 1

Commutative Algebra

The purpose of this chapter is to introduce the basic algebraic conceptsneeded to define more interesting geometric constructs. We do this by firstdefining rings and their homomorphisms, which are both key instruments,as well as ideals and modules. An important result on modules, namely theSnake Lemma, is also proved in this chapter.

A glimpse of what will be done in the following chapters can be seenwhen we define the prime spectrum of a ring, this introduces us to theZariski topology, which we will use to define affine algebraic varieties lateron. The concept of the spectrum of a ring is a special case of studies madeon schemes. We do not approach that subject, but it can be found in [5].

1.1 Rings and Homomorphisms

We begin by defining the algebraic structure we will use the most throughoutall chapters, that is the structure of a ring.

Definition 1.1. A ring R is a set endowed with two binary operations,addition (+) and multiplication (·), such that:

• (R,+) is an abelian group.

• Multiplication is associative, i.e., x · (y · z) = (x · y) · z.

• Multiplication is distributive over addition, i.e., x ·(y+z) = x ·y+x ·z,(y + z) · x = y · x+ z · x.

Remark. Multiplication is commonly written xy instead of x · y.

1

CHAPTER 1. COMMUTATIVE ALGEBRA 2

We say that a ring R is commutative if xy = yx for all x, y ∈ R. Ifthere is an element x ∈ R such that xy = yx = y for all y ∈ R, i.e., x isthe identity element of multiplication, then x is a unity of the ring and it isdenoted by 1R or 1.

A unit is an element x ∈ R, for which there exists a unique elementy ∈ R such that xy = yx = 1. Since y is unique it is called x−1. Unit shouldnot be confused with unity, because the unity is always a unit but not theconverse, i.e., not every unit is a unity.

From now on we will asume all rings as a commutative rings with unity,unless stated otherwise.

Example 1.1. Let R be a ring, the set or formal sums

R [x] ={rnx

n + rn−1xn−1 + · · ·+ r1x+ r0 | ri ∈ R,n ∈ Z+ ∪ {0}}

is also a ring with addition and multiplication defined as follows, let f(x),h(x) ∈ R [x], such that

f(x) = fnxn + fn−1xn−1 + · · ·+ f1x+ f0

andh(x) = hmx

m + hm−1xm−1 + · · ·+ h1x+ h0

then we define the sum of f(x) and h(x) to be

f(x) + h(x) = (fs + hs)xs + · · ·+ (f1 + h1)x+ (f0 + h0)

where s = max {m,n}, and the coefficients fi = 0 for i > n, hi = 0 fori > m. Also, we define their product as

f(x)h(x) = αm+nxm+n + αm+n−1xm+n−1 + · · ·+ α1x+ α0

whereαk =

∑

i+j=k

fihj .

This ring is called the ring of polynomials over R in one variable.

Definition 1.2. The degree of an element f(x) ∈ R[x] is the greatest n ∈Z+ ∪ {0} for which fn 6= 0, and it is denoted by deg(f).

Remark. The degree of the polynomial f(x) ≡ 0 is −∞.

Definition 1.3. A subset S of R is a subring if S is a ring with the sameoperations of R.


Definition 1.4. A ring homomorphism is a mapping f : R1 → R2 betweentwo rings such that:

• f(x+ y) = f(x) + f(y)

• f(xy) = f(x)f(y)

If for every y ∈ R2 there exists x ∈ R1 such that f(x) = y we call f anepimorphism. Let r1, r2 ∈ R1, if r1 6= r2 implies that f(r1) 6= f(r2) we saythat f is a monomorphism. A morphism that satisfies both these conditionsis called an isomorphism.

Example 1.2. Let R be a ring, R [x] the ring of polynomials in one variablewith coefficients in R, and r an element of R. Consider the map

σr : R [x] −→ Rf(x) 7−→ f(r)

where f(r) is defined as

f(r) = fnrn + · · ·+ f1r + f0

i.e., substituting the variable x with r. It is clear that f(r) is an element ofR. Now, let h(x), g(x) ∈ R [x] of degree n and m respectively, then we havethat

σr(h(x) + g(x)) = hnrn + · · ·+ h0 + gmr

m + · · ·+ g0 = σr(h(x)) + σr(g(x))

and

σr(h(x)g(x)) = αm+nrm+n + αm+n−1rm+n−1 + · · ·+ α1r + α0

= (hnrn + · · ·+ h1r + h0) · (gmrm + · · ·+ g1r + g0) = σr(h(x))σr(g(x))

i.e., σr is a homomorphism.

1.2 Ideals and Radicals

Like with groups we wish to define quotient rings, so we define the analogousto a normal subgroup, called an ideal.

Definition 1.5. An ideal is a subgroup I of the group (R,+) such that

RI = {ri | r ∈ R, i ∈ I} ⊆ I.


Definition 1.6. Let R be a ring and I an ideal of R, the set {r + I | r ∈ R}is a ring with the operations:

• (s+ I) + (t+ I) = (s+ t) + I

• (s+ I)(t+ I) = st+ I

we denote this ring by R/I and call it the quotient of R by I.

A zero divisor is an element x ∈ R for which there exists y 6= 0 ∈ R suchthat xy = 0. An element x ∈ R is nilpotent if there exists an n ∈ N suchthat xn = 0.

An integral domain is a ring D in which there are no zero divisors. Afield F is an integral domain in which every non-zero element is a unit, i.e.,a ring with no zero divisors in which every non-zero element is a unit.

Definition 1.7. An ideal I on R is said to be prime if xy ∈ I implies thateither x ∈ I or y ∈ I, and maximal if any ideal J that contains I is eitherequal to I or R, i.e., I ⊆ J ⊆ R⇒ I = J or J = R.

Example 1.3. Let R be an integral domain. Since R is commutative thenR[x] is too, the element f(x) ≡ 1 is its unity, and if we have two non-zeroelements

h(x) = hmxm + hm−1xm−1 + · · ·+ h1x+ h0

k(x) = knxn + kn−1xn−1 + · · ·+ k1x+ k0

then we know that the leading coefficient of h(x)k(x) is hmkn, and since R isan integral domain this product is non-zero, i.e., R[x] is an integral domain.

When working with an integral domain D, it is natural to construct afield in which it can be embedded, such a field is called the field of fractionsof D, and it is constructed as follows. Let S = D \ {0}, define the relation∼ on D × S as

(a, b) ∼ (c, d)⇔ ad− cb = 0.

Now, take KD = D × S/ ∼, and define addition and multiplication as

+ : KD ×KD −→ KD

((a, b), (c, d)) 7−→ (ad+ cb, bd)· : KD ×KD −→ KD

((a, b), (c, d)) 7−→ (ac, bd)

with these two operations (KD,+, ·) is a ring with unity (a, a) and zeroelement (0, a).

First, we will show that it is transitive. Let (a, b) ∼ (c, d) and (c, d) ∼(e, f), then we have that ad− bc = 0 and cf − de = 0, multiplying by cf on


both sides of the first equation we obtain ad(cf)− bc(cf) = 0, replacing cffor de we obtain

ad(cf)− bc(de) = af(cd)− be(cd) = (af − be)cd = 0

and since D is an integral domain we must have that either cd = 0 oraf − be = 0. If cd = 0 then c = 0, for b 6= 0, and together with (a, b) ∼ (c, d)and (c, d) ∼ (e, f) we obtain that a = e = 0, in which case it is clear that(a, b) ∼ (e, f). If af − be = 0, then by definition we have that (a, b) ∼ (e, f).

Now, we will prove that KD is in fact a field. It is clear that KD is anintegral domain, so we must only prove that every non-zero element is aunit, for this let (a, b) be a non-zero element of KD, i.e., a 6= 0, and see thatit is a unit. If a = b then (a, b) ∈ (a, a) and it is clearly a unit. On the otherhand, if a 6= b, consider the element (b, a), since a 6= 0 this element is in KD,multiplying these two elements we obtain

(a, b) · (b, a) = (ab, ab) = (b, a) · (a, b)which means that (b, a) is the inverse of (a, b), therefore every non-zeroelement is a unit.

We will now prove that ∼ is an equivalence relation. From ab − ab = 0we obtain that (a, b) ∼ (a, b), therefore it is reflexive. If we have (a, b) ∼(c, d), then ad − cb = 0 and both ad and cb are elements of D, thus theyhave additive inverses, adding them to both sides of the equation we obtaincb− ad = 0, i.e., (c, d) ∼ (a, b), this proves that the relation is symmetric.

To embed D in KD consider the morphism φ : D −→ KD, which mapsevery element d ∈ D to (d, 1) ∈ KD. To prove this is an embedding wemust prove that it is an injective ring homomorphism. It is clear that it isa ring homomorphism. To prove that it is injective take a, b ∈ D, such thata 6= b, and suppose φ(a) = φ(b), i.e., (a, 1) ∼ (b, 1) which by definition is1a−1b = 0, implying a = b, which is a contradiction, therefore φ is injective.

Remark. The process described above to construct the field of fractions canbe generalized by taking any multiplicatively closed subset S of D insteadof D \ {0}, and changing the relation to

(a, b) ∼ (c, d)⇔ (ad− cb)s = 0 , s ∈ S.The ring obtained is denoted S−1D and is called the localization of D withrespect to S.

Example 1.4. Take the integral domain Z and construct its field of frac-tions, the resulting field is

KZ = {(a, b) | a ∈ Z, b ∈ Z \ {0}} / ∼


with the previously defined operations. Consider the ring homomorphism

ϕ : KZ −→ Q(a, b) 7−→ a

b

and two elements (x, y), (p, q) of KZ. Suppose ϕ((x, y)) = ϕ((p, q)), thenxy = p

q , which implies xq = yp, i.e., xq − yp = 0, therefore (x, y) ∼ (p, q).Thus, ϕ is injective.

Now, consider an element xy ∈ Q, then ϕ((x, y)) = x

y , i.e., ϕ is surjective.Since ϕ is bijective KZ is isomorphic to Q, therefore the field of fractions ofZ is Q.

Definition 1.8. Let X ⊆ R, we denote by 〈X〉 the smallest ideal of R thatcontains X. This ideal is called the ideal generated by X.

Remark. If the set X consists of only one element x, then the ideal 〈X〉 isdenoted by 〈x〉, and it is called the principal ideal generated by x, and if Xis a finite set 〈X〉 is said to be a finitely generated ideal.

Since 〈X〉 is the smallest ideal that contains X, it is easy to see that

〈X〉 =⋂

X⊆II

where the I are ideals of R.

Proposition 1.9. The ideal generated by X ⊆ R is the set

XR = {x1r1 + · · ·+ xnrn | n ∈ N, ri ∈ R, xi ∈ X} .Remark. One may also consider RX and RXR defined as follows:

RX = {r1x1 + · · ·+ rnxn | n ∈ N, ri ∈ R, xi ∈ X}RXR = {r1x1r′1 + · · ·+ rrxnr

′n | n ∈ N, ri, r′i ∈ R, xi ∈ X}

but since we are considering only commutative rings we have that

XR = RX = RXR.

Proof. Let I be an ideal that contains X, then it must contain al finite sumsof elements xr, with r ∈ R and x ∈ X. Since this is true for every ideal thatcontains X we get that

XR ⊆⋂

X⊆II = 〈X〉.

Now, since 〈X〉 is the smallest ideal that contains X, it can be obtained byclosing X under all operations that define an ideal. Since R is a ring withunity XR contains X, and it is closed under the operations that define anideal, therefore 〈X〉 ⊆ XR.


Proposition 1.10. The set N of all nilpotent elements of a ring R (calledthe nilradical of the ring) is an ideal, and R/N has no nilpotent elements.

Proof. If x ∈ N, then for all a ∈ R we have that (ax)n = anxn = an0 = 0.Let x, y ∈ N and consider (x + y)m+n+1, this sum consists of products ofthe form xrys with r+ s = m+n+ 1, which means that we cant have s < nand r < m at the same time, therefore each of these products vanishes and(x+ y) ∈ N.

Let x ∈ R/N be represented by x ∈ R, then xn is represented by xn,that means that xn = 0, then xn ∈ N , i.e., there exists a k ∈ N such that(xn)k = 0, but that means that x ∈ N so x = 0.

Remark. Notice that we used the binomial theorem and the fact that (ax)n =anxn, that is because they are both valid in any commutative ring.

Now, let I be an ideal of R, the set

r(I) = {x ∈ R | xn ∈ I, n ∈ N}

is called the radical of I, and it is an ideal of R, if I = r(I) we say that I isa radical ideal. More generally, we define the radical of any subset S of R as

r(S) = {x ∈ R | xn ∈ S, n ∈ N}

but this is not an ideal in general.

Remark. If π : R → R/I is the natural projection, then r(I) = π−1(NI),where NI is the nilradical of R/I.

Proposition 1.11. The radical of an ideal I is the intersection of all primeideals which contain it.

Proof. Let us consider n as the intersection of all prime ideals of R. If x ∈ N,then xn = 0 ∈ J for some n ∈ N and J a prime ideal, therefore x ∈ n, i.e.,N ⊆ n.

