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Topics in Commutative AlgebraLecturer: Tamás Szamuely

Typed by: Szabolcs Mészáros

January 3, 2016

Remark. This is the live-texed notes of Topics in Commutative Algebra course held by Tamás Szamuely inthe winter of 2015. Any mistakes and typos are my own.

First Lecture, 13th of January

Literature:

• Matsumura: Commutative Ring Theory (useful encyclopedia-type reference book)• Atiyah - MacDonald: Introduction to Commutative Algebra• Eisenbud: Commutative algebra with a view toward Algebraic Geometry• J. de Jong et al.: The Stacks Project (www.math.columbia.edu)

Outline:

1. Krull’s Hauptidealsatz (principal ideal theorem)2. Cohen’s structure theorem for complete local rings)3. Serre’s characterization of regular local rings by homological dimension4. Notion of depths in local rings5. Cohen-Macaulay rings6. Koszul complex

1 HauptidealsatzAssumptions: In the following k is an algebraically closed field and Ank stands for the n-dimensional affine

space. Every ring in this course will be commutative with a unity and all homomorphisms will respectthis unity.

An affine closed set X is the common locus of an ideal in I C k[x1, . . . , xn], i.e. X = V (I). Similarly, wecan define the corresponding ideal I(X) to a closed set X by the property “f vanishes on every point of X”.Hilbert Nullstellensatz states that I(V (I)) =

√I for every ideal ICk[x1, . . . , xn]. Therefore, I ↔ V (I) induces

an (inclusion-reversing) bijection between radical ideals (i.e. I =√I) and closed sets of Ank . In particular,

maximal ideals correspond to points of Ank and prime ideal correspond to irreducible closed subsets of Ank(the ones that cannot be written as the union of two proper closed subsets), sometimes also called (affine)varieties.

Understanding all closed subsets of Ank is, in principle, equivalent to understanding irreducible closed sub-sets of Ank because every closed subset is the union of finitely many irreducible closed subsets. Algebraically,it means that every ideal is the intersection of finitely many prime ideals. This is also a consequence of theexistence of a primary decomposition in general Noetherian rings.

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Szabolcs Mészáros Topics in Commutative Algebra

Definition 1.1. The dimension of a variety is the maximal length of a chain X ) Zn ) Zn−1 ) · · · ) Z1 ofsubvarieties contained in X.

Definition 1.2. Let A be a ring and P ⊆ A be a prime ideal. Then

ht(P ) := sup{r ∈ N | ∃P1 ( P2 ( · · · ( Pr ( P chain of prime ideals in P}

Analogously, the Krull dimension of the ring A is

dim(A) := sup{ht(P ) | P ⊆ A prime}

Remark 1.3. If X = V (I) is an affine variety where I =√I then the coordinate ring of X is AX :=

k[x1, . . . , xn]/I. By this terminology, dim(X) is basically defined as Krulldim(AX).

Theorem 1.4. (Krull’s Hauptidealsatz) Let A be a Noetherian ring and x ∈ A. If P is a minimal primeideal such that x ∈ P then ht(P ) ≤ 1.

Theorem 1.5. (“Zusatz”) Moreover, if x is not a zero-divisor then ht(P ) = 1.

Theorem 1.6. (Generalization of Krull’s Hauptidealsatz) Let A be a Noetherian ring and x1, . . . , xr ∈ A .If P is a prime ideal which is minimal among the ideals with xi ∈ P for all i then ht(P ) ≤ r.

Corollary 1.7. (Of Theorem 1.6) The prime ideals in a Noetherian ring satisfy the descending chain con-dition. (Or in other words, every prime ideal has finite height.)

Reminder: If A is a ring, S ⊆ A is a multiplicatively closed subset (i.e. 1 ∈ S, x, y ∈ S then xy ∈ S)then there exists a localization of A by S. It is a ring A[S−1] together with a ring homomorphismλ : A→ A[S−1] such that for all ring homomorphisms ϕ : A→ R such that ϕ(s) is a unit in R for alls ∈ S there exists a unique factorization: ϕ = ϕ ◦ λ.Important special cases are when S = {1, x, x2, x3, . . . } or when S = A\P where P is a prime ideal.In this latter case, we write AP = A[S−1] where AP is a local ring with a unique maximal ideal PAP .(There is a slight abuse of notation: PAP refers to λ(P )AP .) More generally, I 7→ λ(I)AP induces abijection between ideals contained in P and the ideals of AP .

Lemma 1.8. If A is a Noetherian ring that has exactly one prime ideal P then A is Artinian.

Proof. Generally, in any Noetherian ring, the intersection of the prime ideals is the set of nilpotent elements.Therefore, if x ∈ P then x must be nilpotent. Since A is Noetherian, P is finitely generated so for a generatingsystem y1, . . . , yk there is a big enough exponent N such that yNi = 0 for all i hence all products of k · Nelements in P are zero. Now, we have a finite filtration of A with the ideals A ⊇ P ⊇ P 2 ⊇ P 3 ⊇ · · · ⊇ PN = 0where every quotient is a finite dimensional (since finitely generated) vector space over the field A/P . So weare done because the finite extension of Artinian submodules (such as the above finite dimensional vectorspaces) is also Artinian.

Proof of Theorem 1.4. : We show that if Q ( P then ht(Q) = 0. Replace A by AP so we may assume thatA is local with maximal ideal P . Let Q(n) :=

(Qn : (A\Q)

)= {q ∈ A | ∃s /∈ Q such that sq ∈ Qn}. This

looks a not really well motivated definition but in fact it is the preimage of (QAQ)n by the localizing mapλ : A→ AQ. Its name is the symbolic n-th power of Q.

Now, we can apply the previous Lemma 1.8 on A/(x) since P is minimal over (x) and a maximal idealat the same time. The lemma tells that A/(x) is Artinian. Therefore, the chain (x,Q) ⊇ (x,Q(2)) ⊇ · · · ⊇(x,Q(m)) ⊇ . . . stabilizes at some level n. So if f ∈ Q(n) ⊆ (x,Q(n)) = (x,Q(n+1)) then f = ax+ q for somea ∈ A and q ∈ Q(n+1). Then ax = f − q ∈ Q(n) but x /∈ Q because Q ( P and P is minimal over x. Bydefinition, there exists s /∈ Q such that sax ∈ Qn but then a ∈ Q(n) since sx /∈ Q by the prime property ofQ.

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In summary, we got that Q(n) ⊆ (x)Q(n) +Q(n+1) and the reverse containment is automatic. Therefore,Q(n)/Q(n+1) = P

(Q(n)/Q(n+1)) because x ∈ P and we just proved that every element of Q(n)/Q(n+1) can be

expressed as an element of (x)Q(n)/Q(n+1). So – by Nakayama’s lemma – we get Q(n)/Q(n+1) = 0. In otherwords, (QAQ)n = (QAQ)n+1 as ideals in AQ. So now we can apply Nakayama in AQ where the radical isQAQ what implies that (QAQ)n = 0. Now, we are left with a local ring with a nilpotent maximal ideal. Itmeans that QAQ is the only prime ideal in AQ since the intersection of the prime ideals is exactly the set ofnilpotent elements. Therefore, ht(Q) = 0 and that was the statement.

Lemma 1.9. In a Noetherian ring all minimal prime ideals consist of zero-divisors.

Proof of Theorem 1.5. : The above lemma clearly implies the statement.

Proof of the Lemma 1.9. : Note that there exists only finitely many minimal prime ideals in A. Indeed, wecan look at an irredundant primary decomposition of

√(0) which consists of finitely many minimal prime

ideals, let these be P1, . . . , Pr. If P would be a minimal prime ideal not listed here then by∏Pi ⊆ ∩Pi =√

(0) ⊆ P we get that there exists an i ≤ r such that Pi ⊆ P . Hence Pi = P by minimality. Now, we proceedby induction on r.

If r = 1 then we are done since the√

(0) is a prime ideal that is contained in any other prime ideal. Ifr > 1 then pick an a ∈ P1. We show that a is a zero-divisor and then the statement follows. First, thereexists b ∈ P2 ∩ · · · ∩Pr such that b /∈ P1 because the decomposition is irredundant. Clearly, ab ∈ P1 ∩ · · · ∩Prso (ab)n = 0 for big enough n since P1 ∩ · · · ∩ Pr =

√(0). But bn 6= 0 since b /∈ P1 so there exists an i such

that aibn 6= 0 but ai+1bn = 0 and then a is a zero-divisor.

Proof of the generalized Hauptidealsatz. : We prove by induction on r. The case r = 1 is exactly the Haup-tidealsatz. For r > 1 pick any prime ideal P1 ( P such that there does not exist P ′: P1 ( P ′ ( P (i.e. itis maximal in P which exists by the Noetherian property). We show that there exists y1, . . . , yr−1 ∈ A suchthat P1 is minimal over (y1, . . . , yr−1) so then we can use induction.

We may assume that P is maximal by replacing A by AP . Since P1 ( P and P is minimal above(x1, . . . , xr) there exists an i such that xi /∈ P1. Let’s say that is xr. Then P is a minimal prime ideal suchthat (xr, P1) ⊆ P . then A/(xr, P1) has dimension zero because its only one maximal ideal is a minimal prime(Same argument as before). Therefore, the image of P in A/(xr, P1) is a nilpotent ideal because it is theintersection of every prime ideal, so for all i ≤ r − 1 we have xmi = aixr + yi for some yi ∈ P1, ai ∈ A andbig enough m. It means that the image of P in A/(y1, . . . , yr−1, xr) is nilpotent hence the image of P inA/(y1, . . . , yr−1) is minimal over (xr).

Now, by the Hauptidealsatz ht(P ) ≤ 1 in A/(y1, . . . , yr−1) so ht(P1) = 0 in A/(y1, . . . , yr−1) and thatwas our statement.

Second Lecture, 20th of January

Proposition 1.10. (Converse of Hauptidealsatz) If A is Noetherian and PCA is a prime ideal and ht(P ) ≤ rthen there exists x1, . . . , xr ∈ A such that P is minimal above (x1, . . . , xr).

Lemma 1.11. (Prime avoidance) Let A be any (commutative, unital) ring, I1, . . . , In, J CA ideals such thatall Ij are prime ideal except perhaps In−1 and In. If J 6⊆ Ij for all j then there exists x ∈ J such that x 6∈ Ijfor all j ≤ n. (Equivalently, J ⊆ ∪Ij implies J ⊆ Ii for some i ≤ n.)

Proof. Induction on n: the case n = 1 is clear. For n > 1, assume that J ⊆ ∪Ij . By induction, for i = 1, . . . , nthere exists an xi ∈ J such that xi /∈ Ij for all j 6= i. For n = 2: By the assumption, xi ∈ Ii must holdso x1 + x2 6∈ I1 and also x1 + x2 6∈ I2 and we are done. If n > 2 then I1 is necessarily a prime ideal sox1 +

∏xi 6∈ ∪nj=1Ij and that is a contradiction.

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Proof of Proposition 1.10. : We proceed by induction on r: the case r = 0 is an empty statement. Assumethat ht(P ) = r. To prove the statement, we construct the sequence x1, . . . , xr recursively with the strongerproperty requiring for all 1 ≤ i ≤ r and for all minimal primes above (x1, . . . , xi) to have height at least i(hence exactly i by the Hauptidealsatz 1.4).

First, let x1 ∈ P be an element such that it is not contained in the minimal primes of A which exists byLemma 1.11 since P is not minimal by ht(P ) = r > 0. Then all minimal primes above (x1) have height 1by the Zusatz 1.5 and by noting that the union of minimal primes is the set of zero-divisors. For 1 < i ≤ r,if we have already constructed x1, . . . , xi−1 then choose an xi ∈ P such that it is not contained in theminimal primes above (x1, . . . , xi−1). By Lemma 1.11 such an xi exists else the Lemma claims that one ofthe minimal primes above (x1, . . . , xi−1) would contain P giving the contradiction ht(P ) ≤ i−1 < r using theHauptidealsatz 1.4. The resulting ideal (x1, . . . , xi) indeed has the required property because any minimalprime above (x1, . . . , xi) is not minimal above (x1, . . . , xi−1) by the choice of xi. Hence, it contains a minimalprime above (x1, . . . , xi−1) which has height at least i− 1. Therefore, any minimal prime above (x1, . . . , xi)has height at least (hence equally) i proving the claim.

Proposition 1.12. Let ϕ : A → B a homomorphism of Noetherian rings, Q ⊆ B be a prime ideal, anddenote P = ϕ−1(Q). Then ht(Q) ≤ ht(P ) + dimBQ/PBQ.

Proof. We may replace A by AP , P by PAP , B by BQ and Q by QBQ since it does not change the relationof the mentioned numbers. So we may assume that A and B are local and then we have to prove thatdimB ≤ dimA + dimB/PB. Set r := ht(P ) and s := ht(Q mod PB). By Proposition 1.10 we getx1, . . . , xr ∈ A such that P is minimal above them and similarly, let y1, . . . , ys ∈ B such that Q modulo PBis minimal above y1 . . . , ys modulo PB. Then by a similar argument as before (i.e. a maximal ideal beingminimal prime is nilpotent, see the proof of Theorem 1.4) we get that QN ⊆ PB + (y1, . . . , ys) and similarlyPM ⊆ (x1, . . . , xr) therefore QNM ⊆ (ϕ(x1), . . . ϕ(xr), y1, . . . , ys). This means that Q is a minimal idealabove (ϕ(x1), . . . ϕ(xr), y1, . . . , ys).

So we got that dim(B) = dim(Q) ≤ r + s = ht(P ) + dim(B/PB) where the inequality is a consequenceof the Generalized Hauptidealsatz (Theorem 1.6).

Corollary 1.13. If A is a Noetherian ring then dimA[x] = dimA+1. (In particular, dim k[x1, . . . , xn] = n.)

Proof. The inequality ≥ is easy: a chain of prime ideals P0 ( P1 ( · · · ( Pn in A can be consideredas a chain of primes in A[x] using the natural embedding ϕ : A → A[x] since the factors of the imagesA[x]/A[x]ϕ(Pi) ∼= (A/Pi)[x] are still domains. However, the image of a maximal ideal Pn will not be maximalin A[x] because the factor is not a field (x is not invertible). Therefore, dim(A[x]) ≥ dimA+ 1.

Conversely, it is enough to show that for a maximal ideal QCA[x] we have ht(Q) ≤ ht(A ∩Q

)+ 1. For

this, set P := A∩Q which is a prime ideal in A hence has height at most n. By the previous proposition weknow that

ht(Q) ≤ ht(P ) + dim(A[x]Q/P ·A[x]Q

)So we need to prove that the appearing dimension on the right is at most 1 (in fact it is always exactly one).This can be computed as follows

A[x]Q/P ·A[x]Q ∼=(A[x]/P ·A[x]

)Q∼=((A/P )[x]

)Q∼=

where we can first localize at the smaller multiplicative subset A\P ⊆ A[x]\Q giving

∼=((A/P )P [x]

)Q∼=(κ(P )[x]

)Q

using the notation κ(P ) := (A/P )(0) = (A/P )P ∼= AP /PAP for the residue field at P . Since κ(P )[x] isa one-variable polynomial ring over a field, it has dimension 1 and localizing at Q can only lower it. Thisproves the statement.