Now, let x be a non-nilpotent element and Σ the set of ideals p suchthat, for n > 0 the element xn /∈ p. The set Σ is non-empty because 0 ∈ Σ,and since it is partially ordered by inclusion we can apply Zorn’s lemma,therefore Σ has a maximal element m. Let a, b /∈ m, then the ideals m+ (a),m+ (b) (where (a) and (b) are the ideals generated by a and b respectively)strictly contain m, therefore

xn ∈ m + (a), xm ∈ m + (b)


for some n,m, it follows from this that xn+m ∈ m+(ab), i.e., ab /∈ (m), whichmeans that m is prime and for any element x /∈ m we have that x /∈ N, i.e.,n ⊆ N.

Applying this same process to R/I we obtain that the nilradical of thisring is the intersection of all its prime ideals, and taking into account thatr(I) = π−1(NI) we get that the radical of I is the intersection of all primeideals that contain it.

Let f : R → R′ be a ring homomorphism, if I is an ideal of R thenf(I) does not have to be an ideal of R′. The extension Ie of I is the idealgenerated by f(I) in R′ , i.e., Ie is the set of all sums

∑rif(ii) with ri ∈ R′,

ii ∈ I.On the other hand, if J is an ideal of R′ then f−1(J) is always an ideal

of R, this ideal is called the contraction of J and is denoted by Jc

Proposition 1.12. i) I ⊆ Iec, J ⊇ Jce

ii) Ie = Iece, Jc = Jcec

Proof. i) Let x ∈ I, by definition f(x) ∈ Ie then x ∈ Iec. The proof forJ ⊇ Jce is analogous.

ii) Since I ⊆ Iec, then Ie ⊆ Iece. Now, taking into account that Ie isan ideal of R′ we can apply i) and we get Ie ⊇ Iece.The proof forJc = Jcec is analogous.

Let R be a ring and let P be the set of all prime ideals of R. For anysubset S ⊆ R we define V (S) as the set of all prime ideals that contain S.It is clear that V (∅) = P, and V (1) = ∅ because 1 /∈ p for all p ∈ P.

Let s be the ideal generated by S, if an ideal p is an element of V (S),then p must contain all elements of the form

∑risj , with ri ∈ R and sj ∈ S,

i.e. s ⊆ p, therefore V (S) ⊆ V (s). On the other hand, since S ⊆ s all idealsin V (s) contain S, then V (s) ⊆ V (S). Knowing that r(s) is the intersectionof all prime ideals that contain it, it is clear that V (s) = V (r(s)).

Now, let (Si)i∈I be a family of subsets of R, consider V (∪i∈ISi), it is clearfrom the definition of V (S) that this is the same as ∩i∈IV (Si), because ifp ∈ V (∪i∈ISi), then p ∈ V (Si) for all i ∈ I, therefore p ∈ ∩i∈IV (Si),conversely if p ∈ ∩i∈IV (Si), then p ∈ V (Si) for all i ∈ I, then p ∈ V (∪i∈ISi).

Consider V (I ∩ J), with I, J ideals of R, this is the same as V (r(I ∩ J))and using properties of the radical we obtain that

V (r(I ∩ J)) = V (r(IJ)) = V (IJ) = V (I) ∪ V (J)


the last equality comes from the fact that IJ is the ideal generated by thefinite sums of elements ikjk with ik ∈ I, jk ∈ J , then any ideal p ∈ V (IJ)must contain I or J , and if p ∈ V (I) ∪ V (J) then it contains either I or J ,and since p is prime it contains any product of elements of I with elementsof J .

With these results we have shown that the sets V (S) satisfy the axiomsof closed sets in a topological space. With these sets we can define a topologyon P, the resulting topology is called the Zariski topology, and P togetherwith this topology is called the prime spectrum of R, denoted by Spec(R).

1.3 Modules and Sequences

Definition 1.13. Let R be a ring, an R-module is a set M with

1. A binary operation +, such that (M,+) is an abelian group,

2. An action of R on M , denoted by rm, such that:

a) (r + s)m = rm+ sm, for all r, s ∈ R and m ∈Mb) (rs)m = r(sm), for all r, s ∈ R and m ∈Mc) r(m+ n) = rm+ rn, for all r ∈ R and m,n ∈Md) 1m = m, for all m ∈M

Remark. If R is a field an R-module is an R-vector space.

Definition 1.14. Let M and N be R-modules, a R-module homomorphism(or R-linear mapping) is a mapping f : M → N that satisfies:

1. f(m1 +m2) = f(m1) + f(m2), for all m1,m2 ∈M .

2. f(rm) = rf(m), for all r ∈ R and m ∈M .

The set of R-module homomorphisms from M to N is denoted byHom(M,N), and it can be turned into a R-module defining

(f + g)(x) = f(x) + g(x)

(r · f)(x) = r · f(x)

where r ∈ R and f, g ∈ Hom(M,N). The homorphisms u : M ′ → M andv : N → N ′ induce the morphisms u : Hom(M,N) → Hom(M ′, N) andv : Hom(M,N)→ Hom(M,N ′) defined as

u(f) = f ◦ u, v(f) = v ◦ f


Definition 1.15. Let R be a ring and M an R-module. An R-submoduleis a subgroup M ′ of M closed under the action of R. The group M/M ′ isan R-module and it is called the quotient of M by M ′

If f : M → N is an R-module homomorphism, its kernel is a submoduleof M defined as

ker(f) = {x ∈M | f(x) = 0} ,its image is the submodule of N

Im(f) = f(M),

and the cokernel of f is the quotient module

coker(f) = N/ Im(f).

Definition 1.16. Let M,N be R-modules, their direct sum M ⊕N is theset of pairs (m,n) with m ∈ M , n ∈ N . M ⊕ N can be turned into aR-module by defining

(m1, n1) + (m2, n2) = (m1 +m2, n1 + n2)

r(m1, n1) = (rm1, rn1)

where m1,m2 ∈ M , n1, n2 ∈ N and r ∈ R. If instead of 2 modules wetake a family of modules (Mi)i∈I , we define its direct sum ⊕i∈IMi, whereits elements are the families (mi)i∈I , such that mi ∈ Mi and only a finitenumber mi are non-zero. If we allow all elements to be non-zero we havethe direct product Πi∈IMi.

Remark. If the index set I is finite, the direct product and direct sum arethe same, but in general they are not.

Let X, Y and Z be R-modules, and α ∈ Hom(X,Y ), β ∈ Hom(Y,Z),the following is called a sequence of R-modules and homomorphisms

Xα // Y

β // Z ,

if Im(α) = ker(β) we say that the sequence is exact at Y . Let 0 denotethe R-module consisting only of the 0 element and consider

0 // Xα // Y (1.1)

Yβ // Z // 0 (1.2)


these two sequences are important, because (1.1) is exact if and only ifα is a monomorphism and (1.2) is exact if and only if β an epimorphism. Asequence like (1.3) is called a short exact sequence if it is exact at X, Y andZ, i.e., α is injective, β is surjective and Im(α) = ker(β).

0 // Xα // Y

β // Z // 0 (1.3)

Let us now consider the sequence

· · · // Xi−1α1 // Xi

αi+1 // Xi+1// · · · (1.4)

this sequence is said to be exact if it is exact at each Xi. An exactsequence can be separated into short exact sequences 0 → Yi → Xi →Yi+1 → 0 by taking Yi = Im(αi). Now we will state and proof an importantresult on exact sequences called the snake lemma.

Lemma 1.17 (Snake Lemma). Suppose we have the commutative diagram

0 // X1α1 //

γ1��

X2α2 //

γ2��

X3//

γ3��

0

0 // Y1β1 // Y2

β2 // Y3 // 0

with exact rows, then

0 // ker(γ1)α1 // ker(γ2)

α2 // ker(γ3)δ //

coker(γ1)β1 // coker(γ2)

β2 // coker(γ3) // 0

is exact, with α1, α2 restrictions of α1, α2, and β1, β2 induced by β1,β2.

Proof. First we will prove that the sequences

0 // ker(γ1)α1 // ker(γ2)

α2 // ker(γ3) (1.5)

coker(γ1)β1 // coker(γ2)

β2 // coker(γ3) // 0 (1.6)

are exact. For (1.5) let x0 ∈ Im(α1), that means that x0 ∈ ker(γ1) andx0 ∈ Im(α1), but we already have that Im(α1) = ker(α2), then x0 ∈ ker(α2).


Now, since the diagram commutes we have that (β1◦γ1)(x0) = (γ2◦α1)(x0),but x0 ∈ ker(γ1), then (γ2 ◦ α1)(x0) = 0, hence x0 ∈ ker(γ2). Since we havethat x0 ∈ ker(α2) and x0 ∈ ker(γ2) then x0 ∈ ker(α2).

Now, let x1 ∈ ker(α2), then x1 ∈ ker(α2) and x1 ∈ ker(γ2), like beforefrom this we have that x1 ∈ Im(α1), and since the diagram commutes we alsohave that (β1◦γ1)(x0) = (γ2◦α1)(x0), but x1 ∈ ker(γ2), then (β1◦γ1)(x0) =0, however β1 is injective, then x1 ∈ ker(γ1), which means that x1 ∈ Im(α2).

Since ker(α1) = ker(α1) ∩ ker(γ1) = {0} ∩ ker(γ1) = {0} we have that(1.5) is exact.

Now, for (1.6) we have that β1, β2 are induced by β1, β2, that is βi(x) =βi(x + Im(γi)) = βi(x) + Im(γi+1). Let y3 = y3 + γ3(x3) ∈ coker(γ3), i.e.,y3 + γ3(x3) is a representative of y3. Since the rows of the diagram areexact we have that α2 and β2 are surjective, which implies that there existy2 ∈ Y2, x2 ∈ X2 such that x3 = α2(x2), y3 = β2(y2). Then we have thaty3 = β2(y2), where y2 = y2 + γ2(x2), thus β2 is surjective and the sequenceis exact at coker(γ3)

For the exactness at coker(γ2), let y2 ∈ kerβ2, i.e., β2(y2) = 0, whichmeans that there exists x3 ∈ X3 such that β2(y2) = γ3(x3) for any repre-sentative y2 of y2. Since α2 is surjective there exists x2 ∈ X2 such thatα2(x2) = x3, and considering that the diagram is commutative we have that

β2(y2) = γ3(x3) = (γ3 ◦ α2)(x2) = (β2 ◦ γ2)(x2)then β(y2 − γ2(x2)) ∈ ker(β2) = Im(β1), therefore there exists y1 ∈ Y1 suchthat β1(y1) = y2 − γ2(x2), but clearly γ2(x2) ∈ Im(γ2), then β1(y1) = y2 ∈ker(β2).

Now, let us consider the composition

(β2 ◦ β1)(y + Im(γ1)) = β2(β1(y) + Im(γ2)) = (β2 ◦ β1)(y) + Im(γ3)

but thanks of the exactness of the diagram we have that β2 ◦ β1 = 0, thenIm(β1) ⊆ ker(β2).

The morphism δ is called the connecting or boundary homomorphism,and is defined as follows. Let x3 ∈ ker(γ3), since α2 is surjective, there existsx2 ∈ X2 such that α2(x2) = x3 and γ2(x2) ∈ Y2. Taking into considerationthe commutativity of the diagram we obtain that

(γ3 ◦ α2)(x2) = γ3(x3) = 0 = (β2 ◦ γ2)(x2)then, we have that γ2(x2) ∈ ker(β2) = Im(β1), this implies that there existsy1 ∈ Y1 such that β1(y1) = γ2(x2), and since β1 is injective this y1 is unique.Now we define

δ(x3) = y1 = y1 + Im(γ1)


this map might not be well-defined because α2 is not injective, so we willsee that δ does not depend on the choice of x2 and thus it is well-defined.

For this, suppose x′2 6= x2 such that α2(x′2) = α2(x2) = x3, there exists

y′1 ∈ Y1 such that β1(y′1) = γ2(x

′2), and this y′1 is unique. Our assumption on

the images of x2 and x′2 under α2 gives us that α2(x2−x′2) ∈ ker(α2) = Imα1,therefore there exists x1 ∈ X1 such that α1(x1) = x2−x′2, and we have that

(β1 ◦ γ1)(x1) = (γ2 ◦ α1)(x1) = γ2(x2 − x′2) = β1(y1 − y′1)

which, with the injectivity of β1 gives us that y1 − y′1 = γ1(x1) ∈ Im(γ1),thus y1 = y′1.

Consider x2 ∈ ker(γ2), such that α2(x2) 6= 0. We will calculate (δ ◦α2)(x2) and see that it is 0. Since γ2(x2) = 0, we have that γ2(x2) =β1(0) because β1 is injective and the diagram is commutative, but from thedefinition of the boundary morphism we have that (δ◦α2)(x2) = 0+Im(γ1) =0.

Now consider x3 ∈ ker(δ), we know that x3 = α2(x2) for some x2 ∈ X2,therefore γ2(x2) = β1(y1) with y1 ∈ Im(γ1), then there exists x1 ∈ X1 suchthat γ1(x1) = y1. Then (γ2 ◦α1)(x1) = (β1 ◦ γ1)(x1) = γ2(x2), which meansthat x2 − α1(x1) ∈ ker(γ2), and applying α2 we obtain that

α2(x2 − α1(x1)) = α2(x2)− (α2 ◦ α1)(x1) = α2(x2) = x3

therefore x3 ∈ Im(α2), and Im(α2) = ker(δ).Finally, we will prove that Im(δ) = ker(β1), for this take y = y1 +

Im(γ1) ∈ ker(β1), then β1(y1) ∈ Im(γ2), take x2 ∈ X2 such that γ2(x2) =β1(y1) and let x3 = α2(x2), then

(γ3 ◦ α2)(x2) = (β2 ◦ γ2)(x2) = (β2 ◦ β1)(y1) = 0

which means that x3 ∈ ker(γ3) and δ(x3) = y, then y ∈ Im(δ).Now let x3 ∈ ker(γ3) and x2 ∈ X2 such that α2(x2) = x3, then γ2(x2) =

β1(y1) for some y1 ∈ Y1, so δ(x3) = y1. Then (β1◦δ)(x3) = β1(y1)+Im(γ2) =0, i.e., Im(δ) ⊆ ker(β1). With this the snake lemma is proved.

1.4 Algebras

Let R, S be rings, f : R → S a ring homomorphism and M a S-module,then M can be given an R-module structure by defining the product rm,with r ∈ R and m ∈ M , as f(r)m. We say that M as an R-module isobtained from the S-module M by restriction of scalars.


Definition 1.18. Let R, A be rings and f : R→ A a ring homomorphism,if we define the product

ra = f(r)a

with r ∈ R and a ∈ A, then A has a R-module structure. The ring A withthis R-module structure is called and R-algebra.

Remark. In general the ring A must not be commutative, in that case wesay that f : R→ Z(A), where Z(A) is the center of A.

An algebra is an example of restriction by scalars, by taking the ring Aas an A-module.

Example 1.5. Every ring R with unity is a Z-algebra with the morphismf : Z→ R, defined as n 7→ n · 1R.

The morphism f is said to be finite, and A a finite R-algebra, if A isfinitely generated when considered as an R-module.

1.5 Valuation Rings

Definition 1.19. Let f ∈ R [x] of degree n, we say f is monic if its leadingcoefficient, fn, is 1.

Let R be a ring, and S a subring of R that contains its unity. We saythat r ∈ R is integral over S if r is the root of a monic polynomial f ∈ S [x].For s ∈ S consider the polynomial gs(x) = x − s in S [x]. It is clear thatthis polynomial is monic, and its only root is s, therefore every element ofS is integral over S.

Definition 1.20. The set C of all elements of R that are integral over S iscalled the integral closure of S in R.

Remark. If S = C, then S is integrally closed in R. On the other hand, ifR = C we say that R is integral over S.

It is easy to see that the S-module S[r] is finitely generated if and onlyif r is integral over S. To prove this, let r be integral over S, then thereexist s0, . . . , sn−1 ∈ S such that

rn + rn−1sn−1 + · · ·+ rs1 + s0 = 0

multiplying rk, k ≥ 0, we obtain

rn+k = −(rn+k−1sn−1 + · · ·+ rks0)


therefore, every positive power of r can be written in terms of 1, r, . . . , rn−1,i.e., the S-module S[r] is generated by 1, r, . . . , rn−1.

Now, let S[r] be finitely generated by 1, r, . . . , rk−1, then we can writerk as rk = rk−1sk−1 + · · ·+ rs1 + s0. From this equation we obtain

rk + rk−1s′k−1 + · · ·+ rs′1 + s′0 = 0

where s′i = −si. This is a monic polynomial on S that has r as a root. Thus,r is integral over S.

An integral domain is said to be integrally closed it is integrally closedin its field of fractions, but this need not always happen. An example ofthis ring requires some definitions we have not given yet, thus ti will beconstructed in the following chapter.

Proposition 1.21. Let S ⊆ R ⊆ R′ with R integral over S, and R′ integralover R, then R′ is integral over S.

Remark. This property is called transitivity of integral dependence.

Proof. Let c ∈ R′, since R′ is integral over R we have

cn + rn−1cn−1 + · · ·+ r1c+ r0 = 0

for some ri ∈ R, but R is integral over S, then for every ri we have that S[ri]is finitely generated. From this we obtain that S′ = S[r1, . . . , rn] is finitelygenerated as an S-module, and since S′ ⊆ R, we have that c is integralover S′, i.e., S′[c] is finitely generated as an S-module, and c is integral overS.

Definition 1.22. Let D be an integral domain and KD its field of fractions.D is said to be a valuation ring of KD if for every non-zero element k of KD

we have that k ∈ D and/or k−1 ∈ D.

Remark. Every field is trivially a valuation ring of itself.

Example 1.6. Let R be an integral domain, as we have seen before, thisimplies R[x] is an integral domain too, therefore we can construct its fieldof fractions. Let KR[x] de the field of fractions of R[x], and let

D ={f/g ∈ KR[x]

∣∣ deg(f) ≤ deg(g)}.

It is easy to check that D is in fact an integral domain, and it is a valuationring of KR[x], since if f/g is not in D, that means that deg(f) ≥ deg(g),which in turn means that (f/g)−1 = g/f ∈ D.


Example 1.7. Take the field of rational numbers Q, and the set D of allrational numbers of the form prm

n , with r ≥ 0 and p a fixed prime. Let x

be any rational number, if it is of the form prmn there is nothing to prove.

On the other hand, if x is not of this form, then it must be either mn or

mprn . Since p0 = 1 we have that x = m

n ∈ D, and the inverse of x = mprn is

x−1 = prnm , which is an element of D. Thus, D is a valuation ring of Q.

Proposition 1.23. Let D be a valuation ring of KD, then D is integrallyclosed in KD.

Proof. Let k ∈ KD be integral over D, then we have

kn + dn−1kn−1 + · · ·+ d1k + d0 = 0

for some di ∈ D. If k ∈ D we have nothing to prove. On the other hand, ifk /∈ D then k−1 ∈ D. Multiplying both sides of the equation above by k1−n

we obtaink = −(d0k

1−n + d1k2−n + · · ·+ dn−2k−1dn−1)

then, since all elements on the right-hand side are elements of D, k ∈ D.

Chapter 2

Varieties

In the previous chapter we defined most of the algebraic structures thatwe will need in order to define and study some results on affine algebraicvarieties, but a key concept was left out of the previous chapter due to itsclose relation with Noetherian rings, which are defined here, and that is thering of polynomials in several variables.

This close relationship comes from Hilbert’s Basis Theorem, and thefact that the rings we work on the most, which are fields, are Noetherian.This chapter also revisits the Zariski topology and introduces some basictopological concepts used to define affine varieties.

2.1 Noetherian Rings

Before we speak about affine varieties we must first discuss some propertiesof fields and their rings of polynomials in various variables. We alreadydefined the ring of polynomials in one variable in section 1.1, now we willdefine the polynomial ring in n variables.

Definition 2.1. Let R be a ring, and consider the set

R [x1, x2, . . . , xn] =

{k∑

i=1

fαixαi

∣∣∣∣∣ fαi ∈ R, αi ∈ Nn}

where, if α = (α1, α2, . . . , αn), xα = xα11 xα2

2 · · ·xαnn . This set is called the

ring of polynomials in n variables over R, with addition defined coefficient-wise and multiplication by the use of distributive law and the rules of expo-nents.

17

CHAPTER 2. VARIETIES 18

Remark. One may also define the ring of polynomials in n variables induc-tively as R[x1, x2, . . . , xn] = R[x1, . . . , xn−1][xn]. This definition is equiva-lent to the one given above.

This definition is enough to give the example mentioned in Section 1.5of an integral domain that is not integrally closed over its field of fractions.

Example 2.1. Let K be a field, then R = K[x, y]/〈x2 − y3〉 is an integraldomain because of the ring homomorphism φ : R → K[t2, t3] that mapsy 7→ t2 and x 7→ t3.

R is not integral over its field of fractions KR, because if it were the setof elements in KR that are integral over R would consist only of elementsof R, but the element t3/t2 ∈ KR \ R is a zero of the monic polynomialf(z) = z2 − t2. Thus, R is not integrally closed

An important application of polynomial rings in several variables is toalgebras, as they give us the tools to define finitely generated algebras. Thehomomorphism f : R → A is said to be of finite type, and A is called afinitely generated R-algebra, if there exist finite elements a1, . . . , an in Asuch that every element in A can be written as a polynomial in a1, . . . , anwith coefficients in f(R), i.e., there exists an epimorphism from R[x1, . . . , xn]to A.

Since every ring with unity is a Z-algebra, we say that a ring with unityis finitely generated if it is finitely generated as a Z-algebra.

Definition 2.2. A ring R is said to be a principal ideal domain (PID) ifevery ideal on R is a principal ideal.

Proposition 2.3. Every prime ideal on a PID is maximal.

Proof. Let R be a PID, and I = 〈x〉 a prime ideal on R. Consider an idealJ = 〈y〉 that contains I, then x = ry for some r ∈ R, and since I is primeand ry ∈ I then either r ∈ I or y ∈ I. If y ∈ I, then I = 〈x〉 = 〈y〉 = J .Now, let r ∈ I, which means that r = xk, then x = ry = xky, i.e., y is aunit, which implies that J = R.

With this proposition it is easy to see that if R[x] is a PID then R is afield, simply take the ideal 〈x〉 on R[x], since it is prime it is also maximal,and we have that R[x]/〈x〉 ' R, then R must be a field.

On the other hand, assume that R is a field, then R[x] is an integraldomain. Now, let I be an ideal on R[x], if I = {0} = 〈0〉 there is nothing toprove. If I 6= {0} take f ∈ I of minimum degree. It is clear that 〈f〉 ⊆ I. Letg ∈ I, then there exist q, r ∈ R[x] such that g = qf+r, and deg(r) ≤ deg(f)


or r ≡ 0. But, r = g− qf ∈ I contradicts our assumption on the minimalityof the degree of f , therefore r ≡ 0 and I = 〈f〉.

Proposition 2.4. The ring of polynomials in 2 variables over a field,K[x1, x2], is not a PID.

Proof. As we have previously stated, if K[x1, x2] is a PID then K[x1] mustbe a field, but this cannot be since the element f(x1) = x1 has no inverse inK[x1], i.e., there exists a non-zero element in K[x1] that is not a unit.

Remark. The ring of polynomials in n variables over a field, K[x1, . . . , xn],is not a PID.

We have seen from the previous proposition that the ring of polynomialsin n variables over a field is not a PID, but we will now see that it is in fact aunique factorization domain. For this we need some preliminary definitions.

Definition 2.5. Let R be an integral domain, and x, y be elements of R.We say x and y are associates if x = uy, for some unit u ∈ R. Let x bean element of R, such that x is not a unit and x 6= 0. We say that x is anirreducible, if whenever x = yz for some y, z ∈ R, then either y or z is aunit. An element p ∈ R is said to be prime if p|xy implies that either p|x orp|y, i.e., the ideal 〈p〉 is prime.

Remark. If an element is not irreducible it is said to be reducible.

Now we can define the property of being a unique factorization domain.

Definition 2.6. A unique factorization domain (UFD) is an integral domainR, such that:

1. Every non-unit r 6= 0 can be written as a product of irreducibles of R.

2. This factorization on irreducibles is unique up to permutations andassociates.

Proposition 2.7. In an integral domain prime implies irreducible, and ina unique factorization domain an element is prime if and only if it is irre-ducible.

Proof. Let R be an integral domain, x ∈ R prime, and a, b ∈ R non-zero. Letx = ab, it is clear that since ab ∈ 〈x〉, either a ∈ 〈x〉 or b ∈ 〈x〉. Without lossof generality assume the former, then we have that a = xr for some r ∈ R.This, with our previous assumption on x, gives us that, x = ab = xrb, i.e.,rb = 1 and b is a unit, therefore x is irreducible.


Since a UFD is also an integral domain, it suffices to prove that everyirreducible is a prime in a UFD. Let a, b, x be the same as before, but xnow irreducible, and R a UFD. If x|ab, there must exist y ∈ R suchthat ab = xy, since the factorization into irreducibles is unique, we havethat x is an associate to an irreducible factor of a or b, assume it is a,i.e., a = (ux)a1 · · · an for some unit u, but this means that x|a, then x isprime.

In order to prove that K[x1, . . . , xn] is a UFD, first we need to prove thatevery PID is also a UFD. For this, let R be a PID and r0 ∈ R be a non-zero,non-unit, we will see that it can be written as a product of irreducibles. Ifr0 is irreducible there is nothing to prove.

On the other hand, if r0 is not irreducible we can assume that we haver0 = x1r1, where x1 and r1 are not units, and r1 6= 0. If r1 is not irre-ducible, then we have r1 = x2r2, where x2 and r2 are not units, and r2 6= 0.Continuing with this process we obtain the sequence r0, r1, . . . of non-zeroelements, and because of their construction it is clear that

〈r0〉 ⊂ 〈r1〉 ⊂ · · ·

but, since R is a PID and the ideals 〈ri〉 form an ascending chain, the set∪〈ri〉 is also an ideal, more so it contains every other ideal on the chain, andit is generated by a single element, say r. Then, the chain can be re-writtenas

〈r0〉 ⊂ 〈r1〉 ⊂ · · · ⊂ 〈r〉.This shows us that r is an irreducible factor of r0, i.e., every element has atleast one irreducible factor.