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Remark 1.14. Without the Noetherian property the statement is not true: For a ring A, the polynomialring A[x] can have arbitrary dimension between dimA + 1 and 2 dimA + 1. The point where we rely onNoetherian property is at Proposition 1.12 and its proof.

By an analogous argument, we also get

Corollary 1.15. If A is a Noetherian ring and dimA[[x]] = dimA+ 1. (In particular, dim k[[x1, . . . , xn]] =n.)

Proof. Clearly, an increasing chain of prime ideals in A gives an increasing chain of prime ideals in A[[x]]too by A[[x]]/PA[[x]] ∼= (A/P )[[x]]. Moreover, (A/P )[[x]] is never a field (x is not invertible) so dimA[[x]] ≥dimA+ 1 as in the previous proof. Conversely, for a maximal ideal QCA and P = A ∩Q we have

ht(Q) ≤ ht(P ) + dim(A[[x]]Q/P ·A[[x]]Q

)by Proposition 1.12. Here, we can still compute the ring on the right:

A[[x]]Q/P ·A[[x]]Q ∼=((A/P )[[x]]

)Q∼=((A/P )P [[x]]

)Q∼=(κ(P )[[x]]

)Q

So we only have to observe that dim(κ(P )[[x]]

)= 1 which is true since it is a primary example of a discrete

valuation ring. In details, every element can be written as xn · (c+ x · p) where c ∈ κ(P ) and p ∈ κ(P )[[x]],

but c+ x · p is invertible with inverse 1c

∑∞k=0

(− xp

c

)k, so the only nontrivial ideals are (xn) for n ≥ 1.

Remark 1.16. If A is an integral domain which is finitely generated algebra over a field k then tr.degk(A)denotes the maximal number of elements in A that are algebraically independent over k.

Proposition 1.17. If A is an integral domain then tr.degk(A) = dimA. (Without proof, we would need thegoing up theorem, not only the Hauptidealsatz.)

Remark 1.18. Geometric interpretation of the Hauptidealsatz: Let X ⊆ An be an affine variety over k = kand X = V (I) for some radical ideal in k[x1, . . . , xn]. Then f1, . . . , fr ∈ AX = k[x1, . . . , xn]/I define a closedsubset Z ⊆ X. The terms Zi in the irreducible decomposition of Z = ∪Zi correspond to minimal primeideals above f1, . . . , fr. In this context, Generalized Hauptidealsatz means that dimZi ≥ dim−r for all i(the existence of such i would be much easier).

Corollary 1.19. (Dimension Theorem) Let X,Y ⊆ An be subvarieties such that X ∩Y 6= ∅. Write X ∩Y =∪Zi where Zi’s are the irreducible components. Then dimZi ≥ dimX + dim Y − n for all i.

Remark 1.20. By the notation codimX := n− dimX the claim says that codimZi ≤ codimX + codimY .

Proof. (Sketch!) The intersection X ∩ Y ∼= (X × Y ) ∩ ∆n where ∆n stands for the diagonal in An × An.(Isomorphic in the sense of varieties.) Fortunately, dimX ×Y = dimX + dim Y because of Proposition 1.17.So, by Proposition 1.12 and by the description of ∆n = V (xi − yi | i ≤ n) the statement follows.

2 Regular ringsDefinition 2.1. A Noetherian local ring A with a maximal ideal P is a regular local ring if dimA/P P/P

2 =dimA where A/P is used as a field in dimA/P .

Remark 2.2. The algebraic meaning of regularity is the following: if x1, . . . , xr ∈ P are such that their imagesmodulo P 2 generate P/P 2 then they generated P as an ideal as well (by Nakayama: the elements of P · Pare superfluous in a generator system of the module P ). If the images form a basis in P/P 2 then we callx1, . . . , xr a regular system of parameters. On the other hand, if x1, . . . , xr generates P then r ≥ dimA.(By the generalization of the Hauptidealsatz since P is the maximal ideal so ht(P ) = dimA.) In short, aNoetherian local ring is regular if and only if P is generated by the smallest possible number of elements.

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Remark 2.3. The geometric meaning of regularity is the following: Let X be an affine variety and A be thelocal ring of it at P . Therefore, P/P 2 ∼= T ∗PX the cotangent space at P . So A is regular if and only ifdimT ∗PX = dimX, i.e. if P is a smooth point.

Proposition 2.4. If A is a regular local ring and x1, . . . , xr a regular system of parameters in A thenA/(x1, . . . , xi) is a regular local ring of dimension r − i for all 1 ≤ i ≤ r.

In fact, we prove more:

Proposition 2.5. If A is a Noetherian ring, P is a minimal prime above x1, . . . , xr and ht(P ) = r thenht(P/(x1, . . . , xi)) = r − i in A/(x1, . . . , xi) for all 1 ≤ i ≤ r.

The previous Proposition follows from it as locality is trivially inherited by the factor, the dimension ofP/P 2 decreases by one at each step (i.e. when we factor out one more xi) so we only need that height of Pdecreases by one at a step and not more. And this is the actual statement.

Proof. By the Hauptidealsatz 1.4 we have ht(P/(x1, . . . , xi)) ≤ r − i since P/(x1, . . . , xi) is minimal abovethe images of xi+1, . . . , xr in A/(x1, . . . , xi). Assume that ht(P/(x1, . . . , xr)) = s. By the converse ofHauptidealsatz 1.10, we get elements y1, . . . , ys such that P/(x1, . . . , xi) is minimal above y1, . . . , ys. Now,lift them into A as y1, . . . , ys ∈ P so it is minimal above x1, . . . , xi, y1, . . . , ys, i.e. s ≥ r − i.

Theorem 2.6. A regular local ring is an integral domain.

Proof. We prove by induction on dimA. (Note that regularity implicitly contains the finiteness of dimA.)Let us denote dim(A) = d. If d = 0 then P/P 2 = 0 but P is nilpotent so A = A/P which is a field.Now, assume the proposition for d − 1. Let P1, . . . , Pm be the minimal prime ideals of A. We can applyprime avoidance to P1, . . . , Pm, P

2 and P (here, P 2 is not necessarily prime but it is not needed in the primeavoidance). We know that P 6⊆ Pi and P 6⊆ P 2 so there exists an x ∈ P\P 2 such that x /∈ Pi for all i.(Geometrically, it vaguely corresponds to taking a function that is not vanishing and has nonzero derivativewhere it is not necessary.) Then we can take a regular system of parameters containing x.

Now, we can apply the previous Proposition 2.4 on A/(x) so it is regular and local with dimA/(x) = d−1.By the induction, we know that A/(x) is an integral domain so (x) is a prime ideal. It is of height 1 by theHauptidealsatz 1.4 and the Zusatz 1.5. Then there exists i such that Pi ⊆ (x) by minimality. On the levelof elements, it means that for all y ∈ Pi we have y = ax for some a ∈ A. But x /∈ Pi so a ∈ Pi. Hence, weconclude that Pi = (x)Pi, so even Pi = PPi is true. Therefore, by Nakayama, Pi = 0 what means that zerois a prime ideal, as we stated.

2.1 Regular sequencesDefinition 2.7. Let A be a ring, x1, . . . , xn ∈ A is a regular sequence if

1. xi is not a zero-divisor modulo (x1, . . . , xi−1) for all 1 ≤ i ≤ n.

2. (x1, . . . , xn) 6= A.

Remark 2.8. We will later prove that the property does not depend on the order of the elements in aNoetherian ring. It is not that surprising in the light of Proposition 2.4.Remark 2.9. Geometrically, it corresponds to the existence of a flag of local subvarieties in a local finitelygenerated k-algebra.

Definition 2.10. Let A be a ring and I = (x1, . . . , xn) a finitely generated ideal such that ∩∞n=1In = 0.

Look at the filtration I ⊇ I2 ⊇ I3 ⊇ . . . the associated graded ring is grI(A) = ⊕∞i=0Ii/Ii+1. There exists

an induced multiplication on grI(A).

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There also exists a natural surjective morphism of graded rings (A/I)[t1, . . . , tn]→ grI(A) where we tookthe usual grading on (A/I)[t1, . . . , tn] = ⊕∞n=0{homogeneous polynomials with coefficient in A/Iof degree n}sending ti to xi modulo I.

Theorem 2.11. Let A be a Noetherian local ring with maximal ideal P and x1, . . . , xn a minimal (henceregular) system of generators for P . Then the following conditions are equivalent:

1. A is a regular local ring

2. x1, . . . , xn is a regular sequence

3. The above map k[t1, . . . , tn]→ grP (A) is an isomorphism, where k = A/P .

Third Lecture, 27th of January

Proof. 1) ⇒ 2) was proved last time in Proposition 2.4 and Theorem 2.6. 3) ⇒ 2): By the assumptionfor all i we have isomorphisms k[ti, . . . , tn]

∼=→ grP/(x1,...,xi−1)(A/(x1, . . . , xi−1)) by factoring out the givenisomorphism on both sides. So it is enough to prove the following lemma:

Lemma 2.12. Let A be a ring, I ⊆ A an ideal such that ∩jIj = 0. Assume that grI(A) is an integraldomain, then A is also an integral domain.

To check the assumptions of the lemma observe that k[ti, . . . , tn] is an integral domain. Moreover, inCorollary 3.18 we will prove that ∩m(x1, . . . , xi−1)m = 0 using that (x1, . . . , xi−1) is the maximal ideal ofthe local ring A/(xi, . . . , xn). Therefore, the above lemma can be applied proving that x1, . . . , xn is a regularsequence by definition, this proves the discussed direction of the theorem.Remark 2.13. Note that the originally proposed argument for ∩m(x1, . . . , xi−1)m = 0 using Nakayama’slemma does not work. The reason of it is that P ·∩mPm is not necessarily ∩mPm but may be smaller, this isexactly the content of Corollary 3.18 where this distinction is bridged by the non-trivial Artin-Rees lemma,3.17.

Proof of Lemma 2.12. : Let 0 6= a, b ∈ A. Then by ∩jIj = 0 there exists r and s such that a ∈ Ir\Ir+1

and b ∈ Is\Is+1. Set a = a mod Ir+1 and b = b mod Is+1 as “they live in that degree”. Similarly, letab = ab mod Ir+s+1. Then by the definition of product in grI(A) we get ab = a · b ∈ grI(A) where a · b 6= 0because grI(A) is an integral domain so it implies ab 6= 0.

Now, we prove 2)⇒ 1): For this we need the following lemma:

Lemma 2.14. If A Noetherian local ring and x1, . . . , xr is a regular sequence in A then dimA/(x1, . . . , xr) =dimA− r.

Remark 2.15. Note that here x1, . . . , xr is a regular sequence (i.e. a sequence of non-zerodivisors in theappropriate factors) while in the similarly shaped Proposition 2.4 it was merely a regular system (i.e. arepresenting system for an A/P -basis of P/P 2) but there we assumed regularity of the ring while here weare trying to prove that.

Now, to get direction 2)⇒ 1) of the theorem, apply the lemma with n = r so we get that 0 = dimA/P =dimA− n meaning that A is regular.

Proof of Lemma 2.14. : To prove dimA/(x1, . . . , xr) ≥ dimA − r we use the same method as in the proofof 2.5 (which statement is not directly applicable now, compare the assumptions!). First, note that by theConverse of Hauptidealsatz 1.10 there exists y1, . . . , ys ∈ A/(x1, . . . , xr) such that P/(x1, . . . , xr) is minimalcontaining y1, . . . , ys where s = ht

(P/(x1, . . . , xr)

)= dimA/(x1, . . . , xr). Therefore, PCA is minimal among

prime ideals containing x1, . . . , xr, y1, . . . , ys hence dimA = htP ≤ r + s by the generalized Hauptidealsatz1.6. For the converse, we use that x1 is not a zerodivisor in A so for all minimal prime ideals P ′ ⊇ (x1) we have

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ht(P ′) = 1 by the Zusatz, Theorem 1.5. Therefore dimA/(x1) ≤ dimA− 1. Now, we can use induction on rsince the definition of a regular sequence is also inductive and we get that dimA/(x1, . . . , xr) ≤ dimA−r.

Now, we prove the hard part of Theorem 2.11: 2)⇒ 3). More generally, we prove the following theoremdue to Rees:

Theorem 2.16. (Rees) Let A be a ring and x1, . . . , xn be a regular sequence. The (surjective) graded mapA[t1, . . . , tn]→ grI(A) where I = (x1, . . . , xn) given by ti 7→ xi mod I2 induces an isomorphism

(A/I)[t1, . . . , tn]∼=→ grI(A)

In other words, A[t1, . . . , tn] 3 F 7→ 0 ∈ grI(A) is possible only if it is trivially true, namely if F ∈I[t1, . . . , tn].

Proof. Note that if F is homogeneous element of A[t1, . . . , tn] of degree d then F (x1, . . . , xn) ∈ Id so the mapis indeed graded. The proof will go by a double induction on n and d where d still stands for the degree Fwhich is assumed to be homogeneous. In more details, if we denote by Ψ(n, d) the statement of the theoremfor a fixed n, d ∈ N then to prove Ψ(n, d) we will use Ψ(n− 1, c) for arbitrary c (in fact c ≤ d− 1 is enough)together with Ψ(n, d− 1).

The case n = 0 is the empty statement and the case n = 1 is easy: if F (x1) ∈ (xd+11 ) then F has

coefficients divisible by x1 since x1 is regular i.e. a non-zerodivisor. Now, assume that the theorem is truefor n− 1 (and for all d ∈ N).

Lemma 2.17. xn is not a zerodivisor on A/(x1, . . . , xn−1)j for all j > 0.

Proof. We prove by induction on j (which induction is an inner induction relative to the induction of thetheorem on n). The case j = 1 is just our assumption: x1, . . . , xn is a regular sequence. If j > 1 then assumethat xn ·y ∈ (x1, . . . , xn−1)j . Since (x1, . . . , xn−1)j−1 ⊇ (x1, . . . , xn−1)j 3 xn ·y we can apply the induction onj: xn is not a zero-divisor modulo (x1, . . . , xn−1)j−1 hence y must be in (x1, . . . , xn−1)j−1. This means that(by definition) there exists G ∈ A[t1, . . . , tn−1] homogeneous of degree j − 1 such that y = G(x1, . . . , xn−1),in particular xnG(x1, . . . , xn−1) ∈ (x1, . . . , xn−1)j . The latter fact can be expressed as

A[t1, . . . , tn−1] 3 xnG(t1, . . . , tn−1) 7→ 0 ∈ gr(x1,...,xn−1)(A)

where xn ∈ A but the ti’s are purposely variables. Now, we can apply the induction hypothesis of the theoremon n (in fact Ψ(n−1, j−1)). It claims that the polynomial xnG(t1, . . . , tn−1) has coefficients in (x1, . . . , xn−1).However, xn is not a zerodivisor modulo (x1, . . . , xn−1) so G has coefficients in (x1, . . . , xn−1) too. As Gis homogeneous of degree j − 1 we got that y = G(x1, . . . , xn−1) ∈ (x1, . . . , xn−1) · (x1, . . . , xn−1)j−1 =(x1, . . . , xn−1)j and that was the statement.