Then, we have that r0 = r1k1, for some irreducible r1 and a non-unit k1,if k1 is not irreducible the it can be written as k1 = r2k2, with r2 irreducibleand k2 a non-unit. Repeating this process we obtain the sequence

〈r0〉 ⊂ 〈k1〉 ⊂ · · ·

and, as before, there exists k such that

〈r0〉 ⊂ 〈k1〉 ⊂ · · · ⊂ 〈k〉

thus, k is irreducible, and we have r0 = r1r2 · · · rnk.Now, all that is left is to see that this factorization is unique up to permu-

tations and associates. Let r1r2 · · · rn and k1k2 · · · km be two factorizationsof r0 into irreducible factors, i.e.,

r0 = r1r2 · · · rn = k1k2 · · · km.


If n = 1, then r0 is irreducible and r1 = k1, by induction on n we may assumethat the factorization into irreducibles in fewer that l factors is unique upto permutations and associates. Let r0 = r1r2 · · · rl = k1k2 · · · ks, then r1divides k1k2 · · · ks, i.e., it divides some ki, without loss of generality assumei = 1. Since k1 is irreducible, we have that k1 = ur1, for some unit u ∈ R.From this we obtain

ur1r2 · · · rl = uk1k2 · · · ks,replacing ur1 with k1 we obtain

k1r2 · · · rl = k1(uk2) · · · ks,and finally, cancelling k1 from both sides we get

r2r3 · · · rl = (uk2)k3 · · · ksWhich is a factorization in less that l factors, thus unique up to permutationsand associates.

Theorem 2.8. A ring R is a UFD if and only if R[x] is a UFD.

To prove Theorem 2.8 we first need to state and prove Gauß’s Lemma,as it will give us an important result on the irreducibility of a polynomial inR[x] over KR[x].

Lemma 2.9 (Gauß’s Lemma). Let R be a UFD, and f ∈ R. If f is reduciblein KR[x], then it is reducible in R[x].

Proof. Let G,H ∈ KR[x] such that f = GH, i.e., f is reducible in KR[x].Since the coefficients of G and H are in KR there exists r1, r2 ∈ R such thatr1G = g and r2H = h belong to R[x]. Multiplying the first equation byr = r1r2 we obtain rf = gh.

If r is a unit in R there is nothing to prove. On the other hand, assumer is not a unit in R, and write it as a product of irreducibles, say r =p1 · · · pn. Since p1 is irreducible, by definition the ideal 〈p1〉 is prime, thenthe ideal p1[x], of all polynomials with coefficients in 〈p1〉, is prime in R[x]and R[x]/p1[x] is an integral domain.

Since p1 divides r, we have that rf = 0, i.e., 0 = gh. This means thateither g = 0 or h = 0. Without loss of generality assume g = 0, this impliesthat all coefficients of g are divisible by p1, and that we can cancel p1 fromboth sides of the equation.

This yields the equation r′f = g′h, where r′ has one irreducible factorless than r, and g′ = g

p1. Continuing in this fashion for all factors of r

we obtain f = g(n)h, where g(n) = gp1···pn , which is a factorization of f in

elements of R[x], i.e., f is reducible in R[x] as desired.


This result will prove useful in the proof of Theorem 2.8, as we will seenow.

Proof of Theorem 2.8. Let R[x] be a UFD, and take the subring of R[x]isomorphic to R, i.e., the polynomials of degree 0. Since these polynomialscan be factored uniquely into irreducibles of R[x], these irreducibles mustall be of degree 0, therefore R must be a UFD.

Now, let R be a UFD, and f ∈ R[x] a non-zero polynomial of degreegreater than 0, for if deg(f) = 0 then f ∈ R and it can be factored uniquely.Let c ∈ R be the greatest common divisor of the coefficients of f , then wehave that f = cf ′, where f ′ is primitive, i.e., each pair of coefficients of f ′

are relatively prime.Since c ∈ R, it can be factored into irreducibles, then it suffices to prove

that a primitive polynomial can be factored uniquely into irreducibles. Toprove such a factorization exists we will use induction over the degree ofthe polynomial. If deg(f) = 1 then f is irreducible. Assume this holdsfor any primitive polynomial of degree less that n, and let deg(f) = n. Iff is irreducible we are done, if not, then f can be written as a productof two primitive polynomials of degree less that n, and by our inductivehypothesis they can be factored into irreducibles. Thus, f can be factoredinto irreducibles.

To prove the uniqueness of this factorization, let p1 · · · pn and q1 · · · qmbe two distinct factorizations of f into irreducibles. Consider the the ring ofpolynomials in one variable over the field of fractions of R, namely KR[x],since this ring is a PID it is also a UFD. Because of Lemma 2.9 the pi and qiare irreducible in KR[x], and from the uniqueness of the factorization of f inKR[x] we obtain that m = n, even more so, that the pi and qi are associatesin KR[x], so all that is left to prove is that they are also associates in R[x].

Let g and h be primitive polynomials that are associates in KR[x], thenwe have that g = kh for some unit k ∈ K, if k also is a unit in R there isnothing to prove. On the other hand, let k be of the form a

b , with a, b ∈ Rboth non-zero. Then g = a

bh, from which we obtain bg = ah. Since both aand b are greatest common divisors of the coefficients of h and g respectively,and we have the previous equality, then a and b must be associates in R,i.e., the exists a unit u ∈ R such that a = ub, then g = uh. Thus, thefactorization is unique up to associates.

Corollary 2.10. Let R be a UFD. The ring of polynomials in n variablesover R, R[x1, . . . , rn], is a UFD.


Proof. Applying the previous theorem recursively the corollary is proved.

Now, it is easy to check that the ring of polynomials in n variables overa field K is a UFD. Because, since K is a field then K[x] is a PID, then it isa UFD, applying the previous theorem we can see that every field is a UFD,and applying the corollary we obtain the desired result.

We will now discuss the main focus of this section, that is the propertyof being Noetherian. We have already seen an example of this property withPIDs, so now we will give the formal definition. A ring R is said to beNoetherian if it satisfies the following equivalent conditions:

1. Every non-empty set of ideals has a maximal element.

2. Every ascending chain of ideals in R is stationary.

3. Every ideal in R is finitely generated.

Proof. It is clear that 1 is equivalent to 2. In order to prove that 1 and2 are equivalent to 3 let I be an ideal on R, and let Σ be the set of allfinitely generated ideals contained in I. Clearly the zero ideal is containedin Σ, therefore Σ has a maximal element M = 〈r1, r2, . . . , rn〉. If M 6= I,we can construct the ideal M ′ = 〈r1, r2, . . . , rn, r′〉, with r′ ∈ I and r′ /∈M .Then the ideal M ′ is finitely generated and M ( M ′, but this contradictsour assumption on the maximality of M , therefore M = I and I is finitelygenerated.

Now, assume that every ideal on R is finitely generated, and let I0 ⊆I1 ⊆ I2 ⊆ · · · be an ascending chain of ideals on R, then the ideal

∑∞i=0 Ii is

finitely generated, and contains every Ii. Thus, the chain is stationary.

Remark. Since every ideal in a PID is generated by a single element, everyPID is Noetherian.

Example 2.2. Since the only ideals a field can have are the trivial ideals,it is clear that every field is Noetherian.

Example 2.3. The ring Z is a PID, hence it is also a Noetherian ring.

We will now prove an important theorem regarding Noetherian rings andtheir rings of polynomials, known as Hilbert’s Basis Theorem.

Theorem 2.11 (Hilbert’s Basis Theorem). If R is Noetherian, then thering R [x] is Noetherian.


Proof. Let J be an ideal in R [x] that is not finitely generated, take f0 ∈J , such that deg(f0) ≥ 1, and construct J/〈f0〉. Since J is not finitelygenerated, J/〈f0〉 isn’t either. Now we can choose f1 ∈ J/〈f0〉 of smallestdegree and take a representative f1 ∈ R [x].

Repeating the process described above, we can choose f0, f1, · · · ∈ R [x]inductively, such that fn+1 is of smallest degree in J/Jn, where Jn =〈f0, . . . , fn〉. Suppose deg(fn) = dn and that its leading coefficient is cn ∈ R.It is clear that d0 ≤ d1 ≤ · · · from our assumption on the degree of fn inJ/Jn.

Let In = 〈c0, c1, . . . , cn−1〉, then I1 ⊆ · · · ⊆ In−1 ⊆ In ⊆ · · · . Since R isNoetherian there exists m ∈ N such that cm ∈ Im, i.e., the chain stabilizesform Im onwards. Then, we can express cm as α0c0+α1c1+ · · ·+αm−1cm−1,where αi ∈ R. Now, construct g as

g = fm −m−1∑

i=0

αixdm−difi.

Since all fi are in Jm, then g ∈ J/Jm. But g has degree less that dm, whichcontradicts our assumption on the degree of fm.

Remark. Although rarely stated, the converse of this theorem is also true,i.e., if the ring R[x] is Noetherian then R must be Noetherian. It is easyto prove this by contradiction, let R[x] be Noetherian and assume R is not.Since for every ideal I in R we have that I[x] is an ideal in R[x], if we hadan ascending chain of ideals on R that doesn’t stabilize, there would be anascending chain on R[x] that doesn’t stabilize, contradicting our assumptionthat R[x] is Noetherian.

Corollary 2.12. If R is Noetherian, then the ring R [x1, . . . , xn] is Noethe-rian.

Proof. Applying the previous theorem recursively the corollary is proved.

Corollary 2.13. Let A be a finitely generated R-algebra. If R is Noetherianthen A is too.

Proof. Since there exists an epimorphism form R[x1, . . . , xn] to A, then Ais Noetherian.

Remark. In particular every finitely generated ring, and every finitely gen-erated algebra over a field is Noetherian.


Proposition 2.14. Let S ⊆ R ⊆ R′ be rings such that S is Noetherian,and R′ is a finitely generated S-algebra. If R′ is finitely generated as anR-module, then R is finitely generated as an S-algebra.

Proof. Let r′1, . . . , r′n generate R′ as an S-algebra, and r1, . . . , rm generate

R′ as an R-module. Then, we have that

r′i =∑

xijrj (2.1)

rirj =∑

xijkrk (2.2)

with the xij and xijk in R. Consider R0 to be the S-algebra generated bythe xij and xijk. It is clear that S ⊆ R0 ⊆ R, and from Corollary 2.13 wehave that R0 is Noetherian.

Since any element of R′ can be written as a polynomial in r′i with coef-ficients in S, we can substitute using (2.1) and use (2.2) to see that R′ isfinitely generated as an R0-module. There is a result on modules that tellsus that if M is a Noetherian module, then every sub-module of M is finitelygenerated. Applying this result to our case we get that R is finitely gen-erated as an R0-module, and since R0 is finitely generated as en R-algebrathen R also is.

Remark. The condition on R′ of being finitely generated as an R-modulecan be replaced by R′ being integral over R, since we saw in section 1.5 thatthese two conditions are equivalent.

Let K and F be fields such that F ⊆ K, then we say that K is anextension of F . If every element in K is the zero of a polynomial withcoefficients in F , we say that K is an algebraic extension of F , even moreso, if K is finitely generated as an F -algebra we say that K is a finitealgebraic extension of F .

Proposition 2.15 (Zariski’s Lemma). Let K be a field, and E a finitelygenerated K-algebra. If E is a field, then it is a finite algebraic extension ofK.

Proof. Suppose that E is a finite extension that is not algebraic over K,from Corollary 2.13 we know that E = K[x1, . . . , xr]. We can re-organizethe xi in such a way that x1, . . . , xm are algebraically independent over K,i.e., x1, . . . , xm are not the zero of any non-zero polynomial with coefficientsin K, and the xm+1, . . . , xr are algebraic over F = K[x1, . . . , xm].

Then, E is a finite algebraic extension of F , and applying the previousproposition to K ⊆ F ⊆ E we get that F is a finitely generated K-algebra,


say by f1, . . . , fs, i.e., F = K[f1, . . . , fs]. Each of the fi is of the form gihi

,with gi, hi ∈ K[x1, . . . , xm] and hi 6= 0.

Since there exists an infinite amount of irreducible polynomials inK[x1, . . . , xm], there exists a polynomial h, e.g., h = h1 · · ·hm + 1, thatis prime to each hi, and thus h−1 cannot be expressed as a polynomial inf1, . . . , fs, which contradicts the fact that F is a field. Therefore, E is al-gebraic over K, and since it is finite as a K-algebra it is a finite algebraicextension of K.

2.2 Affine Varieties

Now that we have defined and discussed various properties of Noetherianrings, we may speak about affine varieties.

Definition 2.16. Let K be an algebraically closed field, an affine n-spaceover K, denoted An, is the set of all n-tuples of elements of K. An elementp = (p1, . . . , pn) ∈ An is called a point, and the pi are called the coordinatesof p.

Having defined affine spaces, now we can interpret elements ofK [x1, . . . , xn] as functions from the affine n-space to K by defining f(p) =f(p1, . . . , pn), for f ∈ K [x1, . . . , xn] and p ∈ An. Now, consider T ⊆K [x1, . . . , xn], since all the elements of this subset are polynomials, we mayask for the set of zeros all the elements of T have in common. This set iscalled the zero set of T , denoted by Z(T ), and can be described as

Z(T ) = {p ∈ An | f(p) = 0 ∀f ∈ T} .