Continuation of the proof of Theorem 2.16: Let F ∈ A[t1, . . . , tn] homogeneous of degree d for somed > 0, and assume that F (x1, . . . , xn) ∈ Id+1. We have to show that F ∈ I[t1, . . . , tn]. First, we reduceto the case of F (x1, . . . , xn) = 0 by finding and subtracting an appropriate element of I[t1, . . . , tn] havingthe same evaluation as F . The assumption F (x1, . . . , xn) ∈ Id+1 means that there exists F ∈ A[t1, . . . , tn]homogeneous of degree d+ 1 such that F (x1, . . . , xn) = F (x1, . . . , xn). Now, write F =

∑i tiFi where Fi is

homogeneous of degree d for all i. Then F −∑i xiFi ∈ A[t1, . . . , tn] is homogeneous of degree d (note that

here xi’s are parts of the coefficients, that does the trick) and it is zero on x1, . . . , xn. Since xiFi ∈ I[t1, . . . , tn]by the definition of I = (x1, . . . , xn), it is enough to prove F −

∑i xiFi ∈ I[t1, . . . , tn] to get F ∈ I[t1, . . . , tn]

i.e. the reduction to the case F (x1, . . . , xn) = 0 is indeed possible by replacing F by F −∑i xiFi.

Now, decompose F = G+tnH into a tn-free component G ∈ A[t1, . . . , tn−1] and the remaining tnH wheredegH = d− 1 since F is homogeneous. We will prove that separately, G and H are in I[t1, . . . , tn] hence theclaim. Turning to the image (i.e. evaluating the equality F = G+ tnH on x1, . . . , xn) gives

xnH(x1, . . . , xn) = −G(x1, . . . , xn−1) ∈ (x1, . . . , xn−1)d

8


since G is homogeneous of degree d and F (x1, . . . , xn) = 0 by the previous paragraph. The above Lemma 2.17applied on j = d claims that H(x1, . . . , xn) ∈ (x1, . . . , xn−1)d. This means that we are in a weird situationbecause H(x1, . . . , xn) by nature is only in Id−1 i.e. not that deep by degH = d − 1, but in a bigger ideal.We use both of its deviations: First, by H(x1, . . . , xn) ∈ Id and degH = d− 1 we get H ∈ I[t1, . . . , tn] usingthe induction hypothesis on d (precisely, this is Ψ(n, d− 1)). So the problem of H is solved.

Secondly, to prove G ∈ I[t1, . . . , tn] it seems reasonable to find a way to apply the induction hypothesis(precisely Ψ(n − 1, d)) on G and proving the stronger G ∈ (x1, . . . , xn−1) · A[t1, . . . , tn−1] but we cannoteasily check the assumptions on G. Instead, we take a substitute H ∈ A[t1, . . . , tn−1] of degree d suchthat H(x1, . . . , xn) = H(x1, . . . , xn−1) which exists since H(x1, . . . , xn) ∈ (x1, . . . , xn−1)d (as in the firstparagraph). Then we can interchange H with H in the equation

(G+ xnH)(x1, . . . , xn−1) = F (x1, . . . , xn) = 0

Therefore, by G + xnH ∈ A[t1, . . . , tn−1] and deg(G + xnH) = d we can apply the induction hypothesison n (precisely, this is Ψ(n − 1, d)) so we get that G + xnH ∈ I[t1, . . . , tn−1]. But xnH already hadcoefficients in I by xn ∈ I hence G is also in I[t1, . . . , tn]. Therefore, after the two observations we getF = G+ tnH ∈ I[t1, . . . , tn].

End of the proof of Theorem 2.11: The implication 2)⇒ 3) follows from 2.16 by setting I = P .

3 CompletionsMotivation: When studying algebraic varieties, regular functions are fractions of polynomials and one

comfortable way to study them is to introduce generalized regular functions in the form of power series.So we should study rings of power series with an algebraic treatment.

Definition 3.1. Let A be a ring and I ⊆ A is an ideal. The completion of A with respect to I is A :={(an) ∈

∏∞n=1(A/In) | an = an+1 mod In ∀n}. There exists a natural map a 7→ a mod In. If it is an

isomorphism then A is said to be complete.

Example 3.2.

1. Let A = k[x] and I = (x). Then A/In is the truncated polynomial ring of degree ≤ n − 1. ThenA = k[[x]].

2. More generally, if A is a polynomial ring k[x1, . . . , xn] and I = (x1, . . . , xn) then A := k[[x1, . . . , xn]].The answer for A = k[x1, . . . , xn](x1,...,xn) is the same. This is an example of a regular local ring suchthat completion is the power series ring. However, generally it does not hold for all regular, local ring.

3. For A = Z and I = (p) we get the p-adic integers Zp.

As before, we have grI(A) = ⊕∞n=0In/In+1. In example 1) and 2) we get back the polynomial ring this way.

Remark 3.3. Recall that an inverse system of modules (over a fixed ring) indexed by n ∈ N is given by

• modules Mn for all n ∈ N

• morphisms ϕn : Mn →Mn−1 for all n ∈ N.

Then the inverse limit of this sequence is

lim←Mn =

{(mn) ∈

∞∏n=0

Mn | mn = ϕn+1(mn+1) ∀n ∈ N}

9


Exercise 3.4.

1. Given a chain of submodules M0 ⊇ M1 ⊇ M2 ⊇ . . . one can construct an inverse system whereMn := M/Mn where the maps Mn → Mn−1 are the natural ones. Let us denote the inverse limit byM . There exists a projection pn : M → Mn by forgetting all components except the n-th one. Thesemaps are obviously surjective but they can have kernel Mn := ker(pn). Nevertheless, M/Mn

∼=→M/Mn

is an isomorphism.

2. There is an important special case: when I ⊆ A is an ideal and Mn := InM . In this case, M is calledthe I-adic completion of M . An even more special case is when A = M . This way we get back theprevious definition of a completion of a ring.

Remark 3.5. M can be endowed with a topology in which M i form a basis of neighborhoods of (0). In thecase where Mn = InM it is called the I-adic topology.

Definition 3.6. (mn) ⊆M is a Cauchy-sequence if for all n there exists an N0 such that mi−mj ∈Mn forall i, j ≥ N0. It converges to m ∈M if for all n there exists an i0 such that i ≥ i0 we have m−mi ∈Mn.

Remark 3.7. If A is Noetherian then in M , every Cauchy-sequence is convergent, i.e. it is a completion inthe metric sense using the pseudo-metric

d(x, y) = exp(−max{n ∈ N | x− y ∈Mn}

)which is a metric if ∩nMn = 0. If A is not necessarily Noetherian this completion is not necessarily complete(!). Counterexample: Take A = k[x1, x2 . . . ] the polynomial ring in infinitely many variables over a field as amodule over itself and set I = (x1, x2, . . . ). Then it is easy to check that A = k[[x1, x2, . . . ]] and similarly forM . However, the sequence an =

∑ni=1 x

ii is Cauchy but has no limit. Indeed, if it would have then that could

be only∑∞i=1 x

ii by looking at the finite degree parts but the tails

∑∞i=n+1 x

ii are not elements of IkM for

any k. (see Amnon Yekutieli: On Flatness and Completion for Infinitely Generated Modules over Noetherianrings, Communications in Algebra, 2011)

Proposition 3.8. Slogan: If Mn and Nn generate the same topology then the completions are isomorphic.Precisely: Let M be an A-module with two chains of submodules M0 ⊇ M1 ⊇ M2 ⊇ . . . and N0 ⊇ N1 ⊇N2 ⊇ . . . such that for all n there exists an m such that Nm ⊆Mn and similarly with the reversed order ofN and M (This property is sometimes called “nested sequences”). Then there exists a canonical isomorphismsuch that

lim←M/Mn ∼=→ lim

←M/Nm

Proof. In the special case, when (Nm) is a subsequence of Mn, the statement is obvious since the naturalmap lim←M/Mn → lim←M/Nm forgetting the unnecessary elements gives an isomorphism. (We can definethe inverse explicitly). Generally, there exist strictly increasing maps α, β : N → N such that Nα(n) ⊆ Mn

and Mβ(n) ⊆ Nn. Hence, we get natural projections M/Nα(n) → M/Mn and M/Mβ(n) → M/Nn yieldingmaps

lim←M/Mn → lim

←M/Nα(n) and lim

←M/Nn → lim

←M/Mβ(n)

which can be composed from the right with the previously constructed maps in the case of subsequences. Sowe get two maps between lim←M/Mn and lim←M/Nn and by the canonicity of the construction it is easyto verify that they are inverses of each other.

Proposition 3.9. Let A be a local ring with maximal ideal P . Then A is a local ring with maximal idealP := Ker(A→ A/P ).

10


Proof. First, if m ∈ P then 1 − m is a unit in A because∑∞i=0m

i ∈ M is its inverse. Moreover, sinceA/P

∼=→ A/P is a field, P is a maximal ideal. What is not so trivial is that it does not have any othermaximal ideals. If P ′ ⊆ A is a maximal ideal, m ∈ P\P ′ then (m,P ′) = A by maximality of P ′. This meansthat there exists an a ∈ A and b ∈ P ′ such that 1 = am+ b. However, then 1−am = b ∈ P ′ what contradicts1− am being a unit.

Fourth Lecture, 3rd of February

Assumptions: In the following A will be a fixed Noetherian ring, I ⊆ A is a fixed ideal and M is a fixedA-module. We consider the completion M := lim←M/InM .

Proposition 3.10. Completion is an exact functor. In details: let 0 → M1 → M2 → M3 → 0 be an exactsequence of finitely generated A-modules. Then their completions 0 → M1 → M2 → M3 → 0 also form anexact sequence.

Before proving it, we first list some of its consequences.

Corollary 3.11. There exist canonical isomorphisms A/I ∼= A/I and more generally In/In+1 ∼=→ In/In+1

for all n ∈ N. Therefore, grI(A)∼=→ grI(A) by “bringing the generators into the generators” i.e.

k[x1, . . . , xn] // grI(A)

��k[x1, . . . , xn] // grI(A)

Proof. (of the corollary) Apply the previous proposition with M1 = In+1, M2 = In and M3 = In/In+1 andobserve that In/In+1 ∼= In/In+1. Therefore the stated isomorphisms exist (by I0 = A the first one as well)and gr(.) is just the direct sum of the appropriate terms.

Corollary 3.12. (Corollary of Corollary 3.11) Assume that A is a Noetherian, local ring. Then A is aregular if and only if A is regular ring.

Proof. By Proposition 3.9 we know that A is local, if A is local. The regularity follows directly from theprevious corollary grP (A) ∼= grP (A) and Theorem 2.11 telling that regular local rings are characterized bythe property that the natural map (A/P )[x1, . . . , xn]→ grP (A) is an isomorphism. .

Corollary 3.13. (of Proposition 3.10) If J ⊆ A is any ideal then its I-adic completion (as an A-module)satisfies J ∼= J ·A where J · stands for multiplication with the image of J in A. (Note that A is not necessarilya subring of A.)

Proof. Let’s apply the Proposition 3.10 on the short exact sequence 0→ J → A→ A/J → 0 yielding

0 // J // A // A/J // 0

so we get that A/J ∼= A/J . Now, assume that J = (a1, . . . , an) by Noetherian property. Let’s define a map

ϕ : An → A

(t1, . . . , tn) 7→n∑i=1

aiti

11


This induces a map on the completions ϕ : An → A given by the same formula. The image of ϕ is exactlyAJ = J (we defined it so) hence the sequence

Anϕ // A // A/J // 0

is exact. Therefore, its completion

Anϕ // A // A/J // 0

is also exact. But J exactness= Im ϕdef of ϕ= JA so we got the statement.

Corollary 3.14. If I = (a1, . . . , an) and we complete A at I then A ∼= A[[x1, . . . , xn]]/(x1−a1, . . . , xn−an).

The statement seems a bit counter intuitive when compared to the isomorphism A[x1, . . . , xn]/(x1 −a1, . . . , xn − an) ∼= A. However, it is at least a characterization but not a really nice one.

Proof. Let B := A[x1, . . . , xn] and J = (x1 − a1, . . . , xn − an). Then we get a short exact sequence:

0 // J // B // A // 0

where we mapped B 3 xi 7→ ai for all i. Half-exactness is clear and exactness can be proved by the followingelementary argument: if p ∈ ker(B → A) then∑

i

cixi =: p(x) = p(x)− p(a) =

∑i

ci(xi − ai) = (x− a)∑i

ci(xi−1 + xi−2a+ · · ·+ ai−1)

proving p ∈ J . For n > 1 we can use induction on the number of variables.Now, we complete everything with respect to (x1, . . . , xn) as a B-module, where J ⊂ B is a B-module in

a natural way and A is a B-module by the B → A map we just described. Hence, completing these modulesgives – by the Proposition 3.10 – that A = B/J . But J = JB by the Corollary 3.13 and B = A[[x1, . . . , xn]]by unpacking the definition. The statement follows.

Example 3.15. Let A = Z and I = (p) where p is an arbitrary prime. We observed that the completionwith respect to (p) gives Zp, the p-adic integers. By the proposition we get that

Zp ∼= Z[[x]]/(x− p)

Remark 3.16. The idea of the proof of Proposition 3.10 is that completion is automatically a left exact functor(as it is an inverse limit) and we have a condition, called Mittag-Leffler condition that can decide whethera left exact functor is exact. In principle, we check this condition on the completion functor but in a moredirect language.

Proof. The first step is showing that the following short sequence is exact:

0 // lim←M1/(InM2 ∩M1) // lim←M2/InM2 // lim←M3/I

nM3 // 0

where the reason we took M1/(InM2 ∩M1) is that the natural filtration induced on the “image” of M1 inM2/I

nM2 is (InM2 ∩M1).The second step is to show that for allm > 0 there exists an n > 0 such that InM2∩M1 ⊆ ImM1. Besides,

note that ImM1 ⊆ ImM2 ∩M1 obviously holds so by the second step we will see that the two filtrationsgive the same topology on M1. Therefore, by Proposition 3.8, lim←M1/I

nM1 ∼= lim←M1/(InM2 ∩M1). Inshort, we can replace the first term in the short exact sequence.

12


To prove the first step, we will use the following general statement about inverse limits: let (N1n), (N2

n),(N3

n) be inverse systems of modules such that for all n there exists a commutative diagram of exact sequenceas

0 // N1n

fn //

ϕ1n

��

N2n

gn //

ϕ2n

��

N3n

//

ϕ3n

��

0

0 // N1n−1

fn−1 // N2n−1

gn−1 // N3n−1

// 0

If ϕ2n and ϕ3

n are surjective for all n (hence ϕ1n too) then

0 // lim←N1n

// lim←N2n

// lim←N3n

// 0

is exact. Indeed, the left-exactness is true for every (inverse) limit by general nonsense while exactness atlim←N3

n follows from diagram chasing: if we have a sequence (cn)n∈N ∈∏nN

3n representing an element of the

limit, i.e. ϕ3n(cn) = cn−1 then by induction we can choose a compatible system of preimages (bn)n∈N ∈

∏nN

2n

so that ϕ2n(bn) = bn−1 and gn(bn) = cn as follows. The induction can start by an arbitrary preimage b1.

Now, assume that bn−1 is already given. Take a preimage bn ∈ N2n of cn ∈ N3

n along gn arbitrarily (whichexists by row-exactness). It can happen that ϕ2

n(bn) is not bn−1 so take the difference bn−1 − ϕ2n(bn). This

maps to 0 ∈ N3n−1 because

gn−1ϕ2n(bn) = ϕ3

nfn(bn) = ϕ3n(cn) = cn−1 = gn−1(bn−1)

Therefore, we can take a preimage of the difference an−1 ∈ N1n−1 along fn−1 and also its (arbitrarily chosen)

preimage an ∈ N1n along ϕ1

n. Then bn := bn+fn(an) has the required properties: ϕ2n(bn) = bn−1 and gn(bn) =

cn. This proves the first step since the surjectivity assumption on the three sequences M1/(InM2 ∩M1),M2/I

nM2 and M3/InM3 are trivially satisfied so the exactness of the first sequence of the proof follows.

Moreover, step two follows from the following statement:

Lemma 3.17. (Artin-Rees Lemma) Let A be a Noetherian ring, I and ideal and M be a finitely generatedA-module. Let M1 ⊆M be a (necessarily finitely generated) submodule. Then there exists a k > 0 such thatInM ∩M1 = In−k(IkM ∩M1) for all n > k.

One can see the second step by choosing M1 = M1 and M = M2, moreover, choosing n so large thatn− k > m.

Proof. The key of the lemma is a clever application of the Hilbert Basissatz to the following special ring.Consider BI(A) := ⊕∞n=0I

n and BI(M) = ⊕∞n=0InM . The former is a graded ring and the latter is a graded

module over this ring. Clearly, In(IkM) ⊆ In+kM for all n, k > 0. Moreover, M is a finitely generatedA-module so BI(M) is a finitely generated BI(A)-module using the same generators. Besides, a finite set ofgenerators of I together with 1 ∈ I0 generates BI(A) as an A-algebra. Therefore, it is Noetherian by theHilbert Basissatz.

Let B1 := ⊕∞n=0(InM ∩M1). It is a BI(A)-submodule of the finitely generated module BI(M) henceit is also finitely generated by the Noetherian property proved in the previous paragraph. It means thatthere exists a k > 0 such that all generators are contained in ⊕kn=0(InM ∩M1). Let m1, . . . ,mr be thesegenerators, where we may assume that each mi is homogeneous (if one is not then we replace it with itshomogeneous components). So for all 1 ≤ j ≤ r there exists an α(j) ≤ k such that

mj ∈ Iα(j)M ∩M1

Now, assume that m ∈ InM ∩M1 for some n > k. Then, by the above notation it has a decomposition

m =r∑j=1

in−α(j)j mj

13


where i1, . . . , ir ∈ I (so the exponents are just representing the degrees of the coefficients). Now, let usexpress this m as

m =r∑j=1

in−kj (ik−α(j)j mj) ∈ In−k(IkM ∩M1)

by the choice of mj and α(j). This is exactly the required form.

The statement follows.

Corollary 3.18. (Krull’s Intersection theorem) If A is a Noetherian local ring with maximal ideal P thenthe intersection ∩nPn = (0).

Proof. Set N := ∩nPn. It is enough to prove PN = N since then N = 0 by Nakayama. Clearly, PN ⊆ Nand for the converse we use the Artin-Rees lemma. The Artin-Rees Lemma gives a k such that

N = P k+1 ∩N = P (P k ∩N)

by the choice of N = M1 and P = M . However, the right hand side is already contained in PN so we gotthe claim.

Remark 3.19. Note that this corollary is already used in proving Theorem 2.11 about regular sequences butthis does not yield a circular argument since in the proof of Corollary 3.18 we only used the Nakayama lemmaand the Artin-Rees lemma 3.17 (hence the Hilbert Basissatz), nothing more.

Corollary 3.20. Under the assumption of Krull’s Intersection theorem, A→ A is injective.

Proof. The claim depends only on the fact that the kernel of this map is exactly ∩nPn which is now zero bythe referred statement.

4 Cohen Structure TheoremTheorem 4.1. Let A be a complete Noetherian local ring that contains a subfield k ⊆ A that maps iso-morphically onto A/P via A → A/P . Then A is a quotient of some power series ring k[[x1, . . . , xn]]. If,moreover, A is regular then A ∼= k[[x1, . . . , xn]].

Remark 4.2.

1. The power series ring is regular since it is complete and its associated graded ring is the polynomialring (see Theorem 2.11).

2. For the most natural local regular rings, i.e. the local rings at a point of a variety over k = k satisfythe above assumption.

Lemma 4.3. Let ϕ : A → B a homomorphism of complete local rings (with maximal ideals PA and PB)such that ϕ(PnA) ⊆ PnB for all n ∈ N. If the induced homomorphism grϕ : grPA(A) → grPB (B) is injective(resp. surjective) then ϕ too.

Proof. Consider the following diagram:

0 // PnA/Pn+1A

//

grn(ϕ)��

A/Pn+1A

//

ϕn+1

��

A/PnA//

ϕn

��

0

0 // PnB/Pn+1B

// A/Pn+1B

// A/PnB // 0

14


where ϕn is induced by ϕ on the factor by ϕ(PnA) ⊆ PnB . By induction we can prove that all the ϕn areinjective. For n = 0 we can apply the hypothesis since ϕ0 = gr0(ϕ) : A/PA → B/PB . For the induction step,assume that ϕn is injective. Then in the above diagram grn(ϕ) and ϕn are injective so ϕn+1 is injective tooby the Snake-lemma. The injectivity of ϕ follows by the left-exactness of the inverse limit.

In the case of surjectivity, one has to be more careful since we want to lift the elements in a compatibleway. The proof is diagram chasing, see reference.

Proof of Theorem 4.1. : Let t1, . . . , td be a system of generators of the maximal ideal P ⊆ A. Then thereexists a unique k-algebra homomorphism λ : k[[x1, . . . , xd]]→ A by xi 7→ ti, because for all n there exists

k[[x1, . . . , xd]]/(x1, . . . , xd)n ∼= k[x1, . . . , xd]/(x1, . . . , xd)n → A/Pn

because the polynomial ring is a free k-algebra, these together build λ. To prove surjectivity of λ we wouldlike to use the previous Lemma so we turn to the induced map gr(λ) : gr(x1,...,xd)(k[[x1, . . . , xn]])→ grP (A).Since A/P ∼= k where P = (t1, . . . , td) and the λn’s for n > 0 are surjective too by t1, . . . , td being a system ofgenerators for P , we get that gr(λ) is surjective too. Therefore, by Lemma 4.3 we get that the above definedλ is surjective.

The second part of the statement states that if A is regular then λ is also injective. Again, we would liketo use the previous lemma. By the assumption, we can choose d = dimA in t1, . . . , td by the definition ofregularity. Therefore,

λ : k[[x1, . . . , xd]]→ A

is surjective with kernel the prime ideal Q. (It is indeed a prime since by Theorem 2.6 a regular local ring isan integral domain). However, dimA = dim k[[x1, . . . , xd]] = d hence htQ = 0 so it is zero as k[[x1, . . . , xd]]is an integral domain.

Corollary 4.4. Assume now that A is a regular local ring and there exists a k ⊆ A subfield mapping ontoA/P . Then there exists an injection A→ k[[t1, . . . , td]].

Proof. We know by Corollary 3.20 that A→ A is injective and A is regular by Corollary 3.12 if the originalA was. However, then by the previous Theroem 4.1 we know that A is isomorphic to a power series ring,proving the statement.

Remark 4.5. In the d = 1 case, there is also an elementary proof. Assume that b ∈ A and P = (t) (by d = 1).Then write b = a0 + b1t where a0 ∈ k and b1 ∈ A. Now, we can iterate this on b1 = a1 + b2t, then on b2 andso on. So we get a power series expansion for b. One can check that it gives an injective homomorphism.

Fifth Lecture, 10th of February

4.1 Equi-characteristic caseGoal: The condition k ⊆ A in Theorem 4.1 is superfluous if the characteristic of A is the same as of the

residue field k.

Lemma 4.6. (Hensel’s lemma) Let A be a complete local ring with maximal ideal P and residue field A/P =:k. Let f ∈ A[t] and f be the image of f in k[t]. Assume that a ∈ k is such that f(a) = 0 but f ′(a) 6= 0. Thenthere exists a unique a ∈ A such that a = a mod P and f(a) = 0. In other words, the following diagram canbe completed if f ′(a) 6= 0:

f ∈ A[t] // //

��

k[t] 3 f

��0 ∈ A // // k 3 0

15


Proof. Since A is complete with respect to P , it is enough to construct elements an ∈ A/Pn for all n such thata1 = a, an−1 = an mod Pn and f(an) = 0 in A/Pn. (This is called successive approximation.) Therefore, itis enough to prove the following:Claim 4.7. Let B be an arbitrary ring with ideal I such that I2 = 0 and f ∈ B[t] with a b ∈ B such thatf(b) ∈ I but f ′(b) is a unit in B. Then there exists a unique c ∈ B such that f(c) = 0 and c = b mod I.

The claim implies Lemma 4.6: by the choice B = A/Pn, I = Pn−1/Pn and an := b being any lift of an−1to A/Pn. Then to continue the induction all the conditions of the claim are clear except maybe that f ′(b) isa unit in A/Pn. But this is also true because an maps to a mod P/Pn hence f ′(an) maps to f ′(a) in P/Pnwhich is nonzero hence it is invertible by locality.

In fact, instead of the claim, we will prove an even more general fact:

Proposition 4.8. Let B be an R-algebra (i.e. a ring and an R-module with a compatible way) with an idealI such that I2 = 0. Let S = R[T ]/(f) which is naturally an R-algebra as R ⊆ S. If for a map λ : S → B/Ithere exists a preimage b ∈ B of λ(T ) such that f ′(b) is a unit in B then there is a unique map of R-algebrasλ completing the following diagram commutatively:

Sλ //

∃!λ

!!

B/I

B

OO

Remark 4.9. If one wants to remove the word R-algebras and unpack the definition “morphism of R-algebras”then the statement is equivalent to the following diagram

Sλ //

∃!λ

!!

B/I

R

OO

// B

OO

since for a unital ring A an R-algebra structure is the same as having a fixed ring-morphism R→ A. Similarly,the morphism of R-algebras is just the compatibility of a ring morphism with these fixed ones.

The previous claim is a special case of this proposition by the choice R = B, µ = idB and λ : B[T ]/(f)→B/I mapping T to b+ I. Then c := λ(b) satisfies the stated conditions in the previous claim.

Proof of Proposition 4.8. : Choose b as in the Proposition. To define λ it is sufficient to find h ∈ I such thatf(b+ h) = 0 since then λ(T ) := b+ h gives a well-defined map with (/I) ◦ λ = λ. (In fact, there is no otherway to define the image of T in the form of b+h to make the diagram commutative.) If, moreover, we provethe uniqueness of h then that proves uniqueness of λ as well.

By the “Taylor-formula” for polynomials, write

f(b+ h) = f(b) + f ′(b)h

which has no more terms since h2 ∈ I2 = 0. Therefore, h = (f ′(b))−1(−f(b)) which makes sense since f ′(b) isa unit. This choice of h satisfies the statement. Moreover, since we expressed h by known terms, h is uniquehence λ too.

This proves Hensel’s lemma 4.6.

16


Remark 4.10. Grothendieck defines an R-algebra S to be formally étale if for all R-algebra B and I ⊆ Bsuch that I2 = 0 and for every R-algebra map λ : S → B/I, there exists a unique map λ : S → B that makesthe following diagram commutative:

Sλ //

∃!λ

!!

B/I

R

OO

// B

OO

i.e. as an R-algebra morphism λ can be lifted to B. If it only exists but not uniquely then it is called formallysmooth. So, in these terms, we have proven that S = R[T ]/(f) is formally étale if (f, f ′) = (1) in R[T ]. (Infact, we proved a little more since our assumption was somewhat weaker.)

Proposition 4.11. (Generalization of Proposition 4.8, not proven) Let S = R[T1, . . . , Tn]/(f1, . . . , fn) anddefine Jac(f1, . . . , fn) := det[∂ifj ]. If this latter maps to a unit in S then S is formally étalé over R.Similarly, if S = R[T1, . . . , Tn]/(f1, . . . , fm) such that the maximal minors of [∂ifj ] map to units in S thenS is formally smooth.

Corollary 4.12. Let L | K be a (not necessarily finite) separable algebraic field extension, B a K-algebrawith an ideal I such that I2 = 0. Then there exists a unique λ : L → B such that the following diagramcommutes:

Lλ //

∃!λ

!!

B/I

K

OO

// B

OO

The same holds if instead of I2 = 0 we assume that B is complete with respect to I.

Proof. If L | K is a finite separable extension then L = K[T ]/(f) where f ′ is a unit modulo (f). So thestatement in this case is the direct consequence of Proposition 4.8. In the general case L = ∪i∈ILi whereLi | K is a finite separable extension for all i and λ|Li lifts to λ on Li uniquely (!). Hence they “glue together”since on the intersection they have to coincide.

Remark 4.13. If L | K is not finite then this is an example of a formally smooth algebra that is not of finitetype.

Definition 4.14. A complete local integral domain is called equi-characteristic if charA = char k wherek = A/P , the residue field.

Remark 4.15. Note that A is equi-characteristic if and only if A contains a field. Indeed, if charA = chark =p 6= 0 then we have the prime field Fp (the generated subfield of 1). Similarly, if charA = chark = 0 thenZ ⊆ A and for this we have Z ∩ P = (0) by the assumption. Therefore, every nonzero m ∈ Z is a unit henceQ ⊆ A. The converse is even more straightforward. This field in A could be always smaller than k. Thefollowing theorem states that it is not.

Theorem 4.16. (Cohen) If A is an equi-characteristic complete local domain with residue field k then thereexists a subfield in A mapping isomorphically onto k via A→ A/P ∼= k. (I.e. A→ A/P is a retraction.)

Combining this theorem with Theorem 4.1 gives

Corollary 4.17. (Cohen Structure Theorem) If A is equi-characteristic complete local domain with residuefield k then A is a quotient of some power series ring k[[t1, . . . , tn]]. If, moreover, A is regular of dimensiond then A ∼= k[[t1, . . . , tn]].

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Proof of Theorem 4.16. : First, assume that charA = chark = 0. Then we want to factor through on A theidentity map of k. We can choose a maximal subfield k′ ⊆ k for which such a factorization exists. Indeed,there exists Q ⊆ k and we can use Zorn’s lemma since the supremum of a chain is its union.