It is easy to see that if I is an ideal in K [x1, . . . , xn] generated by T ,then Z(T ) = Z(I). Since K is a field its ring of polynomials in n variablesis Noetherian, therefore every ideal is finitely generated, then Z(T ) can beexpressed as the intersection of zeros of finitely many polynomials.

Definition 2.17. Let Y ⊆ An. Y is called an algebraic set if there existsT ⊆ K [x1, . . . , xn] such that Y = Z(T ).

With these algebraic sets we will define a topology on An, for this weneed to prove the following proposition.

Proposition 2.18. The following statements are true for algebraic sets:

1. The union of two algebraic sets is again an algebraic set.


2. The intersection of a family of algebraic sets is an algebraic set.

3. An and ∅ are algebraic sets.

Proof. 1. Let Y1 = Z(T1) and Y2 = Z(T2), then Y1 ∪ Y2 = Z(T1 · T2). Ifp ∈ Y1 ∪ Y2, then p ∈ Y1 or p ∈ Y2, so p is a zero of T1 · T2. On theother hand, let p ∈ Z(T1 · T2) and assume that p /∈ Y1, this impliesthat there exists f1 ∈ T1 such that f1(p) 6= 0, but for all f ∈ T2 wemust have that (f1 · f)(p) = 0, therefore f(p) = 0 for all f ∈ T2, i.e.,p ∈ Y2.

2. Let Y = {Yi | i ∈ I} be a family of algebraic sets such that Yi = Z(Ti).Then ∩Yi = Z(∪Ti), so ∩Yi is an algebraic set.

3. Let 1, 0 ∈ K [x1, . . . , xn], such that 1(p) = 1 and 0(p) = 0, then Z(1) =∅ and Z(0) = An.

Proposition 2.18 shows us that algebraic sets satisfy the axioms of closedsets in a topological space. Given an affine n-space An, the topology definedby taking algebraic sets as closed sets is called the Zariski topology on An.This topology is of great importance, because it is used to define affinevarieties, but before we can define them we need another topological concept.

Definition 2.19. Let X be a topological space, and Y ⊆ X. Y is said tobe irreducible if it cannot be expressed as the union of 2 closed subsets ofX.

With this concept out of the way, we can now define the main object ofthis section.

Definition 2.20 (Affine variety). An affine algebraic variety, or affine va-riety, is an irreducible closed subset of An with the Zariski topology. Anopen subset of an affine variety is a quasi-affine variety.

We can establish a relationship between ideals on K [x1, . . . , xn] and sub-sets of An, by defining the ideal of Y ⊆ An as

I(Y ) = {f ∈ K [x1, . . . , xn] | f(p) = 0 ∀p ∈ Y } .

Theorem 2.21 (Hilbert’s Nullstellensatz). Let K be an algebraically closedfield, J be an ideal in K [x1, . . . , xn], and f ∈ K [x1, . . . , xn] a polynomialthat vanishes at Z(J). Then, there exists r ∈ Z+ such that f r ∈ J , i.e.,I(Z(J)) = r(J).


Remark. In this section we will give two standard proofs of the Nullstellen-satz, and in the next section we will again prove this theorem using Grobnerbases.

First proof of Theorem 2.21. It is clear that r(J) ⊆ I(Z(J)). To prove theother inclusion let f ∈ I(Z(J)), introduce the variable x0 and consider thering K [x0, x1, . . . , xn], on this ring consider the ideal X generated by a setof generators of J , say {f1, . . . , fq} (which are finite due to Theorem 2.11),and x0f − 1, i.e. X = 〈f1, . . . , fq, x0f − 1〉.

Since f ∈ I(Z(J)), it vanishes at any point where all fi are zero, then atany point a ∈ An+1 where all fi are zero, we have that a0f(a)−1 = −1. Thismeans that Z(X) = ∅, and X cannot be a proper ideal in K [x0, x1, . . . , xn].Thus, X = K [x0, x1, . . . , xn], and we have

1 = g1f1 + · · ·+ gqfq + gq+1(x0f − 1)

for some gi ∈ K [x0, x1, . . . , xn]. Now, let y = x−10 and consider the ringK [y, x1, . . . , xn], here we can consider the gi as polynomials in xi and y−1,and we have that gq+1(x0f −1) = y−1gq+1(f −y). From this we obtain that

1 = g1f1 + · · ·+ gqfq + y−1gq+1(f − y)

but the gi are polynomials in y−1, so taking N sufficiently large to clear alldenominators we get

yN = j1f1 + · · ·+ jqfq + jq+1(f − y)

and taking f = y we obtain the desired result and the theorem is proved.

Proposition 2.22. Let I1, . . . , In be ideals, and P a prime ideal that con-tains ∩ni=1Ii, then P contains one of the Ii. If P = ∩ni=1Ii, then P = Ii forsome i.

Proof. Assume that Ii * P for all i, i.e., there exists xi ∈ Ii such thatxi /∈ P . Since

∏Ii ⊆ ∩Ii, we have that

∏xi ∈

∏Ii ⊆ ∩Ii, and

∏xi /∈ P .

Therefore, ∩Ii * P .For the second part of the theorem assume that P = ∩Ii, then P ⊆ Ii

for all i. Thus, P = Ii for some i.

Corollary 2.23. Let Y be an algebraic set, then Y is irreducible if and onlyif its ideal is prime.


Proof. Let P be a prime ideal, since prime ideals are radical ideals we havethat I(Z(P )) = P . Suppose that Z(P ) = Y = Y1 ∪ Y2, then I(P ) =I(Y1∪Y2) = I(Y1)∩I(Y2), and from Proposition 2.22 we get that P = I(Y1)or P = I(Y2). Thus, Y = Y1 or Y = Y2.

Now, let Y be irreducible and take fg ∈ I(Y ), then Y ⊆ Z(fg) =Z(f)∪Z(g). From this we get that Y = (Y ∩Z(f))∪ (Y ∩Z(g)), but sinceY is irreducible we have that either Y = Y ∩ Z(f) or Y = Y ∩ Z(g), i.e.,either Y ⊆ Z(f) or Y ⊆ Z(f). Therefore, since the ideal reverses inclusions,either f ∈ I(Y ) or g ∈ I(Y ).

Definition 2.24. Let Y be an algebraic set in An. We define the affinecoordinate ring of Y as A(Y ) = K[x1, . . . , xn]/I(Y ).

Remark. Since the ideal of an affine variety is prime, the affine coordinatering of a variety is an integral domain.

Definition 2.25. Let R be a ring, we define the set Max(R) to be the setof all maximal ideals in R.

Second proof of Theorem 2.21. Let f ∈ K[x1, . . . , xn] such that f /∈ r(J),we will see that f /∈ I(Z(J)). Since f /∈ r(J), we know from Proposition1.11 that there exists a prime ideal P that contains J but not f . ConsiderR = K[x1, . . . , xn]/P and g the image of f in this quotient ring. It is clearthat g is non-zero since f /∈ P .

Consider the ideal 〈g〉 in R, and R = R〈g〉 = Rg = R[1/g] the localization

of R with respect to this ideal. R has a maximal ideal M , therefore R/M isa field and a K-algebra finitely generated by 1/g and the images of the xi.Then, the K-algebra homomoprphism

φ : K[x1, . . . , xn]→ K[x1, . . . , xn]/J → R→ R→ K

is surjective and its kernel is a maximal ideal N that contains J .Let V be an affine variety. For each p ∈ V let mp denote the ideal

{f ∈ A(V ) | f(p) = 0}, which is maximal in A(V ). Let xi be the image ofthe xi in A(V ), and consider the mapping m : V → Max(A(V )) defined asm(p) = mp. This map is clearly injective, and it is easy to see that it alsois surjective.

Now, consider the morphism from A(V ) onto K that maps f to f(p),clearly its kernel is mp. The polynomials xi − pi ∈ A(V ) vanish at p =(p1, . . . , pn), thus they are in mp. Even more so, the morphism

σp : K[x1, . . . , xn]→ K


is surjective and its kernel is the maximal ideal 〈x1−p1, . . . , xm−pn〉, whichcontains I(V ). Therefore, mp = 〈x1 − p1, . . . , xm − pn〉.

Thus, the maximal ideal N is of the form mp = 〈x1 − p1, . . . , xm − pn〉,for some p ∈ An, and so φ(xi) = pi, and φ = σp. Since φ(J) = 0 andφ(mp) = 0, we have that p ∈ Z(J). Considering that g is a unit in R, itsimage in K also is a unit, thus g(p) = φ(f) 6= 0, and f /∈ I(Z(J)).

2.3 Grobner Bases

The main purpose of this section is to give an alternative proof of Hilbert’sNullstellensatz. With this in mind, we first need to give some preliminarydefinitions on elements of R[x1, . . . , xn]. From now when we refer to an orderon the monomials of R[x1, . . . , xn] we will be referring to the lexicographicorder on the xi, i.e., x1 ≥ x2 ≥ · · · ≥ xn.

Definition 2.26. Let f be an element of R[x1, . . . , xn]. The leading mono-mial of f , denoted lm(f), is the monomial term of maximal order in f . Theleading monomial of f ≡ 0 is 0. The leading term of f , denoted by lt(f),is a monic monomial on R[x1 . . . , xn] such that lm(f) = r lt(f), for somer ∈ R.

For an ideal I in R[x1, . . . , xn] we define LM(I) = {lm(f) | f ∈ I}, andLT(I) = {lt(f) | f ∈ I}. The multidegree of a monomial xα1

1 xα22 · · ·xαn

n is then-tuple α = (α1, . . . , αn), and we define the multidegree of a polynomial f ∈K[x1, . . . , xn], denoted by ∂(f), as the multidegree of its leading monomial.

Remark. If S is a subset of R[x1, . . . , xn], we define LM(S) as the idealgenerated by the set {lm(f) | f ∈ S}, i.e., LM(S) = 〈lm(f) | f ∈ S〉.

Definition 2.27. Let R be a Noetherian ring, and I an ideal inR[x1, . . . , xn]. A set G of non-zero polynomials, not necessarily finite, issaid to be a Grobner basis for I if it satisfies that for all f ∈ I we havef = hig1 + · · · + hrgr, with hi ∈ R[x1, . . . , xn] and gi ∈ G, such thatlt(f) = max(lt(hi) lt(gi)).

Remark. The fact that every element of I can be written as a sum of elementsin G implies that G ⊆ I.

Proposition 2.28. If the set G is a Grobner basis for I, then I = 〈G〉.

Proof. Since G ⊆ I, it is clear that 〈G〉 ⊆ I. To prove the other inclusiontake f ∈ I, from the definition we have that f = h1g1 + · · ·+ hrgr for somehi ∈ R[x1, . . . , xn], i.e., f ∈ 〈G〉, and I ⊆ 〈G〉 as desired.


Theorem 2.29. Let I be an ideal in K[x1, . . . , xn], and G = {g1, . . . , gr} aset of non-zero polynomials in I. Then, LM(G) = LM(I) if and only if Gis a Grobner basis for I.

Proof. Let G be a Grobner basis for I, since G ⊆ I it is clear that LM(G) ⊆LM(I). Let f ∈ I, and S = {i ∈ Z+ | lt(f) = lt(hi) lt(gi)}, then the sum

∑

i∈Slm(hi) lm(gi)

is the leading monomial of f . Since all the elements of the sum are elementsof G, it is clear that lm(f) ∈ LM(G).

A proof of the converse can be found in [1, p. 208].

Now, since the ring K[x1, . . . , xn] is Noetherian, the following corollaryfollows from the theorem.

Corollary 2.30. Let I be an ideal in K[x1, . . . , xn]. Then, I has a finiteGrobner basis.

Proof. Since LM(G) and LM(I) are both ideals in a Noetherian ring, theyare finitely generated. Assume that LM(G) = 〈g1, . . . , gr〉, from Theorem2.29 we know that {g1, . . . , gr} is a finite Grobner basis for I, even more so,from Proposition 2.28 we also know that I = 〈g1, . . . , gr〉.

Theorem 2.31. If G = {g1, . . . , gr} is a Grobner basis in K[x0, x1, . . . , xn],with x0 < xi for 1 ≤ i ≤ n, then G is a Grobner basis in K[x0][x1, . . . , xn].

Proof. Throughout this proof the notation for the leading monomial andterm will be different, lmx0 and ltx0 will denote the leading monomial andterm in K[x0][x1, . . . , xn], and lm and lt the leading monomial and term inK[x0, x1, . . . .xn]. Let I = 〈g1, . . . , gr〉, we need to show that LMx0(I) =〈lmx0(g1), . . . , lmx0(gr)〉.

It is clear that 〈lmx0(g1), . . . , lmx0(gr)〉 ⊆ LMx0(I). To prove the otherinclusion take f ∈ I, since G is a Grobner basis we have that f = h1g1 +· · ·+ hrgr, but the hi can be re-written as elements of K[x0][x1, . . . , xn] as a

sum of ai,jxki,j0 hi,j , where ai,j ∈ K and the hi,j are monic and have only one

term. Then, we have that

f =∑

ai,jxki,j0 hi,jgi =

∑Ci,j .

Without loss of generality assume that

lt(Ci,j) > lt(Ci,j+1) > lt(Ci+1,j).


Considering the gi as elements of K[x0][x1, . . . , xn], we can write their lead-ing term lt(gi) as αixi, with αi ∈ K[x0] and xi ∈ K[x1, . . . , xn]. Since x0

is the smallest in the ordering of the xi, and lt(Ci,j) = xki,j0 hi,j lt(gi) =

xki,j0 hi,jαixi, we have that

hi,j xi ≥ hi,j+1xi ≥ hi+1,j xi+1.

Choose the least L such that hL,j xL > hL+1,j xL+1, i.e., hl,j xl = h1,1x1 forl ≤ L. Then, we have that

lmx0(f) =

(L∑

l=1

al,jxkl,j0 αl

)h1,1x1 =

L∑

l=1

al,jxkl,j0 hl,j(αlxl),

which is a sum of elements of 〈lmx0(g1), . . . , lmx0(gr)〉 as desired.

Definition 2.32. Let I be an ideal in K[x1, . . . , xn]. A strong Grobner basisfor I is a set Γ ⊆ I \ {0}, such that for every m ∈ LM(I) there exists g ∈ Γthat satisfies lm(g)|m.

It is clear form the definition that every strong Grobner basis is a Grobnerbasis, even more so, given any Grobner basis in a ring of polynomials withcoefficients over a PID it is possible to construct a strong Grobner basis. Adetailed explanation of this construction can be found in [1, p. 251].

Theorem 2.33. G is a Grobner basis in K[x0, x1, . . . , xn], with x0 < xi for1 ≤ i ≤ n, if and only if it is a strong Grobner basis in K[x0][x1, . . . , xn].

Proof. The proof of this theorem is very similar to that of Theorem 2.31,and can be found in [1, p. 254].

Proposition 2.34. Let R be a PID, and I an ideal in R[x1, . . . , xn]. Then,I has a finite strong Grobner basis Γ such that I = 〈Γ〉. In particular, if Ris a field then I = R[x1, . . . , xn] if and only if Γ ∩R 6= ∅.

Proof. The first part of the theorem follows from Corollary 2.30 and Theo-rem 2.33. As for the second part, if Γ ∩ R 6= ∅ then I clearly is the wholering, because I = 〈Γ〉. Assume that Γ ∩ R = ∅, then the polynomial f ≡ r,with r ∈ R, does not belong to I, thus I 6= R[x1, . . . , xn].

Proposition 2.35. Let R1 and R2 be rings, I an ideal in R1[x1, . . . , xn],and Γ a strong Grobner basis for I. Consider φ : R1 → R2 an epimorphism,such that φ(g) is neither a zero divisor nor zero for all g ∈ LM(Γ). Then,for all h ∈ φ(I) there exists f ∈ I ∩ φ−1(h) such that lt(f) = lt(h).


Remark. Note that we use φ for both the morphism between R1 and R2 andits natural extension to a morphism between R1[x1, . . . , xn] andR2[x1, . . . , xn], since there is no risk of confusion as it is made clear bythe element in which it is being applied.

Proof. Let h ∈ φ(I), and assume that for all f ∈ φ−1(h) we have thatlt(h) 6= lt(f). From these f choose fm of minimal leading term. Since φis surjective, and lt(fm) 6= lt(h) we must have that ∂(fm) > ∂(h), then

φ(lm(fm)) = 0. Now, construct f ′ = fm − lm(fm)lm(g) g, for some g ∈ Γ. From

our conditions on φ we know that φ(lm(g)) is not a zero divisor nor zero,

then φ( lm(fm)lm(g) ) = 0, this means that φ(f ′) = h and lm(f ′) < lm(fm), which

contradicts our assumption on the minimality of lm(fm).

Corollary 2.36. Let R1 and R2 be rings, I an ideal in R1[x1, . . . , xn], andΓ a strong Grobner basis for I. Consider φ : R1 → R2 as before, then φ(Γ)is a strong Grobner basis for φ(I).

Proof. Since f ∈ I ∩ φ−1(h), we know that there exists g ∈ Γ such thatlm(g)| lm(f), but we also know that lt(h) = lt(f), thus lm(φ(g)) = φ(lm(g))divides lm(h) and φ(Γ) is a strong Grobner basis for φ(I).

To prove Hilbert’s Nullstellensatz with Grobner bases what we will dois prove the weak form of this theorem with Grobner bases, and then provethat the weak form implies the strong form.

Theorem 2.37 (Weak form of Hilbert’s Nullstellensatz). Let K be an al-gebraically closed field, and I a proper ideal in K[x1, . . . , xn]. Then, thereexists a point p ∈ An that satisfies f(p) = 0 for all f ∈ I.

To prove this theorem we will use two auxiliary lemmas, in which K andI are as above, and for k ∈ K we define the morphism εk as

εk : K[x1, . . . , xn] −→ K[x2, . . . , xn]f(x1, x2, . . . , xn) 7−→ f(k, x2, . . . , xn)

.

Lemma 2.38. Assume that I ∩ K[x1] = 〈p〉, for some p ∈ K[x1] of pos-itive degree. Then, there exists k ∈ K, with p(k) = 0, such that εk(I) 6=K[x2, . . . , xn].

Proof. In order to prove this lemma, we will first prove the following threestatements.


i. Let f1, f2 ∈ K[x1], and G = {g1, . . . , gr} ⊂ K[x1, . . . , xn], such that(f1, f2) = 1, then 〈f1f2, G〉 = 〈f1, G〉 ∩ 〈f2, G〉.

To prove this consider the equation q1f1 + q2f2 = 1, which has a solutionand is an element of 〈f1, G〉+ 〈f2, G〉, then

〈f1, G〉 ∩ 〈f2, G〉 = 〈f1, G〉〈f2, G〉 = 〈f1, f2, f1G, f2G〉 = 〈f1, f2, G〉.

Applying this argument repeatedly we obtain

〈kΠi=1

(x1 − αi)ci , G〉 =

k⋂

i=1

〈(x1 − αi)ci , G〉.

ii. Let I be an ideal in K[x1, . . . , xn]. I = K[x1, . . . , xn] if and only ifr(I) = K[x1, . . . , xn].

It is clear that if I = K[x1, . . . , xn], then r(I) = K[x1, . . . , xn]. Now, consideran ideal I in K[x1, . . . , xn] such that r(I) = K[x1, . . . , xn], this means that1r ∈ I for some r ∈ Z+, but 1r = 1 for all r, then 1 ∈ I and K[x1, . . . , xn].

iii. Let K be an algebraically closed field, f ∈ K[x1], and G ⊆K[x1, . . . , xn]. Suppose that 〈f,G〉 6= K[x1, . . . , xn], then there existsk ∈ K such that f(k) = 0 and 〈(x1 − k), G〉 6= K[x1, . . . , xn].

Let 〈f,G〉 6= K[x1, . . . , xn], from i. we know that 〈(x1 − k)c, G〉 6=K[x1, . . . , xn] for some k ∈ K such that f(k) = 0 and c ∈ Z+. It is easyto see that r(〈(x1 − k), G〉) ⊆ r(〈(x1 − k)c, G〉). Then, due to ii., we have〈(x1 − k), G〉 6= K[x1, . . . , xn].

Now, taking into account that

K[x1, . . . , xn]/〈(x1 − k), G〉 ' K[x2, . . . , xn]/εk(〈G〉),

it is clear that 〈(x1−k), G〉 6= K[x1, . . . , xn] if and only if εk(〈(x1−k), G〉) 6=K[x2, . . . , xn], which proves the lemma.

Lemma 2.39. Let I ∩ K[x1] = ∅. Then, there exists q ∈ K[x1] such thatεk(I) 6= K[x2, . . . , xn], for all k ∈ K that satisfy q(k) 6= 0.

Proof. Consider K[x1, x2, . . . , xn] as K[x1][x2, . . . , xn]. Let Γ be a finitestrong Grobner basis for an ideal I in K[x1][x2, . . . , xn], and q ∈ K[x1] bethe product of the leading coefficients of all elements of Γ. If q(k) 6= 0,we have from proposition 2.35 that εk(Γ) is a strong Grobner basis forεk(I). Since Γ ∩ K[x1] ⊆ I ∩ K[x1], we have that Γ ∩ K[x1] = ∅. Then,εk(Γ) ∩ εk(K[x1]) = εk(Γ) ∩ K = ∅ and by proposition 2.34 we have thatεk(I) 6= K[x2, . . . , xn].


With these lemmas we can now proceed to prove the weak form ofHilbert’s Nullstellensatz.

Proof of Theorem 2.37. Let I be a proper ideal in K[x1, . . . , xn]. By lem-mas 2.38 and 2.39 we have that there exists k1 ∈ K such that εk1(I) = I1 6=K[x2, . . . , xn]. Now, applying the two lemmas again we know that there ex-ists k2 ∈ K such that εk2(εk1I) = εk2(I1) = I2 6= K[x3, . . . , xn]. Continuingwith this process we obtain a point p = (k1, . . . , kn) ∈ An such that f(p) = 0for all f ∈ I.

Third proof of Theorem 2.21. As we have stated before, it is clear that r(J)⊆ I(Z(J)). Let f ∈ I(Z(J)), and consider the ideal 〈(x0f − 1), J〉 inK[x0, x1, . . . , xn]. It is clear that Z(〈(x0f − 1), J〉) = ∅, which by The-orem 2.37 means that 〈(x0f − 1), J〉 = K[x0, x1, . . . , xn], and since 1 ∈K[x0, x1, . . . , xn] we have that

1 = (x0f − 1)h0 +

r∑

i=1

gihi

for gi ∈ J and hi ∈ K[x0, x1, . . . , xn]. Taking x0 = f−1 and multiplying byfN , where N is the greatest power of x0 in the hi, we obtain

fN =

r∑

i=1

giji

where ji = fNhi ∈ K[x1, . . . , xn]. Thus, fN ∈ r and f ∈ r(r).

2.4 Primary Decomposition

Definition 2.40. Let I be a proper ideal in R. We say that I is primary ifxy ∈ I implies that either x ∈ I or yn ∈ I, for some n ∈ Z+, i.e., every zerodivisor in R/I is nilpotent.

Remark. It is clear from the definition that every prime ideal is a primaryideal.

Proposition 2.41. Let I be a primary ideal. Then r(I) is the smallestprime ideal that contains I.

Proof. We know from proposition 1.11 that r(I) is the intersection of primeideals, and thus an ideal, so all that is left to prove is that r(I) is prime.


Let x, y ∈ R such that xy ∈ r(I), then (xy)m ∈ I for some m ∈ Z+. SinceI is primary we have that either xm ∈ I or ymn ∈ I, then x ∈ r(I) ory ∈ r(I).

Remark. If I is a primary ideal, and r(I) = P , we say that I is P -primary.

Example 2.4. Let R = K[x1, x2] and I = 〈x1, x22〉. It is clear that R/I 'K[x2]/〈x22〉, and that the zero divisors of this ring are the multiples ofx2. These elements are all nilpotent in R/I, hence 〈x1, x22〉 is primary andr(〈x1, x22〉) = 〈x1, x2〉 = P .

Notice that in the previous example we also have that P 2 ⊂ I ⊂ P ,thus this example shows us that a primary ideal need not be a power of aprime ideal. The following example will show us that the converse is nottrue either, i.e., the power of a prime ideal not always is a primary ideal.

Example 2.5. Consider the ring R = K[x1, x2, x3]/〈x1x2 − x23〉, and let xidenote the image of xi in R. Take the prime ideal I = 〈x1, x3〉 in R. Wehave that x1x2 = x3

2 ∈ I2, but x1 /∈ I2 and x2 /∈ r(I2). Thus, I2 is notprimary even though I is prime.

Let I, J be ideals in R, their ideal quotient is the set

(I : J) = {r ∈ R | rJ ⊆ I} .

It is clear that the sum of two elements of (I : J) is again an element of(I : J). Consider r ∈ R, j ∈ J , and x ∈ (I : J), it is easy to check thatxr ∈ (I : J), since xrj = r(xj). Thus, (I : J) is a ideal. If J = 〈x〉 we write(I : x) instead of (I : 〈x〉).

Example 2.6. Consider the ideals 〈x1x3, x2x3〉 and 〈x3〉 in K[x1, x2, x3],then we have that

(〈x1x3, x2x3〉 : 〈x3〉) = {f ∈ K[x1, x2, x3] | f〈x3〉 ⊆ 〈x1x3, x2x3〉}

= {f =∑αix

n1x

m2 x

r3 | f〈x3〉 ⊆ 〈x1x3, x2x3〉}

= {f =∑αix

n1x

m2 | f〈x3〉 ⊆ 〈x1x3, x2x3〉}

and this is the same as 〈x1, x2〉, thus (〈x1x3, x2x3〉 : 〈x3〉) = 〈x1, x2〉.

Lemma 2.42. Let I be a P -primary ideal, and x ∈ R. Then:

1. If x ∈ I, then (I : x) = 〈1〉 = R.


2. If x /∈ I, then (I : x) is P -primary, i.e., r(I : x) = P .

3. If x /∈ P , then (I : x) = I.

Remark. We will use r(I : J) to denote r((I : J)) in order to avoid acumbersome notation.

Proof. 1. Since x ∈ I, it is clear that 1x ∈ I. Thus (I : x) = 〈1〉.

2. First we will see that (I : x) is primary. Let yx0 ∈ (I : x), withy /∈ r(I : x), then yx0x ∈ I, but because of our condition on y wemust have that x0x ∈ I, i.e., xm0 ∈ (I : x) for some m ∈ Z+. Thus,(I : x) is primary.