Assume, indirectly, that k′ ( k. If there exists x ∈ k transcendental over k′ then lift it to x ∈ A. Thenk′[x]∩P = (0) else it would be generated by an element f(x) hence x is algebraic which is a contradiction. Itmeans that all nonzero elements of k′[x] are units so k′(x) ⊆ A but that is a contradiction since k′(x)→ k(x)is a larger subfield. So there are no transcendental elements over k′. In other words, k | k′ is algebraic ofcharacteristic zero hence it is also separable. Therefore we can apply Corollary 4.12 what gives a contradiction(because there is a lifting).

Now, assume that charA = chark = p > 0. The proof follows from the following lemma the same way asthe characteristic zero case:

Lemma 4.18. Let B be a ring with an ideal I such that I2 = 0 and let L be a field of characteristic p > 0.Then every nonzero map λ : L→ B/I lifts to λ : L→ B. In particular, L is a formally smooth Z-algebra.

Proof. In characteristic p > 0, Lp ⊆ L is a subfield by the binomial theorem. Define λp : Lp → B as follows:Given an element a ∈ L, lift λ(a) to b ∈ B and set λp(ap) := bp. First, observe that this is well-defined: If b′is another lift of λ(a) then b− b′ ∈ I so bp − (b′)p is the same as (b− b′)p ∈ Ip ⊆ I2 = 0. If L is perfect then– by definition – Lp = L so we are done.

If L is not perfect then analogously to the proof in zero characteristic, we can find a maximal subfieldL′ ⊆ L such that L′ ⊇ Lp and λ|L′ lifts to L′ → B. We show that L′ = L. Assume that it is not, and pickan α ∈ L\L′. Then αp ∈ Lp ⊆ L′ and hence xp − αp is the minimal polynomial over L′ (by straightforwardgeneral nonsense about non-separable field extensions). Moreover, if we lift λ(α) to β, then βp = λp(αp) bythe definition of λp so α 7→ β induces a map L′(α) = L′[x]/(xp −αp)→ B. That contradicts the maximalityof L′.

The theorem follows.

4.2 Mixed characteristic caseQuestion. What about complete local rings of mixed character, i.e. when charA = 0 and char k = p > 0.

Step1: Given a field k of characteristic p > 0 there exists a complete discrete valuation ring A0 of charac-teristic 0 (called the Cohen ring) with residue field k and maximal ideal (p). (We will only prove thisfor the case of perfect residue field.)

Step2: Let A be a complete local domain of characteristic zero with maximal ideal P and residue fieldA/P =: k of characteristic p > 0. If A0 is a Cohen ring with residue field k then idk lift to ahomomorphism A0 → A. (This was a project for grade.)Remark 4.19. The map A0 → A must be injective since it has only one nontrivial prime ideal (p) butA has characteristic zero so it is impossible to have (p) as the kernel.

Step3: For A0 ⊆ A as in the previous step, there exists a surjective homomorphism A0[[x1, . . . , xn]] �A. If, moreover A is regular and p /∈ P 2 then there exists a map for n = d = dimA0 − 1 andA ∼= A0[[x1, . . . , xn]]. By the presence of Step 2, the proof is the same as in Theorem 4.17: wefind a minimal system p, t1, . . . , td of generators of P (here, p /∈ P 2 is crucial) and we define mapsA0[[x1, . . . , xn]]/(p, x1, . . . , xn)n → A/Pn...

Sixth Lecture, 17 of February

18


Motivation: There are some situations when we want to deduce similar theorems in characteristic zero thatare true in characteristic p and vice versa. For such purposes the construction below can be useful,where the residue field of the local ring has characterstic p and the field of fractions is characteristiczero.

Definition 4.20. (Serre, Lazard) An integral domain of characteristic p > 0 is perfect if and only if x 7→ xp

is an isomorphism. A strict p-ring is a ring A complete with respect to the ideal pA ⊆ A such that p is nota zero-divisor and ∩npnA = (0).

Remark 4.21. In the definition we do not assume that pA is maximal.

Theorem 4.22.

1. Given a perfect ring B of characteristic p, there exists a strict p-ring A with A/pA ∼= B.

2. The described A is unique up to unique isomorphism, i.e. given strict p-rings A1 and A2 such thatBi = Ai/pAi is perfect for i = 1, 2 then every isomorphism B1

ϕ→ B2 lifts to a unique isomorphismϕ : A1 → A2 such that / ◦ ϕ = ϕ ◦ /.

Remark 4.23. In case B is a field, the construction will give a complete discrete valuation ring A withA/pA ∼= B. For our purposes, this is the interesting case, but we discuss the theorem for rings.

Example 4.24.

1. Zp is a strict p-ring with Zp/pZp ∼= Fp

2. Given a ring R, p a prime. Let R[xp−∞ ] be the R-algebra generated by the indeterminates x−pi with re-lations (xp−(i+1))p = xp

−i . In other words, this is the union of the polynomial rings R[x, xp−1, . . . , xp

−n ].If R = Fp then this is a perfect ring. Indeed, there are no nilpotents so x 7→ xp has kernel zero andany element has a p-th root by substituting xp

−1 . Moreover, if we complete Z[xp−∞ ] with respectto pZ[xp−∞ ] then we get a strict p-ring with “residue field” (i.e. factor at the ideal generated by p)Fp[xp

−∞ ].

3. We can generalize the second example to arbitrary number of indeterminates, even for infinite I.Perform a similar construction with R[xα, . . . , xp

−n

α | α ∈ Λ] where {xα | α ∈ Λ} is a family of variables.After completing and applying the above Theorem we get a strict p-ring Z[{xα}, p−∞] with residue ringFp[{xα}, p−∞].

Remark 4.25. If B is a perfect ring, {bα | α ∈ Λ} are elements of B then it makes sense to substitute xα 7→ bαin F ∈ Fp[{xα}, p−∞] since in a perfect ring we can take (unique) p-th root, so this way we get an elementof B.

Lemma 4.26. (Key Lemma) Given a strict p-ring A, the natural map A→ A/pA has a unique multiplicativesection, i.e. a map A/pA ρ→ A such that π ◦ ρ = idA/pA and ρ(a · b) = ρ(a) · ρ(b). Moreover, ρ(a) is theunique element of A such that a = ρ(a) modulo pA and ρ(a) ∈ ∩∞n=0A

pn .

Definition 4.27. This ρ(a) is called the Teichmüller representative of a.

Proof. It is enough to show that for given a there exists a unique ρ(a) satisfying a = ρ(a) modulo pA andρ(a) ∈ ∩∞n=0A

pn . Indeed, the existence of the map A/pA ρ→ B follows from the existence of ρ(a) for each a,the multiplicativity follows from the uniqueness and the retractive property is exactly a = ρ(a) modulo pA.

So first, given a ∈ A/pA we prove that there exists an ai ∈ A/pi+1A such that a = ai modulo p(A/pi+1A)and ai is in the image of Api ↪→ A → A/pi+1A. Indeed, since A/pA is perfect, there exists an x ∈ A such

19


that a = xpi modulo pA. Moreover, (x+py)pi = xp

i modulo pi+1A for arbitrary y ∈ A (i.e. x is not unique).However, this means that xpi modulo pi+1A does not depend on x. Let us denote this element by ai. Sincexp

i ∈ Api−1 ⊆ Api it follows that (xpi modulo piA) = (ai modulo piA/pi+1A). The right hand side is ai−1 soxp

i modulo piA is also ai−1 by uniqueness. Therefore, we can find a sequence (ai) ∈ lim←A/pi+1A. Denotethis element by ρ(a). By construction, we know that ρ(a) = a modulo pA.

We only need that ρ(a) ∈ Apn for all n ∈ N. So fix n > 0: So let bn ∈ A/pA be the unique element suchthat (bn)pn = a (such element exists since the ring is perfect). As before, there exists a bn ∈ A such thatbn = bn modulo pA and (bn modulo pi+1A) comes from Ap

i . But then bpnn modulo pi+1A also comes fromAp

i for all i and it maps to a modulo pA. The uniqueness of ai implies bpnn = ρ(a).

Corollary 4.28. Every a ∈ A can be uniquely written as

a =∞∑i=0

ρ(ai)pi

for some sequence ai ∈ A/pA.

Proof of Theorem 4.22. Part 2): Consider the multiplicative retractions Bi → Ai given by the Lemma 4.26.We have the diagram

B1

ρ1

��

ϕ // B2

ρ2

��A1 // A2

which must be commutative since the property explained in the Lemma 4.26 is kept by the map ϕ. So theonly possibility for such a lift of ϕ is

∞∑i=0

ρ1(ai)pi 7→∞∑i=0

ρ2(ai)pi

since for finite quotients A/piA the image is given and ∩piA is zero because it is a p-strict ring. What westill have to show is that it is a homomorphism.

We will show this by a universal construction: Consider families of variables {xi}i∈N and {yi}i∈N. Definex :=

∑xip

i and y := yipi ∈ Z({xi, yi}, p−∞). Note that xi = ρ(xi) and yi = ρ(yi) so there exist sj ∈

Fp[{xi, yi | i ∈ N}, p−∞] such that

x+ y =∞∑i/=0

ρ(si)pi

and similarly for the product x·y. Therefore, given a, b ∈ A1 there exists a homomorphism Z({xi, yi}, p−∞)→A1 such that x 7→ a and y 7→ b: If a =

∑ρ1(ai)pi, b =

∑ρ1(bi)pi then let xi 7→ ρ(ai) and yi 7→ ρ(bi) so by

completion we indeed get such a homomorphism. Hence, by substitution

a+ b =∑

ρ1(sj({ai}, {bi}

)pj

Now, we need to check that ϕ(a+ b) = ϕ(a) + ϕ(b). Let’s write them all up by “power series”:

ϕ(a) =∑

ρ2(ϕ(ai))pi

ϕ(b) =∑

ρ2(ϕ(bi))pi

20


ϕ(a+ b) =∑

ρ2

(ϕ(sj({ai}, {bi})

))pj =

∑ρ2

(sj({ϕ(ai)}, {ϕ(bi)}

))pj

because ϕ is a ring homomorphism. So we got additivity. Similar argument works, for multiplication.Part 1): Observe that every perfect ring of charactersitic p is a quotient of some Fp[{xp

−α

α }α∈Λ]. This isthe residue ring of Z[{xα}α∈Λ, p

∞] so it is enough to prove that:

Lemma 4.29. Assume given a surjective homomorphism ϕ : B1 → B2 of perfect rings if characteristicp > 0. If there exists a strict p-ring A1 such that B1 ∼= A1/pA1 then there exists a strict p-ring A2 such thatB2 ∼= A2/pA2 such that the following diagram commutes:

B1

ρ1

��

ϕ // B2

ρ2

��A1 // A2

Proof. Consider

I ={ ∞∑i=0

ρ1(ci)pi | ci ∈ Ker(ϕ)}

where ρ1 : B1 → A1 is the multiplicative retraction. One can check (as in the proof of the second part)that I ⊆ A1 is an ideal, using that Ker(ϕ) is an ideal. Define A2 := A1/I. Now, A2/pA2

∼=→ B2 since ϕ issurjective and we defined I in that way. Also ρ1 induces a multiplicative retraction ρ2 : B2 → A1. Moreover,the filtration {piA1} induces a filtration {piA2} in the A2. Therefore, we may write every a2 ∈ A2

a2 =∞∑i=0

ρ(ai)pi ai ∈ B2

hence A2is a strict p-ring.

Theorem 4.22 follows.

Remark 4.30. Witt vectors: Assume that k is a perfect field and let A be a strict p-ring with residue fieldk. Consider the map A → kN bringing every element to (more or less) the sequence of “coefficients” of itspower series: ∑

ρ(ai)pi 7→ piai

set theoretically, this is a bijection. So we can define a ring structure on kN making it a strict p-ring withresidue field k. (not proven) One can also show: in k N addition, multiplication are given by polynomials inZ[{xi}, {yi}].

5 DepthDefinition 5.1. Let A be a ring, M and A-module, and x = (x1, . . . , xn) ∈ An is an M -regular sequence if

1. xi is not a zero-divisor on M/(x1, . . . , xi−1)M for all i.

2. M/(x1, . . . , xn)M 6= 0

Remark 5.2. If M 6= 0 is finitely generated and A is local with maximal ideal P 3 x1, . . . , xn then 2) isautomatic by Nakayama’s lemma. Conversely, by 2) we have xi ∈ P for all i.

21


Remark 5.3. Let A be a Noetherian, local ring andM a finitely generated module. If {x1, . . . , xn} is a regularsequence for M then every permutation of xi’s is too. (This is a not really hard exercise, it is enough toprove it for transpositions.)

Definition 5.4. Let A be a Noetherian ring and M 6= 0 a finitely generated A-module. If I ⊆ A is an idealthen

I−depth(M) := max length of an M -regular sequence contained in A

If A is local, depth(M) := P−depth(M) where P is the maximal ideal.

5.1 Reminder: Associated primesAssumptions: In the following, A will be a Noetherian ring and all modules are assumed to be finitely

generated.

Definition 5.5. A P ⊆ A is an associated prime ideal of M if there exists m ∈M such that

P = Ann(m) = {x ∈ A | xm = 0}

The set of associated primes is denoted by Ass(M).

Proposition 5.6. If M 6= 0 then a maximal element of {ann(m) | m ∈ M\{0}} is a prime ideal. Inparticular, Ass(M) 6= ∅.

Proof. If P = Ann(m) is maximal then take xy ∈ P . Assume that y /∈ P then ym 6= 0 but P ⊆ Ann(ym) soP = Ann(ym) by maximality. Therefore, x ∈ P since xym = 0.

Proposition 5.7. If 0 // M ′ // M // M ′′ // 0 is an exact sequence of A-modules then

Ass(M) ⊆ Ass(M ′) ∪Ass(M ′′)

(This is an easy exercise.)

Corollary 5.8. If M is a finitely generated A-module then there exists a finite filtration 0 ⊆ M1 ⊆ M2 ⊆· · · ⊆Mn = M such that Mi/Mi+1 ∼= A/Pi for some Pi ⊆ A prime ideal. Moreover, Ass(M) ⊆ {P1, . . . , Pn}.In particular |Ass(M)| <∞.

Notation: Supp(M) := {P ⊆ A prime | P ⊇ Ann(M)}

Proposition 5.9. (without proof) Clearly, Ass(M) ⊆ Supp(M) but these sets also have the same minimalelements.

Proposition 5.10. If I is an ideal consisting of zero-divisors on M then there exists an associated primeP ∈ Ass(M) such that I ⊆ P . Consequently,⋃

p∈Ass(M)

P = {zero− divisors} ∪ {0}

Proof. For all x ∈ I there exists m ∈ M such that x ∈ Ann(m) implies x ∈ P for some P ∈ Ass(M). Butthen I ⊆

⋃{P | P ∈ Ass(M)} where |Ass(M)| <∞ so I ⊆ P for some of the P ’s.

Seventh Lecture, 24th of February[I was missing from this lecture, it is copied from Attila’s notes. The statements are listed without proof.]

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Remark 5.11. Reminder on Ext: It is a functor: for fixed M ∈ A−Mod one has

A−Mod 3 N → ExtiA(M,N) ∈ A−Mod

a covariant functor that takes short exact sequences into long exact sequences. It can be computed byprojective resolutions of M . Similarly, for fixed N ∈ A−Mod one has

A−Mod 3M → ExtiA(M,N) ∈ A−Mod

is a contravariant functor.In particular, the multiplication by an x ∈ A map M ·x→M gives a map ExtiA(M,N)→ ExtiA(M,N).

Proposition 5.12. If A is a Noetherian local ring with residue field k,

depth(M) = min{i | ExtiA(k,M) 6= 0}

for all modules M finitely generated over A. Moreover, all maximal M -regular sequences have lengthdepth(M).

5.2 Cohen-Macauley ringsLemma 5.13. Let A be a Noetherian local ring, M a finitely generated module, x1, . . . , xi an M -regularsequence. Then this sequence extends to an M -regular sequence x1, . . . , xi, xi+1 if and only if HomA(k,Mi) =0 where Mi = M/(x1, . . . , xi)M .

Corollary 5.14. Let A be a Noetherian local ring and M a finitely generated module. If x1, . . . , xr is aregular sequence for M then depth(M/(x1, . . . , xr)M) = depth(M)− r.

Proposition 5.15. Let A be a Noetherian local ring and M a finitely generated module. If Q ∈ Ass(M)then depth(M) ≤ dimA/Q.

Definition 5.16. dimM := dim(A/Ann(M)

)Corollary 5.17. Let A be a Noetherian local ring and M a finitely generated module. Then depth(M) ≤dimM .

Definition 5.18. Let A be a Noetherian local ring and M a finitely generated module. Then M is calledCohen-Macauley (CM) if depth(M) = dimM . Similarly, a ring A is called Cohen Macauley if depth(A) =dimA. More generally, a Noetherian (not necessarily local ring) A is Cohen-Macauley, if AQ is Cohen-Macauley for all maximal ideals QCA.

Remark 5.19. We will see later that if a local ring is CM then its localizations are also CM hence the definitionis unambiguous and we can also say prime ideals in the above definition.

Proposition 5.20. If A is a Noetherian local ring, M is a finitely generated CM-module, Q ∈ Ass(M)then dimA/Q = dimM = depthM . Moreover, all Q ∈ Ass(M) are minimal prime ideals in A containingAnn(M).

Proposition 5.21. Let A be a Noetherian local ring, M a finitely generated module, x1, . . . , xr an M -regularsequence. Then Mr = M/(x1, . . . , xr) is CM if and only if M is CM.

Example 5.22. For CM rings:

1. A regular local ring is CM.

23


2. A Noetherian ring of dimension zero is CM. (Indeed, then every localization has a nilpotent maximalideal.)

3. A Noetherian integral domain of dimension 1 is CM. (Indeed, then the maximal ideals consists ofnon-zerodivisors hence 1 ≤ depth(A) ≤ dim(A) ≤ 1.)

4. We will see later that if k is a field then k[x1, . . . , xn]P is a regular local ring for every prime hencek[x1, . . . , xn] is CM.

5. One can show that if A is CM then A[x1, . . . , xn] is also CM hence most interesting CM rings are ofthe form A/(x1, . . . , xr) where A is regular local and or a polynomial ring and x1, . . . , xr is a regularsequence.

6. Counterexample for CM property: A = k[x, y]/(xy, y2) is of dimension 1 and it is not CM. (Indeed, themaximal ideal in A(x,y) consists of zerodivisors hence it has depth zero but dimension 1 so not CM.

Definition 5.23. A Noetherian ring A satisfies the unmixedness theorem if for all ideals I = (x1, . . . , xr) ofheight r for any r, all prime ideals P ∈ Ass(A/I) are minimal. (Here, ht(I) = min{htP | P ⊇ I}.

Eighth Lecture, 3rd of March

Definition 5.24. Let A be Noetherian and I ( A be an ideal. The ideal is called unmixed if all associatedprimes of I are minimal.

A satisfies the unmixedness theorem if all ideals I = (x1, . . . , xr) of height r are unmixed, for all r ≥ 0.

Theorem 5.25. A Noetherian ring is Cohen-Macauley (CM) if and only if the unmixedness theorem holdsin A.

Corollary 5.26. (Macauley, 1916) The unmixedness theorem holds in k[t1, . . . , tn].

Corollary 5.27. (Cohen, 1946) The unmixedness theorem holds in a regular local ring.

Lemma 5.28. (Lemma 0) If A is a Cohen-Macauley ring then AQ is also Cohen-Macauley for all primeideals Q ⊆ A. Moreover, depth(AQ) = Q−depth(A).

Proof. By the assumptions, we know that

dim(AQ) ≥ depth(AQ) ≥ Q−depth(A)

so we have to prove that dim(AQ) ≤ Q−depth(A). We will do it by induction on Q−depth(A): if it is zerothen all elements of Q are zero-divisors so Q ⊆ Q′ ∈ Ass(A). But A is Cohen-Macauley so Q′ is minimalhence Q = Q′. Therefore, AQ has dimension zero.

Now, assume that Q−depth(A) > 0 and pick an x ∈ Q which is not a zero-divisor. Set A1 := A/(x). Weknow that

Q−depth(A1) ≤ Q−depth(A)− 1

Since A1⊗A AQ ∼= AQ/xAQ which is a nonzero ring as x is not a unit, we get dimAQ/xAQ = Q−depth(A1)by induction. By Proposition 5.21: dim(A1 ⊗A AQ) = dim(AQ)− 1. Therefore,

Q− depth(A) ≥ Q− depth(A1) + 1 = dim(A1 ⊗A AQ) + 1 = dim(AQ)

and that proves the other inequality.

Lemma 5.29. (Lemma 1) Let A be a Cohen-Macauley local ring, Q ⊆ A be a prime ideal with ht(Q) = r.Assume that Q is minimal above (x1, . . . , xr). Then x1, . . . , xr is a regular sequence.

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Remark 5.30. Locality is not needed in the proof, but the “permutation-invariance” of the regular sequenceis only true in local rings so it is better to assume for a clearer statement.

Proof. By the previous lemma, AQ is Cohen-Macauley so we may assume that Q is the maximal ideal of A.It is enough to show that x1 is a non-zerodivisor. Then we can use induction on A/(x1) where the heightof the image of Q is lower so we can find a regular sequence (by induction) that can be pulled back andextended by x1.

If it is a zero-divisor then there exists a Q′ ∈ Ass(A) such that x1 ∈ Q′. We know that, A Cohen-Macauley so dim(A) = dim(A/Q′). But in A/Q′ the ideal Q/Q′ is minimal above the images of x2, . . . , xrso dim(A/Q′) ≤ r − 1 by the Hauptidealsatz.

Lemma 5.31. (Lemma 2) Let A be a Noetherian ring and P ⊆ A a prime ideal with ht(P ) = r. Then thereexist x1, . . . , xr ∈ P such that ht

((x1, . . . , xi)

)= i for all 1 ≤ i ≤ r.

Remark 5.32. Recall ht(I) = min{ht(P ) | P ⊇ I}.

Proof. We proceed by induction on i: if i = 1 then choose an x1 ∈ P that is not contained in any of theminimal prime ideals. Such an x1 exists by the Prime Avoidance Lemma 1.11. Therefore, ht

((x1)

)= 1 by

the Hauptidealsatz (it has height at most 1 but it is not contained in a minimal prime).Assume that we have already constructed x1, . . . , xi−1. By the Prime Avoidance Lemma 1.11, there must

exists an xi ∈ P not contained in any of the minimal primes above (x1), (x1, x2), . . . , (x1, x2, . . . , xi−1) sincethese have height at most i− 1 (by induction) hence they cannot contain P .

By the Hauptidealsatz we get ht((x1, . . . , xi)

)≤ i, but, in fact, equality holds because a minimal prime

above (x1, . . . , xi) is not minimal above (x1, . . . , xi−1) by our choice.

Proof of Theorem 5.25. ⇒: Let I = (x1, . . . , xr) with ht(I) = r. We have to show that all elements ofAss(A/I) are minimal. Assume that P ∈ Ass(A/I) is not minimal. By localizing at P we may assume thatP is maximal. Notice the by Lemma 0, 5.28 we did not lose the Cohen-Macauley property and the set ofassociated primes behaves well under localization. If Q ⊆ P is a minimal associated prime of I = (x1, . . . , xr)then it is a minimal prime above I because the minimal associated primes are minimal prime ideals (seeProposition 5.9). Therefore, ht(Q) = r. Then we can apply Lemma 1, 5.29, implying that x1, . . . , xr is aregular sequence. So A/I is Cohen-Macauley by Proposition 5.21 so all associated primes are minimal.

Conversely, assume that A is Noetherian and the unmixedness theorem holds. Assume, moreover, thatP ⊆ A is a prime ideal with ht(P ) = r. Now, choose x1, . . . , xr ∈ P as in Lemma 2, 5.31. The unmixednesstheorem implies that all associated primes of A/(x1, . . . , xi) are of height i. Therefore, they do not containxi+1 by the higher height of (x1, . . . , xi+1). This means that xi+1 is not a zero-divisor modulo (x1, . . . , xi)by Proposition 5.10, hence x1, . . . , xr is a regular sequence and depth(AP ) ≥ r = dim(AP ). This proves thatAP is Cohen-Macauley for all prime ideals P so we are done by definition.

Remark 5.33. Every quotient of a Cohen-Macauley ring is catenary, i.e. if P ⊆ Q are prime ideals then allchains of prime ideals between P and Q have the same length. (For this, it is enough to prove that Q beingminimal over P have dimQ = dimP + 1. One can use Lemma 1, 5.29 to prove this.)

6 Homological dimensionDefinition 6.1. If A is a ring, M is an A-module then the projective dimension of M is

pd(M) := inf{i | ∃0→ Pi → · · · → P0 →M → 0, Pj is projective for all j ≤ i

}Similarly,

gldim(A) := sup{pd(M) | M is an A-module}where these can be infinite too.

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Proposition 6.2. The following are equivalent for an A-module M :

1. pd(M) ≤ d,

2. Exti(M,N) = 0 for all A-modules N and i > d,

3. Extd+1(M,N) = 0 for all A-modules N ,

4. If 0→Md → Pd−1 → · · · → P0 →M → 0 is exact and Pi are projective then Md is projective.

Proof. (Sketch) The implications 4)⇒ 1)⇒ 2)⇒ 3) are straightforward by the definition of projective coverand the Ext-groups. For 3) ⇒ 4): Recall that if P is projective then Exti(P,N) = 0 for i > 0. Using theassumption Extd+1(M,N) = 0 we get that Ext1(Md, N) = 0 for all N by dimension-shift. Therefore, Md isprojective since the functor Hom(Md, .) is then exact.

Lemma 6.3. Let A be a ring and M , N be A-modules. Then Exti(M,N) = 0 for all M , N if and only ifExti(M,N) = 0 for all N and all finitely generated M .

Corollary 6.4. gldim(A) = sup{pd(M) | M is a finitely generated A-module}.

Remark 6.5. Recall that Exti(M,N) can be calculated by taking injective coresolution 0 → N → Q0 →Q1 → . . . .

Another important fact is Baer’s criterion saying that an A-module Q is injective if and only if everydiagram

I⊆ //

��

A

Q

can be completed in a commutative way with a morphism A→ Q.

Proof of Lemma 6.3. The implication ⇒ is trivial. So for ⇐ take an injective coresolution 0 → N → Q0 →Q1 → . . . where we know that Extj(M,Qj) = 0 for all modules M and for all j > 0. Now, we truncate thecoresolution:

0→ N → Q0 → Q1 → · · · → Qi−2 → Ni−1 → 0

then – as in the previous proof – we have Exti(M,N) ∼= Ext1(M,Ni−1). Then Ext1(M,Ni−1) = 0 forall M . This is equivalent to saying Hom(., Ni−1) is an exact functor which is equivalent to Ni−1 beinginjective. By Baer’s criterion, this is equivalent to Ext1(A/I,Ni−1) = 0 using the short exact sequence0→ I → A→ A/I → 0. This latter is equivalent to saying that Exti(A/I,N) = 0.

Proposition 6.6. Let A be a Noetherian local ring with maximal ideal P and M a finitely generated A-module. Then pd(M) ≤ d if and only if Tord+1(M, k) = 0 where k = A/P .

Corollary 6.7. If A is a Noetherian local ring with residue field k then gldim(A) = pd(k).

Proof of Corollary 6.7. This follows from the previous three propositions: the global dimension can bechecked on finitely generated (first component) ones, so it is enough to check their pd. That is equiva-lent to Tord+1(M,k) = 0 for all M . However, this latter is clearly equivalent to pd(k) ≤ d by the symmetryof the Tor functor.

Remark 6.8. Recall that Tori(M,N) = Hi(P• ⊗A N) where P• → M is a projective resolution. This wayfrom a short exact sequence 0→M1 →M2 →M3 → 0 we get a long exact sequence of Tor’s.

Observe that if P is projective then Tori(P,N) = 0 for all N and i > 0 by the above definition.

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Proposition 6.9. If x ∈ A then Tor1(A/(x)

) ∼= {m ∈M | xm = 0} is called the x-torsion of M .

Proof. Consider the exact sequence 0 → A → A → A/(x) → 0 where the A’s are clearly projective. Thismeans that

0 = Tor1(A,M)→ Tor1(A/(x),M

)→M →M →M/(x)→ 0

So the statement follows.

Proof of Proposition 6.6. The implication⇒ is clear. For⇐ we use induction on d where the d = 0 case is thehard one. We have to prove that if Tor1(M, k) = 0 thenM is projective. So take a basis ofM/PM ∼= M⊗Ak.Now, lift this basis to a set of elements m1, . . . ,mr ∈M . Consider the short exact sequence

0 N // Arf // M // 0

where f : (a1, . . . , ar) 7→∑aimi. Now tensor it with k:

0 = Tor1(M, k) // N ⊗A k // krf // M ⊗A k // 0

where –by construction – kr → M ⊗A k is an isomorphism. Therefore N ⊗A k ∼= N/PN = 0 so N = 0 byNakayama’s lemma. Hence M is free, which is even stronger then the d = 0 step.

For the induction step assume that d > 0 and consider the same exact sequence:

0 N // Arf // M // 0

By dimension shift (i.e. Ar is projective in this sequence) we get that Tord(N, k) ∼= Tord+1(M ;k) so pd(M) ≤pd(N) + 1. This completes the induction noting that N is also finitely generated by the Noetherian propertyof the ring.

Plan (for next time) We will prove a theorem of Serre: If A is a Noetherian local ring. Then it is regularif and only if its global dimension is finite. For this theorem we will use Theorem 6.6 i.e. it is enoughto check the pd of k. Afterward, we will prove the theorem of Auslander and Buchsbaum: If A is aNoetherian local rings and M is a finitely generated A-module then depth(A) = depth(M) + pd(M).

Ninth Lecture, 10th of MarchRecall the following corollary of Nakayama’s Lemma:

Theorem 6.10. Over a Noetherian local ring, finitely generated projective modules are free.