Now, to prove that (I : x) is P -primary, let y ∈ (I : x), then xy ∈ I,but from our assumption on x we must have that y ∈ r(I) = P .Then, we have that I ⊆ (I : x) ⊆ P , and taking radicals we obtainP ⊆ r(I : x) ⊆ P . Thus, (I : x) is P -primary as desired.

3. It is clear that I ⊆ (I : x). To prove (I : x) ⊆ I consider y ∈ (I : x),then y(rx) ∈ I for all r ∈ R. Taking r = 1 we obtain yx ∈ I,then y ∈ r(I) = P . Repeating the process from 2 we obtain thatr(I : x) = P .

Now, consider y ∈ (I : x) such that y /∈ I, then y(rx) ∈ I, and since Iis primary we must have that (rx)n = rnxn ∈ I. Since x /∈ P then rn

must be in I. Taking r = 1 we obtain that I = 〈1〉, which contradictsthe fact that I is a proper ideal in R.

If for an ideal I in R we have that

I =

n⋂

i=1

Ii

with n ∈ N and all the Ii primary, we say that this expression is a primarydecomposition of I. If such a decomposition exists for an ideal, we say thatit is decomposable.

A ring in which every ideal is decomposable is called a Lasker ring inhonor of Emanuel Lasker, who in 1905 proved that polynomial rings areLasker rings. Years later, Emmy Noether proved a generalization of Lasker’sresult, known as the Lasker-Noether Theorem (Theorem 2.49), which statesthat every Noetherian ring is a Lasker ring.


This theorem is of great importance because it generalizes the Funda-mental Theorem of Arithmetics to the ideals of the factorization of everyelement in a UFD. A classic example of this is the ring Z[

√−5], which is an

integral domain that is not a UFD since the element 6 has more that onedecomposition, namely 2 ·3 and (1−

√−5)(1+

√−5). But it can still happen

that the ideal generated by this element can factor into the intersection ofprimary ideals, and the Lasker-Noether theorem guaranties us that at leastthis happens in all Noetherian rings.

Definition 2.43. Let I be a decomposable ideal, with I = ∩Ii. If all ther(Ii) are distinct and ∩j 6=iIj * Ii we say that the primary decomposition isminimal or irredundant.

Now we will prove a form of uniqueness on the minimal primary decom-position of an ideal I.

Theorem 2.44. Let I be a decomposable ideal in R, with ∩ni=1Ii = I aminimal primary decomposition, and Pi = r(Ii). Then, the Pi are the primeideals in the set {r(I : x) | x ∈ R}, and thus are independent of the decom-position of I.

Proof. For any x ∈ R we have that (I : x) = (∩ni=1Ii : x) = ∩ni=1(Ii :x), it follows that r(I : x) = ∩ni=1r(Ii : x), and taking into account ourcharacterization of the ideals (I : x) we obtain that r(I : x) = ∩ni=1Pi. Ifr(I : x) is prime, then by Proposition 2.22 it must be one of the Pi. Thus,every prime in {r(I : x) | x ∈ R} is one of the Pi.

Now, since the decomposition is minimal, we have that there existsxi /∈ Ii such that xi ∈ ∩j 6=iIj , and from lemma 2.42 we have that r(Ii :xi) = Pi. Hence, for every Pi there exists an xi such that Pi = r(Ii, xi) ∈{r(I : x) | x ∈ R}.

The prime ideals Pi in theorem 2.44 are said to be associated with I.Let PI be the set of all ideals associated with I. The minimal elements ofPI are called the isolated prime ideals associated with I. All others idealsin PI are called embedded prime ideals.

These names come from considering the ring R to be K[x1, . . . , xn]. Inthis case the ideal I has an associated algebraic set, namely the zero set Z(I).The zero sets of the isolated ideals are varieties contained in Z(I) that arenot subsets of other varieties, thus the name isolated. On the other hand,the zero sets of embedded ideals are varieties embedded into the isolatedvarieties, therefore the name embedded.


Proposition 2.45. Let I be a decomposable ideal, and I = ∩ni=1Ii a minimalprimary decomposition with r(Ii) = Pi. Then

n⋃

i=1

Pi = {r ∈ R | (I : x) 6= I} .

In particular, if the zero ideal is decomposable, then the union of all primeideals associated with 0 is the set of zero divisors of R.

Proof. If I is decomposable, then 0 is decomposable in R/I, thus we willonly prove the proposition for the zero ideal. We know that the set of zerodivisors D can be expressed as ∪x 6=0r(0 : x), but for each r(0 : x) we havethat r(0 : x) = ∩Pi ⊂ Pj for some j, thus D ⊂ ∪Pi.

On the other hand, we have that each Pi is of the form r(0 : x) for somex ∈ R, therefore D ⊃ ∪Pi and D = ∪Pi as desired.

Now, we will study primary decomposition of ideals in rings that haveproven to be very important, Noetherian rings. We will prove that in factevery proper ideal of a Noetherian ring has a primary decomposition.

Definition 2.46. A ideal I is said to be irreducible if for any pair of idealsI1, I2 such that I = I1 ∩ I2, either I1 = I or I2 = I.

Lemma 2.47. In a Noetherian ring every ideal can be written as the inter-section of finite irreducible ideals.

Proof. Assume this does not hold for a non-empty set of ideals Σ, since thering is Noetherian this set has a maximal element M that is reducible, thenthere exist ideals I, J such that M = I ∩ J . This implies that M ⊆ Iand M ⊆ J , i.e., neither I nor J belong to Σ, and by definition they arethe intersection of finite irreducible ideals, meaning that so is M , whichcontradicts our assumption.

Lemma 2.48. In a Noetherian ring every irreducible ideal is primary.

Proof. Let I be an irreducible ideal in R, and consider R′ = R/I, we willshow that every zero divisor is nilpotent. Assume xy = 0 with both x andy non-zero elements of R′, and consider the chain of ideals

(0 : x) ⊆ (0 : x2) ⊆ · · · ,

since the ring is Noetherian the chain stabilizes at some point, say at n,i.e., (0 : xn) = (0 : xn+1) = · · · . Now, we will see that 〈xn〉 ∩ 〈y〉 = 〈0〉.


Let z ∈ 〈xn〉 ∩ 〈y〉, this means that z ∈ 〈y〉 and by definition we have thatzx = 0, even more so, since z is also in 〈xn〉, then it can be written as rxn

for some r ∈ R′, this means that rxn+1 = 0, i.e., r ∈ (0 : xn+1) = (0 : xn),then z = rxn = 0.

Since 〈y〉 6= 0, we must have that 〈xn〉 = 0, which implies that x isnilpotent. Thus, every zero divisor in R′ is nilpotent as desired.

Theorem 2.49 (Lasker-Noether Theorem). Every Noetherian ring is aLasker ring.

Proof. This clearly follows from Lemmas 2.47 and 2.48.

Chapter 3

Dimension

The main purpose of this chapter is to construct the example given byNagata in [7] of a Noetherian ring with infinite dimension, in order to dothis we first define the Krull dimension of a ring, which we use to re-definethe dimension of affine varieties in algebraic terms instead of topologicalones. Another important result stated in this chapter is the Dimensiontheorem, although we exclude the part that uses the Hilbert polynomial asit requires more tools that the ones we have provided so far. Should thereader want the complete theorem with its proof, it can be found in [2].

3.1 Krull Dimension

An important geometric concept is that of dimension, in this section we willgive the definition of the dimension of a ring as well as that of dimensionof an affine variety, and some of their properties. Take P0, . . . , Pn all primeideals, then

P0 ⊂ P1 ⊂ · · · ⊂ Pnis called a prime chain of length n.

Definition 3.1. Let P be a proper prime ideal. The height of P , denotedby ht(P ), is the supremum of the length of all chains that end in P , i.e., thesupremum of the n such that P0 ⊂ P1 ⊂ · · · ⊂ Pn = P .

With this definition we now have the necessary tools to define the di-mension of a ring.

Definition 3.2. The Krull dimension of a ring R, denoted dimR, is thesupremum of the heights of all prime ideals in R, i.e.,

dimR = sup {ht(P ) | P is a prime ideal in R}

41

CHAPTER 3. DIMENSION 42

Remark. If P is a prime ideal in R there is a one to one correspondencebetween prime ideals contained in P and prime ideals in RP , then we havethat

ht(P ) = dimRP .

Example 3.1. Since the only proper ideal a field has is the zero ideal,which is a prime ideal in the case of fields, the only prime chain is the oneconsisting this ideal, i.e., the only possible prime chain has length 0. Thus,the dimension of any field is 0.

On topological spaces there already exists a concept of dimension, andsince algebraic varieties are subsets of the affine space An, which is a topo-logical space, we can endow them with a dimension.

Definition 3.3. Let X be a topological space. The length of the chain

X0 ⊂ X1 ⊂ · · · ⊂ Xn,

with the Xi irreducible closed subsets of X, is n. The dimension of X asa topological space is defined as the supremum over the length of chains ofirreducible closed subsets of X.

The dimension of an algebraic variety V , denoted dimV , is its dimensionas a topological space. Our previous definition of the dimension of a varietyis purely topological, there is also an algebraic definition to this dimension,and we will prove that they are equivalent, but in order to define it we firstneed some preliminary concepts.

Definition 3.4. A topological space is said to be Noetherian if every de-scending chain of closed subsets stabilizes.

Proposition 3.5. The affine n-space with the Zariski topology is a Noethe-rian topological space.

Proof. Let X0 ⊇ X1 ⊇ · · · be a chain of closed subsets of An, then I(X0) ⊆I(X1) ⊆ · · · is a chain of ideals in K[x1, . . . , xn], thus it stabilizes and thechain of closed subsets does too.

Proposition 3.6. Let X be a Noetherian topological space, then every closedsubset Y of X can be written as the finite union of irreducible closed subsetsof X, i.e.,

Y = Y1 ∪ · · · ∪ Yr,with all the Yi irreducible. Even more so, if we have that Yi * Yj when i 6= j,then this decomposition is unique.


Proof. Let Σ be the set of all non-empty closed subsets of X for whichthe proposition does not hold, and assume that it is non-empty. SinceX is Noetherian Σ has a minimal element, say Y0. By definition Y0 isnot irreducible, then there exist Y1 and Y2 closed subsets of X such thatY0 = Y1 ∪ Y2.

Due to the minimality of Y0, we have that both Y1 and Y2 are expressibleas a union of finite irreducible closed sets, but that would mean that Y0 isalso expressible as a finite union of irreducible closed sets, contradicting ourassumption on Y0. Thus Σ is empty and every closed subset of X can bewritten as the union of finite irreducible closed sets.

To prove the unicity, let Y = Y1 ∪ · · · ∪ Yr and assume that we haveeliminated all sets such that the resulting ones satisfy Yi * Yi when i 6= j.Suppose that Y = Y ′1 ∪ · · · ∪ Y ′s is another decomposition that satisfies thecondition, then Y ′1 ⊆ Y = Y1 ∪ · · · ∪ Yr, but Y ′1 is irreducible, hence itmust be contained in one of the Yi. Without loss of generality, assume thatY ′1 ⊆ Y1. Applying the same process to Y1 we obtain that Y1 ⊆ Y ′i , but fromour condition on the Y ′i we must have that i = 1, i.e., Y1 = Y ′1 . Repeatingthis process inductively we get that both decompositions are equal, andsince we chose any two decompositions this implies that the decompositionis unique.

This proposition gives us the following corollary, which will be useful togeneralize the concept dimension to any algebraic set.

Corollary 3.7. Every algebraic set can be written uniquely as a union ofdistinct affine varieties.

Lemma 3.8. Let V be an affine variety, then its dimension is the same asthe Krull dimension of A(V ).

Proof. Consider a chain of subvarieties of V , then the chain of their ideals isa prime chain in K[x1, . . . , xn] that starts at I(V ). Considering this primechain as a chain in A(V ) it is a prime chain that starts at 0, thus the Krulldimension coincides with the dimension of V as a topological space.

Corollary 3.7 shows us that the affine varieties that make up an affineset are unique. Then, with Lemma 3.8 we can define the dimension of anyalgebraic set as follows.

Definition 3.9. Let Y be an algebraic set in An. We define the dimensionof Y , denoted by dimY , as the Krull dimension of its affine coordinate ringA(Y ), i.e., the greatest dimension of the algebraic varieties that make up Y .


Since the varieties that make up an algebraic set are unique, the dimen-sion of algebraic sets is well defined. The following theorem is an importantresult on the dimension of local Noetherian rings, the proof of this theoremneeds more concepts of dimension theory, hence we will only state it andnot prove it. A proof of it can be found in [2, p. 121].

Theorem 3.10 (Dimension Theorem). For any local Noetherian ring Rwith maximal ideal M the following integers are equal:

• The maximum length of prime chains in R.

• The least number of generators of an M -primary ideal.

Corollary 3.11. Let R be a Noetherian ring, and x1, . . . , xr elements ofR. Then, every minimal ideal P associated with 〈x1, . . . , xr〉 has height lessthat r.

Proof. Consider the ring RP , and the image of 〈x1, . . . , xr〉 via φ : R→ RP ,which is a P e-primary ideal. Thus, by Theorem 3.10 we have that

ht(P ) = dimRP ≤ r.

Theorem 3.12 (Krull’s Hauptidealsatz). Let R be a Noetherian ring, andr a non-zero element in R that is not a unit. Then, every minimal primeideal P associated with 〈r〉 has height 1.

Proof. From Corollary 3.11 we have that ht(P ) ≤ 1. If ht(P ) = 0, then P isassociated with the zero ideal, which by Proposition 2.45 means that everyelement in P is a zero divisor, but this contradicts our assumption on x, andthus ht(P ) = 1.

The following example is of great importance and there are at least twopossible ways of proving it. One of them uses Hilbert’s polynomial anddimension theory, which we have not properly introduced, so we will proveit the other way, which uses the Dimension Theorem and the fact that thecoordinate ring of An is K[x1, . . . , xn].

Example 3.2. The dimension of the affine n-space, An, is n

Theorem 3.13. Let R be a Noetherian ring, then

dimR[x] = 1 + dimR.


We will first prove a result on rings that are not necessarily Noetherian,and then we will proceed to prove this theorem.

Proposition 3.14. Let R be a ring. Then, 1 + dimR ≤ dimR[x] ≤ 1 +2 dimR.

Proof. Let P0 ⊂ P1,⊂ · · · ⊂ Pr be a prime chain of length r in R, thenP0[x] ⊂ P1[x],⊂ · · · ⊂ Pr[x] ⊂ Pr + 〈x〉 is a prime chain of length r + 1 inR[x]. All the ideals in the second chain are prime since the Pi are primeand Pr + 〈x〉 is the kernel of the epimorphism from R[x] onto R/Pr. Thus,1 + dimR ≤ dimR[x] for any ring.

Consider the homomorphisms φ : R → R[x] and φ∗ : Spec(R[x]) →Spec(R). Then, the fiber of an ideal P over φ∗, defined as (φ∗)−1(P ), canbe identified in the following way

(φ∗)−1(P ) ' Spec(k(P )⊗R R[x]),

where k(P ) = RP /PRP is the residual field of the local ring RP . Moreprecisely, with the multiplicatively closed subsets S = R \ P of R and T =(R/P ) \ {0} of RP , it can be proven that

k(P )⊗R R[x] = (RP /PRP )⊗R R[x] ' (R/P )P ⊗R R[x]

' T−1(R/P )⊗R R[x] ' T−1(R/P )⊗R/P R/P ⊗R R[x]

' T−1(R/P )⊗R/P (R[x]/PR[x]) ' T−1(R[x]/PR[x])

' S−1(R[x]/PR[x]) ' R[x]P /PR[x]P = k(P )[x].

Thus, there exists a homomorphism between the set of ideals in R[x] thatproject onto P and Spec(k(P ) ⊗R R[x]). Since k(P ) ⊗R R[x] ' k(P )[x],and k(P )[x] is a PID, it follows that dim k(P ) = 1. Thus, a prime chainthat ends at P [x] in R[x] has at most length 1, then a prime chain of lengthr in R is the contraction of a chain of length 1 + 2r in R[x]. Therefore,dimR[x] ≤ 1 + 2 dimR.

With this result on any ring, all that is left to prove is that in theNoetherian case we also have that dimR[x] ≤ 1 + dimR.

Proof of Theorem 3.13. Consider a prime ideal P in R[x], and let P ′ = P c,which is an ideal in R. Since P ′ is prime we can localize with respect toit to obtain an ideal PP ′ in R[x]P ′ and P ′P ′ in RP ′ , even more so, sincedimRP ′ = ht(P ′) we have that ht(P ) = ht(PP ′) and ht(P ′) = ht(P ′P ′). Sowe may assume R is local with maximal ideal P ′.


If ht(P ′) = m, then by Theorem 3.10 there exists a P ′-primary ideal Iin R generated by m elements. Even more so, I[x] is P [x]-primary since

r(I[x]) = r(r(I) + r(I〈x〉)) = r(P ′ + r(I) ∩ r(〈x〉))= r(P ′ + P ′ ∩ 〈x〉) = r(P ′ + P ′〈x〉) = r(P [x]) = P [x],

and every zero divisor in R[x]/I[x] ' (R/I)[x] is nilpotent, since all zerodivisors in R/I are nilpotent.

Furthermore, since I[x] is generated in R[x] by the m generators of Iwe have that ht(I) ≤ m, even more so, from every prime chain P0 ⊂ P1 ⊂· · · ⊂ Pr = P ′ in R we can construct the prime chain P0[x] ⊂ P1[x] ⊂ · · · ⊂Pr[x] = P in R[x]. Thus, ht(P ) ≥ m.

Now, we will show that ht(P ) ≤ 1 + ht(P ′), we do this by induction onht(P ′). If ht(P ′) = 0, then the prime chain begins and ends at P ′, and itis the contraction of a chain of length 1 in R. Assume that this holds forht(P ′) < m, and let ht(P ′) = m, we will show that ht(P ) ≤ 1 + m. Todo this consider all primes ideals I ⊆ P , then ht(I) ≤ m, this means everychain that ends in P has length less than 1 +m.

To prove this consider I ′ = Ic an ideal properly contained in P ′ = P c,then it has height strictly less that m, thus we have that ht(I) ≤ m. IfI ′ = P ′, then P ′[x] ⊂ I ( P , and since the longest prime chain in R[x] overP ′ has two primes, it follows that I ′ = P ′[x] has height m.

Corollary 3.15. For any field K we have that dimK[x1, . . . , xn] = n.

Proof. This follows from Theorem 3.13 and the fact that dimK = 0.

Proposition 3.16. A Noetherian integral domain R is a UFD if and onlyif every prime ideal of height 1 is principal.

Proof. Suppose the condition holds, i.e., every prime ideal of height 1 isprincipal. Let r be an irreducible element in R, and let P be a minimalprime ideal associated with 〈r〉. By Theorem 3.12 we have that ht(P ) = 1,and thus by hypothesis it is principal. Therefore, P = 〈p〉 for some p ∈ P ,and r = mp. But, by the irreducibility of r, m must be a unit and 〈r〉 = P .Since we did this for any irreducible element of R, we have that R is a UFD.

To prove the converse let P be a prime ideal of height 1 in R, and pan irreducible element in P . Clearly 〈p〉 is prime, since R is a UFD, and〈p〉 ⊆ P . Furthermore, since ht(P ) = 1 we must have that 〈p〉 = P . Thus,P is principal.


Proposition 3.17. A variety V in An has dimension n−1 if and only if it isthe zero set of a single non-constant irreducible polynomial in K[x1, . . . , xn].

Proof. Since K[x1, . . . , xn] is a UFD, then f is prime and 〈f〉 is a primeideal. Furthermore, by Theorem 3.12 we have that ht(〈f〉) = 1, and since

ht(P ) + dimK[x1, . . . , xn]/P = dimK[x1, . . . , xn]

for any prime ideal P , we have that dimZ(f) = dimK[x1, . . . , xn]/〈f〉 =n−1. Conversely, if V is a variety of dimension n−1 its ideal I(V ) has height1. Therefore, due to Proposition 3.16, I(V ) is principal, and necessarilygenerated by an irreducible polynomial f since it is prime. Thus, I(V ) =〈f〉.

3.2 Infinite Dimensional Noetherian Rings

Given the fact that all ascending chains in a Noetherian ring stabilize, onewould think that the dimension of every Noetherian ring is finite, but in thissection we will construct an example of a Noetherian ring of infinite Krulldimension. This example is known as Nagata’s example as it was him whofirst constructed it in [7].

Definition 3.18. The ring of polynomials in infinite variables over a ringR is defined as

R[x1, x2, . . . ] =⋃

n∈NR[x1, . . . , xn].

Consider P a prime ideal in R, then the set R \ P is a multiplicativeclosed and we can localize R with respect to P . We denote the localizationof R with respect to R \ P by RP .

Lemma 3.19. Consider R = K[x1, . . . , xd, y1, y2, . . . ], then

R〈x1,...,xd〉 ' K′[x1, . . . , xd]〈x1,...,xd〉,

where K′ is the field of fractions of K[y1, y2, . . . ].

Proof. We know that K[y1, y2, . . . ] ⊆ R, even more so, K[y1, y2, . . . ] \ {0} ⊆R \ 〈x1, . . . , xn〉, thus K′ ⊆ R〈x1,...,xd〉. Then, K′[x1, . . . , xn]〈x1,...,xn〉 ⊆R〈x1,...,xd〉.

Lemma 3.20. Let R be a ring such that RM is Noetherian, for every M ∈Max(R). Assume that every non-zero element of R belongs to finitely manyelements of Max(R), then R is Noetherian.


Proof. To prove this we will see that every ideal I in R is finitely generated.If I = 0 there is nothing to prove. On the other hand, assume there exists anon-zero element a in I, and consider the set Maxa(R) = {M1, . . . ,Mr} ofall maximal ideals that contain a.

Let Si = R \Mi. Since RMi is Noetherian for every i ∈ {1, . . . , r}, thenall the S−1I = IMi are finitely generated, i.e., there exist a1, . . . , an in Isuch that IMi =

∑ni=1RMiai for all i.

Now, we will see that I = 〈a, a1, . . . , an〉, to prove this consider x ∈ Iand the ideal

J = (〈a, a1, . . . , an〉 : x) .

It is clear that a ∈ J , then if M is a maximal ideal such that M ∈ MaxJ(R)it must be equal to one of the Mi. Therefore, IMi = 〈a, a1, . . . , an〉Mi , sox = r

m with r ∈ 〈a, a1, . . . , an〉 and m ∈ R \Mi. This implies that thereexists t ∈ R \Mi such that tmx = tr.

From this we have that tm ∈ J ⊆ Mi and tm ∈ R \ Mi, which is acontradiction. Thus, MaxJ(R) is empty, i.e., J = R and, since 1 ∈ J , x ∈〈a, a1, . . . , an〉. This proves that I ⊆ 〈a, a1, . . . , an〉, and the other inclusionis trivial since a and the ai are all in I, therfore I = 〈a, a1, . . . , an〉.

Let R be the ring K[x1, x2, . . . ]. Consider integers di such that d0 = 0,1 ≤ d1, and di < di+1. We define the ideals Pn in R as

Pn = 〈xdn−1+1, xdn−1+2, . . . , xdn〉,

for n ≥ 1. Now, let S = R \ (∪Pi), this is clearly a subset of R, andsince all the Pi are prime, it is multiplicatively closed and we can constructR′ = S−1R.

Lemma 3.21. With the same notation as above, let I ⊆ R. If I ⊆ ∪∞i=1Pi,then I = Pi for some i ∈ N.

Proof. Let f ∈ R, and f = f1 + · · ·+ fr a decomposition of f in monomials,it is clear that f ∈ Pi if and only if fj ∈ Pi for all j. Take g ∈ ∪∞i=1Pi, ifg ∈ ∪i∈JPi for some finite subset J of N there is nothing to prove.

Assume g ∈ Pm with m /∈ J , and let g = g1 + · · ·+gs be a decompositionof g in monomials. Since m /∈ J , we have that fi 6= kgj for all k ∈ K, then

f + g = f1 + · · ·+ fr + g1 + · · ·+ gs

is a decomposition in monomials. If f + g ∈ ∪∞i=1Pi, then f + g ∈ Pi forsome i, and g ∈ ∪i∈JPi. Thus, I ⊂ ∪i∈JPi with J a finite subset of N.


Proposition 3.22 (Nagata’s Example). The ring R′ is Noetherian and itsdimension is sup {dn − dn−1}.

Proof. By Lemma 3.20 we have that R′ is Noetherian, and from Lemma3.21 we have that

Spec(R′) = {P ∈ Spec(R) | P ⊆ Pn} =

∞⋃

i=1

Spec(R′Pi),

so Max(R′) = {Pi | i ≥ 1}. We know that the rings R′Piare Noetherian of

Krull dimension di − di−1 from Lemma 3.19 and the fact that

R′Pi= RPi ' K[x1, . . . , xdi−di−1

]〈x1,...,xdi−di−1〉,

therefore ht(Pi) = di−di−1, and dimR′ = sup {dn − dn−1} is as desired.

Bibliography

[1] William Wells Adams and Philippe Loustaunau. An introduction toGrobner bases. Number 3 in Graduate Studies in Mathematics. AmericanMathematical Soc., 1994.

[2] Michael Francis Atiyah and Ian Grant Macdonald. Introduction to com-mutative algebra. Addison-Wesley Reading, 1969.

[3] David Steven Dummit and Richard M Foote. Abstract algebra. JohnWiley & Sons Inc., 3 edition, 2004.

[4] Lev Glebsky and Carlos Jacob Rubio-Barrios. Una prueba sencilla delteorema de los ceros de Hilbert usando bases de Grobner. LecturasMatematicas, 34(1):77–82, 2013.

[5] Robin Hartshorne. Algebraic Geometry. Number 52 in Graduate Textsin Mathematics. Springer Science & Business Media, 1977.

[6] J Peter May. Munshi’s proof of the Nullstellensatz. The American math-ematical monthly, 110(2):133–140, 2003.

[7] Masayoshi Nagata. Local rings, volume 13 of Interscience Tracts in Pureand Applied Mathematics. Interscience, 1962.

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ponti cia universidad javeriana facultad de ciencias

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