Theorem 6.11. (Auslander-Buchsbaum) If A is a Noetherian local ring and M is a finitely generated A-module with pd(M) <∞ then depth(A) = depth(M) + pd(M).

Remark 6.12. As we will see later, in the regular case, depth(A) = gldim(A) hence depth(M) = gldim(A)−pd(M). This is the reason why depth is sometimes called homological codimension. However, this makessense only when the ring is regular.

Proposition 6.13. Let A be any (commutative, unital) ring, x ∈ A be a nonzero divisor and M is anA/(x)-module such that pdA/(x)(M) <∞. Then pdA(M) = pdA/(x)(M) + 1.

Proof. We proceed by induction on pdA/(x)(M). If it is zero then M is a projective over A/(x). Since x is anon-zerodivisor, we have a short exact sequence

0 // A·x // A // A/(x) // 0

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This is a projective resolution of A/(x) over A hence pdA(A/(x)) ≤ 1. But pdA(A/(x)) 6= 0 because if itwere zero then A/(x) would be projective over A so it is a direct summand of a free A-module but thatis impossible since x is a zero-divisor ring element on the module A/(x) but it is a non-zerodivisor on themodule A so on the module Aα as well. Hence, pdA(A/(x)) = 1 so pdA(F ) = 1 for any free A/(x)-moduleF . Therefore, pdA(M) = 1 as M is a direct summand of a free A/(x)-module.

Now, for the inductive step, take an exact sequence of A/(x)-modules

0 // K // P // M // 0

where P is projective over A/(x). Then by turning to homologies, we get two exact sequences (one byconsidering the above as an exact sequence of A-modules, the other by considering them as A/(x)-modules):

Exti(P,N) // Exti(K,N) // Exti+1(M,N) // Exti+1(P,N)

Here, over A/(x) we have Exti(P,N) = 0 for all i > 0 hence ExtiA/(x)(K,N) ∼= Exti+1A/(x)(M,N) for all i > 0.

While, over A we have Exti(P,N) = 0 for all i > 1 by pdA(P ) = 1 (see the first step of the proof). HenceExtiA(K,N) ∼= Exti+1

A (M,N) for all i > 1. Using the first equation, we get

pdA/(x)(M) = pdA/(x)(K) + 1 = pdA(K) (6.1)

by the induction hypothesis. Therefore, by the equation interpreted over A, we get

pdA(M) = pdA(K) + 1

provided that pdA(M) > 1 since we only have the isomorphism in second equation for i > 1. Therefore,putting together the two equations, we proved the proposition for the case pdA(M) > 1.

So now we only have to show that pdA(M) = 1 and pdA/(x)(M) > 0 is impossible since all the othercases are handles by the above induction. So suppose indirectly that pdA(M) = 1 and consider again theexact sequence of Ext’s:

Exti(P,N) // Exti(K,N) // Exti+1(M,N) // Exti+1(P,N)

but now the beginning of it. By the projectivity of P over A/(x) we have

ExtiA(P,N) = 0 (i = 2, 3)

and Ext3A(M,N) = 0 by pdA(M) = 1. Then, by the exact sequence we get Ext2

A(K,N) = 0 for all N ,therefore pdA(K) ≤ 1 where pdA(K) = pdA/(x)(M) as we have seen in equation 6.1 (where no assumptionlike pdA(M) > 1 was used). So – by assuming pdA/(x)(M) > 0 – we reduced the case for pdA(M) =pdA/(x)(M) = 1.

So now choose an exact sequence of A-modules

0 // C // F // M // 0

where F is free. Then – since pdA(M) = 1 – we get that C is projective by Proposition 6.2. So this is, infact, a projective resolution of M . Tensor this sequence with A/(x) yielding

0 = TorA1 (F,A/(x)) // TorA1 (M,A/(x)) // C/xC // F/xF // M/xM // 0

where the first must be zero because F is free, moreover, M/xM = M since M is an A/(x)-module hencexM = 0. Therefore, by pdA/(x)(M) = 1 ≤ 2, we get Tor1(M,A/(x)) is projective over A/(x) by Proposition6.2 since C/xC and F/xF are already projective over A/(x). However, in Proposition 6.9, we have seen thecharacterization of the Tor groups as Tor1(M,A/(x)) = {m ∈ M | xm = 0} which is M in our case. So wegot that M is projective what is a contradiction.

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Corollary 6.14. If A is a Noetherian local ring, x ∈ A is a non-unit non-zerodivisor on A and gldim(A/(x)) <∞ then gldim(A) = gldim(A/(x)) + 1.

Proof. Apply the above proposition with M = k so – by Corollary 6.7 – we are done.

Proposition 6.15. Let A be a ring, M be an A-module and x a non-zerodivisor on M . Then

pdA(M) ≥ pdA/(x)(M/xM)

If, moreover, A is a Noetherian local ring, M is finitely generated and x ∈ A is a non-unit non-zerodivisoron A then there is equality above.

Proof. We may assume that d := pdA(M) <∞. We will do induction on d: For d = 0, M is projective overA so M/xM is projective over A/(x). For d > 0, choose an exact sequence

0 // K // F // M // 0

with F being free. Then – as we have already seen in a previous argument – pdA(K) = d − 1. Therefore,pdA/(x)(K/xK) ≤ d− 1 by induction. So apply ⊗AA/(x) on the sequence above:

TorA1 (M,A/(x)) // K/xK // F/xF // M/xM // 0

where TorA1 (M,A/(x)) = {m ∈ M | xm = 0} = 0 by Proposition 6.9. Therefore, pdA/(x)(M/xM) =pdA/(x)(K/xK) + 1 ≤ d.

Now, assume that A is Noetherian, local, M is finitely generated and x is a non-unit. We prove byinduction on n = pdA/(x)(M/xM). If n = 0 then M/xM is projective over A/(x) however then it is also freeby Theorem 6.10. We claim that if M/xM is free over A/(x) then M is free over A. This is enough to provecase n = 0 since then there are zeros on both sides of the equation in the statement.

So let m1, . . . ,mr be a free A/(x)-generating system of M/xM so by Nakayama’s lemma, it is also agenerating system over A since x ∈ P for the maximal ideal P . Now, assume that

∑aimi = 0 for some

ai ∈ A. We know that ai ∈ (x) since module (x) there is no nontrivial relation. So there exists a′i ∈ A suchthat a′i · x = ai. However, we know that x is a non-zerodivisor on M so

∑a′i ·mi = 0 since we can cancel x.

Now, we can continue this process “forever” getting the sequence of equations

0 =∑

a′imi =∑

a′′imi = . . .

where x · a′′i = a′i and so on. Therefore, ai ∈ ∩n(xn) ⊆ ∩nPn which is zero by Krull’s Intersection Theorem,Corollary 3.18.

For the inductive step, assume that n > 0 and consider the exact sequence

0 // K // F // M // 0

where F is free. We can tensor it with A/(x) getting the exact sequence

TorA1 (M,A/(x)) // K/xK // F/xF // M/xM // 0

where TorA1 (M,A/(x)) is zero again by Proposition 6.9 and F/xF is free over A/(x). So, if we interpret thissequence over A/(x) then by n = pdA/(x)(M/xM) > 0 we get

pdA/(x)(M/xM) = pdA/(x)(K/xK) + 1

Moreover, if we interpret it over A then we get pdA(M) = pdA(K) + 1. By the first part of the theorem, wehave pdA(M) ≥ n > 0 hence, we are done by the induction.

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Proof of Theorem 6.11. : We proceed by a double induction on depth(A) and depth(M). As a first step,we prove that M is free provided that depth(A) = 0. By this, depth(M) = 0 follows because if all of theelements of the ideals of A are zero-divisors on A then they are zero-divisors on a free module as well.

So, assume on the contrary that M is not free hence it is not projective since A is local (see Theorem6.10). So 0 < pd(M) <∞ and we can choose a projective resolution of minimal length:

0 // Pd // Pd−1α // . . . // P0 // M // 0

where all the Pi’s are finitely generated by Lemma 6.3. Here, we can take K := Kerα which has pd(K) = 1.Let us denote the maximal ideal of A by P and let k := A/P . Pick a k-basis of K/PK u1, . . . , um and lift itto u1, . . . , um ∈ K. It also generates K by Nakayama’s lemma. So let F be a free A-module of rank m andtake the short exact sequence

0 // P // Ff // K // 0

where f(a1, . . . , am) 7→∑aiui. Here, P is projective by pd(K) = 1 hence P is free as A is Noetherian, local.

By the construction F/PF ∼= K/PK, therefore, P ⊆ PF . By depth(A) = 0, P consists of zero-divisors soP ∈ Ass(A) so there exists an a ∈ A such that pa = 0 for all p ∈ P hence aP ⊆ aPF = 0. This contradictsthe assumption that P is free.

As a second step, we assume that depth(A) > 0 and depth(M) = 0. In this case, we know that P ∈ Ass(M)but there exists a non-zerodivisor x ∈ P . Therefore, there exists an m ∈ M such that P = Ann(m). So wecan take a short exact sequence

0 // K // Fε // M // 0

where F is a finitely generated free A-module. Choose u ∈ F such that ε(u) = m. Then Pu ⊆ K sincePm = 0. Therefore, P (xu) ⊆ xK. Also, xu ∈ K as x ∈ P but xu /∈ xK as u /∈ K and x is a non-zerodivisoron A hence on F as well. This means thatK/xK 6= 0. But all elements of P are zero-divisors onK/xK by thedefinition of K (i.e. by the above short exact sequence) hence depth(K/xK) = 0. Also, x is a non-zerodivisoron K because it is a sub of the free F ⊇ K. Therefore, by Proposition 6.15, pdA(K) = pdA/(x)(K/xK) butM is not free as depth(M) = 0 but depth(A) > 0. This means that pd(M) > 0 so pdA(M) = pdA(K) + 1.

On the other hand, we can apply the induction and obtain

depth(A/(x)) = depth(K/xK) + pdA/(x)(K/xK) = pdA/(x)(K/xK)

where depth(A/(x)) = depth(A)− 1 by Proposition 5.14, and depth(K/xK) = 0 as we have seen. Therefore,by the previous paragraph, pdA/(x)(K/xK) = pdA(K) = pdA(M) − 1 so we are done with the case whendepth(A) > 0 and depth(M) = 0.

Finally, we prove the general case when depth(M) > 0 and depth(A) > 0: Then P /∈ Ass(M)∪Ass(A) bythese positivity assumptions. By Prime Avoidance (Lemma 1.11) and the finiteness of the set of associatedprimes, we get that there exists an x ∈ P non-zerodivisor on both M and A. We know that depth(A/(x)) <depth(A) by Proposition 5.14 so we may use induction:

depth(A/(x)) = depth(M/xM) + pdA/(x)(M/xM)

where depth(A/(x)) = depth(A) − 1, depth(M/xM) = depth(M) − 1 and pdA/(x)(M/xM) = pdA(M) byProposition 6.15.

Theorem 6.16. (Serre) A Noetherian local ring A is regular if and only if gldim(A) < ∞. In this case,gldim(A) = dim(A) = depth(A).

Remark 6.17. The equality dim(A) = depth(A) is not new, we have just seen that.

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Proof. First, we prove ⇒: We do induction on d := dim(A): If d = 0 then A is a field that clearly has globaldimension zero. If d = depth(A) > 0 then let x ∈ P be a non-zerodivisor. Then gldim(A/(x)) = gldim(A)−1by Corollary 6.14 provided that gldim(A/(x)) <∞ which is true by the induction. Therefore, gldim(A) <∞.By induction, we know that dim(A/(x)) = gldim(A/(x)) so the same holds for A.

For⇐ we use induction on gldim(A). If it is zero then this means that all A-modules are projective hencefree (since the ring is Noetherian local, see Theorem 6.10). In particular, the module k := A/P is free whichis a field so A = k, so the statement holds.

If gldim(A) > 0 then depth(A) 6= 0 by the proof of Case I in Theorem 6.11 of Auslander-Buchsbaum.So let x ∈ P a non-zerodivisor and x /∈ P 2. Such an element exists: apply the Prime Avoidance Lemma1.11 for the associated primes, P 2 (which is not necessarily a prime ideal but it is not needed in the lemma)and P . Assume that we know that gldim(A/(x)) < ∞ (we will get back to this point). Then by Corollary6.14, gldim(A/(x)) = gldim(A) − 1 so we can apply the induction on A/(x) hence it is regular. Therefore,P modulo (x) is generated by a regular sequence y1, . . . , yr. Lift yi to yi ∈ P : then x, y1, . . . , yr is a regularsequence generating P . Hence A is regular.

It is left to prove that gldim(A/(x)) <∞ if gldim(A) <∞. This statement is equivalent to pdA/(x)(k) <∞ by Proposition 6.6 where k := A/P . So consider the following exact sequence:

0 // P/(x) // A/(x) // k // 0

Since A/(x) is free hence projective, it is enough to prove that pdA/(x)(P/(x)

)<∞. By the second part of

Proposition 6.15,pdA/(x)

(P/xP ) = pdA(P ) <∞

The problem is that P/(x) ≤ A/(x) is not necessarily the same as P/xP .To finish the proof, we claim that the exact sequence

0 // (x)/xP // P/xP // P/(x) // 0

splits. This is enough since then P/(x) is a direct summand of P/xP and the projective dimension is additiveso if pd(P/xP ) <∞ then pd(P/(x)) <∞.

As x ∈ P\P 2 there exist x2, . . . , xr ∈ P such that x, x2, . . . , xr modulo P 2 is a basis of P/P 2. Then(x) ∩

((x2, . . . , xr) + P 2) ⊆ xP since if not then there exists a y ∈ (x2, . . . , xr) + P 2 such that y = xu where

u is a unit. (This is exactly the negation of the previous statement since A\P is the set of units.) However,this means that x = u−1y ∈ (x2, . . . , xr) + P 2 what contradicts the choice of (x2, . . . , xr). Now, consider thefollowing maps:

P/(x) =→((x) + (x2, . . . , xr) +P 2)/(x)

∼=→((x2, . . . , xr) +P 2)/((x)∩

((x2, . . . , xr) +P 2))→ P/xP → P/(x)

where the big composition is the identity as one can check. This means exactly that P/xP can be retractedto P/(x) and that was the goal.

Tenth Lecture, 17th of March

Corollary 6.18. (Of Serre’s Theorem 6.16) Let A be a regular local ring and Q ⊆ A be a prime ideal. ThenAQ is regular too.

Proof. Since gldim(A) <∞we have a projective resolution (Pi)i≤d of the A-module A/Q where Pi is in factfinitely generated and free because A is Noetherian, local. Now, we can tensor this projective resolutionwith AQ what stays exact (because either AQ is flat or because Pi is free hence the syzygies are split-exact).Moreover,

A/Q⊗A AQ ∼= AQ/QAQ

the residue field of AQ. Hence we got a finite free resolution the residue field of AQ over AQ, in particularpdAQ(AQ/QAQ) <∞. By Proposition 6.7 it is equivalent to gldim(AQ) <∞.

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Definition 6.19. A Noetherian ring A is regular if all localizations AP (for all prime ideals P ⊆ A) areregular local maps.

Remark 6.20.

• It is enough to require that AP is regular for P being maximal, since the others are further localizationsof them.

• Regular rings are Cohen-Macauley.

Example 6.21. Regular rings:

1. A Dedekind domain (integrally closed Noetherian domain of dimension 1) is regular since all localizationare discrete valuation domains that are clearly regular.

2. If X is a smooth affine variety over a field then the coordinate ring AX is regular. The converse is alsotrue provided that the field is perfect.

Proposition 6.22. If A is a regular ring then A[t] is regular too.

Corollary 6.23. If A is a field (or Dedekind ring) then A[t1, . . . , tn] is regular.

Proof of Proposition 6.22. : Let P ⊆ A[t] be a maximal ideal and take Q = P∩A. Then A[t]P is a localizationof AQ[t] where AQ is regular. Hence, we can assume that A is local with maximal ideal Q. Then P mapsto (f) ⊆ k[t] modulo Q. Now, lift this f to f ∈ A[t]. Then clearly, P = (Q, f) where f is not a zero-divisormodulo Q, hence depthA[t](P ) ≥ depthA(Q) + 1. Moreover,

dimA[t]P = ht(P ) ≥ depth(P ) ≥ depth(Q) + 1 = ht(Q) + 1 = dim(A) + 1

where we used regularity of A at the penultimate equality. However,

dimA[t]P = ht(P ) ≤ ht(Q) + 1 = dimA+ 1

Therefore, we got that dimA[t]P = dim(A) + 1 while depth(A[t]) = depth(A) + 1 by Proposition 5.14 hence– by the regularity of A – we get A[t]P is regular.

To prove Auslander Buchsbaum’s theorem, we first prove the following:

Theorem 6.24. If A is a Noetherian ring of finite dimension d then A is regular if and only if gldim(A) ≤ d.

Corollary 6.25. (Hilbert syzygy theorem) gldim(k[t1, . . . , td]

)≤ d.

Remark 6.26.

• Actually, gldim(k[t1, . . . , td]

)= d what can be proved by induction: d = 0 is clear and for the induction

step one can use Corollary 6.14.

• In fact, over a k[t1, . . . , td] every finitely generated projective module is free. It was a conjecture of Serre,solved by Quillen and Suslin. Consequently, every finitely generated module has a finitely generatedfree resolution.

Proof of Theorem 6.24. ⇐: If P ⊆ A is maximal then gldim(AP ) ≤ gldim(A) ≤ d by the argument usingthe tensoring with AP on the projective resolution of A/P . This direction of the statement follows.

For ⇒: let M be a finitely generated module over A and take a truncated projective resolution of M :

0 // Md// Pd−1 // . . . // P1 // P0 // M // 0

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Here, all the Pi’s are finitely generated by Noetherian property. We need to prove that Md is projectivetoo. If P ⊆ A is maximal then we can tensor the whole sequence by AP getting again an exact sequence byflatness of AP . Here, Pi ⊗AP is projective over AP since it is a direct summand of a free.

The ring AP is regular of dimension d since A is regular (hence Cohen-Macauley) of dimension d too anddepth is preserved there. Therefore, gldim(AP ) ≤ d hence MA⊗AAP is free over AP . The statement can bededuced from the following proposition:

Proposition 6.27. Let A be a Noetherian ring, M a finitely generated A-module. Then M is projective ifand only if M ⊗A AP is free for all P ⊆ A prime. In fact, it is enough to investigate maximal P ’s.

Lemma 6.28. Let A be a Noetherian ring:

1. A finitely generated A-module is zero if and only if M ⊗A AP = 0 for all maximal ideals P .

2. A morphism ϕ : M1 →M2 of finitely generated modules is surjective if and only if ϕ⊗ idAP is surjectivefor all maximal ideals P .

Proof. 1)⇒ 2): Apply a) to the cokernel of ϕ and use the right exactness of the tensor product.Direction ⇒ of 1) is trivial, while to get ⇐, assume that there exists a maximal ideal m 6= 0 such that

Ann(m) ⊆ A is a proper ideal. Therefore, there exists a maximal P ⊆ A containing Ann(m) but then m 6= 0in M ⊗A AP .

Proof of the proposition by the lemma. : Direction ⇒ is clear by Theorem 6.10 since a localization of aprojective is projective (it is still a summand of a free). To get ⇐, take an exact sequence 0 → K → F

f→M → 0 for a free generator F of M . Then it is enough to show that this sequence splits, for which it isenough to show that the dual map

Hom(M,f) : Hom(M,F )→ Hom(M,M)

is surjective. Indeed, if s ∈ Hom(M,F ) is a preimage of idM then this s must be a section of the sequence0→ K → F →M → 0 by definition.

So take the localization of Hom(M,F )→ Hom(M,M) giving

Hom(M ⊗A AP , F ⊗A AP ) ∼= Hom(M,F )⊗A AP → Hom(M,M)⊗A AP ∼= Hom(M ⊗A AP ,M ⊗A AP )

Here, we know that M ⊗A AP is projective hence Hom(M ⊗A AP , F ⊗A AP )→ Hom(M ⊗A AP ,M ⊗A AP )is surjective because it is equivalent to the splitting of

0→ K ⊗A AP → F ⊗A APf⊗AAP−→ M ⊗A AP → 0

so we can apply part b) of Lemma 6.28 on Hom(M,f).

Theorem 6.29. (Auslander - Buchsbaum) A regular local ring is a unique factorization domain.

Remark 6.30. In a unique factorization domain, every prime ideal of height 1 is principal. Therefore, if S isa smooth variety over a field, then A is a local ring at P , and Y ⊆ X is of codimension 1. Hence, Y definesa prime ideal of height 1 in A so it is principal. This means that locally, around P , it is defined by oneequation.

Lemma 6.31. (Nagata) If A is a Noetherian integral domain and x ∈ A is a prime element and A(x) is aunique factorization domain then A is a unique factorization domain.

Recall the following proposition:

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Proposition 6.32. A Noetherian integral domain A is a unique factorization domain if and only if everyheight 1 prime ideal in it is principal.

Proof. If A is a unique factorization domain and P is a prime ideal of height 1 then for all a ∈ P\{0} wehave a = u ·

∏pαii so there exists i such that pi ∈ P by primeness. But then (pi) is a prime ideal contained

in P hence (pi) = P by ht(P ) = 1. Conversely, if every height 1 prime ideal is principal then we want toconclude that every irreducible element is prime. So assume that p is irreducible and take a minimal prime Pcontaining P . Then – by the Hauptidealsatz – ht(P ) ≤ 1 but it is nonzero hence ht(P ) = 1. The assumptionsays that P = (b) for some b ∈ A. Therefore P = (p) because p was irreducible.

Proof of Lemma 6.31. By this proposition, it is enough to prove that every height 1 prime ideal is principal.So take a P ⊆ A prime ideal with ht(P ) = 1. If x ∈ P then P = (x) since x is a prime element and ht(P ) = 1so we are done. The other case is x /∈ P . In this case, there exists a p ∈ P such that PAx = pAx for somep ∈ A since Ax is a unique factorization domain. We may assume that (p) ⊆ A is maximal with this property.Then p /∈ (x), because otherwise there exists an a ∈ A such that p = ax. Here, a ∈ P as x /∈ P but p ∈ Pand (a) ) (p), hence pAx = x−1pAx = aAx which is a contradiction.

So PAx = pAx i.e. for all y ∈ P there exists a ∈ A and m,n > 0 such that yxn = p · a

xm since A is adomain. This means that xly ∈ (p) for big enough l. Therefore, it is enough to show that xy ∈ (p) impliesy ∈ (p). If xy = ap then a ∈ (x) as (x) is a prime ideal and p /∈ (x). Therefore, a = bx for some b hencexy = ap = bxp so y = bp because A is a domain.

Lemma 6.33. (Kaplansky) If A is a domain and I ⊆ A is an ideal such that I ⊕An ∼= An+1 as A-modulesthen I is principal i.e. I ∼= A as A-modules. (proof: next time)

Lemma 6.34. (Serre) If A is a ring and P is a projective A-module such that there exists a finite freeresolution of length n then P is stably free, i.e. there exists free modules F and F ′ such that P ⊕F ′ ∼= F . If,moreover, A is Noetherian and P is finitely generated then we may find finitely generated F and F ′.

Proof. The resolution is denoted as

0 // Fnϕn // Fn−1

ϕn−1 // . . . // F1ϕ1 // F0

ϕ0 // P // 0

As P is projective, ϕ0 has a retraction so F0 ∼= P ⊕ Im(ϕ1) hence Im(ϕ1) is projective. We can iterate thisgiving that Fi ∼= Im(ϕi)⊕ Im(ϕi+1) for all i hence

P ⊕ F := P ⊕n⊕i=1

Im(ϕi) ∼= P ⊕⊕i odd

Fi ∼=⊕i even

Fi =: F ′

so the statement holds.

Proof of Theorem 6.29. : Let A be a regular local domain of dimension d. We proceed by induction ondim(A): the case dim(A) = 0 is clear. Pick an x ∈ P\P 2. We know that A/(x) is again regular, local byProposition 2.4. Therefore, by Theorem 2.6, it is again an integral domain. This means that (x) is a primeideal. Therefore, by Lemma 6.31, it is enough to prove that Ax is a unique factorization domain, i.e. byLemma 6.32, every prime ideal Q ⊆ Ax of height 1 is principal.

If M is a maximal ideal of Ax then (Ax)M is also a localization of A by a prime ideal hence it is alsoregular by Corollary 6.18. Also, dim(Ax)M < dimA as x /∈ M so M ∩ A is not maximal. Therefore, byinduction, (Ax)M is a unique factorization domain, hence, Q(Ax)M is principal since it keeps its height.Therefore, Q(Ax)M is a free module of rank 1 over (Ax)M . However, it is true for all localizations at amaximal ideal, i.e. Q(Ax)M is a locally free module. This means – by Proposition 6.27 – that Q is projectiveas an Ax-module.

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On the other hand, there exists Q′ ⊆ A such that Q = Q′Ax by definition. By the regularity of A weknow that every finitely generated A-module has a finite free resolution since A is local with finite globaldimension. In particular, Q′ has a finite resolution too. Therefore, Q also has a finite free resolution sincewe can tensor the resolution of Q′ with Ax. Therefore, by Lemma 6.34, Q is stably free as an Ax-module,i.e. there exists an m and n such that Q⊕ (Ax)m ∼= (Ax)n. We claim that m = n− 1 as we can tensor thisequation by (Ax)M yielding

Q(Ax)M ⊕ (Ax)mM ∼= (Ax)nMwhere Q(Ax)M is a free module of rank 1 so n = m − 1 by picking a module-basis. Hence, we can applyLemma 6.33 to complete the proof.

Eleventh Lecture, 24th of March

7 Koszul complexDefinition 7.1. Let A be a ring, M is an A-module and n ≥ 0. Then the n-th exterior power of M is:

Λ0M := A; Λ1M := M

ΛnM := M⊗n/〈m1 ⊗ · · · ⊗mn | ∃1 ≤ i < j ≤ n : mi = mj〉

It is characterized by the following universal property: for all A-modules N and for all n-linear maps ϕ :M × · · · ×M → N such that ϕ(m1, . . . ,mn) = 0 if mi = mj for some i 6= j there exists a factorization

M × · · · ×M

ϕ''

f // ΛnM

��N

where f is the natural surjection.

Notation: the image of m1 ⊗ · · · ⊗mn in ΛnM is denoted by m1 ∧ · · · ∧mn.

Remark 7.2.

1. Alternative definition: Consider the tensor algebra T (M) = ⊕M⊗n which is a graded A-algebra withthe usual (tensor-)multiplication. Then we can take

Λ(M) = T (M)/(m⊗m | m ∈M)

where the latter is a generated two-sided ideal. The n-th homogeneous component of Λ(M) is the sameas the above ΛnM .

2. There is a multiplication ΛnM ⊗ ΛmM → Λn+mM , the induced multiplication from Λ(M).

3. Moreover, all A-module map M → N extends to a map Λ(M) → Λ(N) of graded A-algebras. In,particular, there exists maps ΛnM → ΛnN for all n ≥ 0 since the map is graded.

4. In ΛnM we have the relation

m1 ∧ · · · ∧mi ∧mi+1 ∧ · · · ∧mn = (−1) ·m1 ∧ · · · ∧mi+1 ∧mi ∧ · · · ∧mn

35


5. If M and N are A-modules then

Λn(M ⊗N) ∼=⊕i+j=n

ΛiM ⊗ ΛjN

what property follows from the analogous property for the tensor product.

6. If M ∼= An is free of rank r then ΛnM is free of rank(rn

). If e1, . . . , er is a basis of M then{

ei1 ∧ · · · ∧ ein | 1 ≤ i1 < i2 < · · · < in ≤ r}

is a basis of ΛnM . The idea for this is that one can take the determinant as a nonzero multilinear mapand reduce an indirectly existing relation to a relation det = 0 which is a contradiction hence the aboveelements are independent.

Proposition 7.3. If A is a domain I1, . . . , In, J1, . . . , Jn ⊆ A ideals such thatn⊕i=1

Ii ∼=n⊕i=1

Ji

as A-modules then I1 · · · · · In ∼= J1 · · · · · Jn as A-modules.

Corollary 7.4. Lemma 6.33 where I1 = I, Ii = A for i ≥ 2 and Ji = A for all i.

Proof of Proposition 7.3. First, we prove that

I1 · · · · · In ∼= Λn(I1 ⊕ · · · ⊕ In

)/T

where T is a torsion submodule. This is enough since the right hand side is the same for both Ii’s and Ji’sby the assumption.

For this, take K = Frac(A) and let us denote M = I1 ⊕ · · · ⊕ In. Then M ⊗A K ∼= Kn since I ⊗K = Kfor all ideals I. Then

ΛnM ⊗A K ∼= Λn(M ⊗A K)

hence there exists a mapΛnM ϕ //

α7→α⊗1 &&

K

ΛnM ⊗K

OO

by the universal property. The kernel of this map ϕ is exactly T by construction.We claim that in this case Im(ϕ) = I1 · · · · · In ⊆ K proving the statement. Let e1, . . . , en be the standard

basis of Kn. If mi ∈M = ⊕ni=1Ii then we can write

mi =∑i,j

aijej aij ∈ Ij

hence we can rearrange the elements of ΛnM as

m1 ∧ · · · ∧mn = det((aij)) · e1 ∧ · · · ∧ en ∈ ΛnKn = K

Conversely, I1 · · · · · In = 〈det((aij)) | aij ∈ Ij〉 ⊆ K.

36


Definition 7.5. Let A be a ring, M and A-module, f : M → A an A-linear map. Then the Koszul complexK(f) of f is defined as

. . . // ΛnMdnf // Λn−1M

dn−1f // . . . // Λ2M

d2f // M

df // A // 0

where df = f and

dnf (m1 ∧ · · · ∧mn) =n∑i=1

(−1)i+1 · f(mi) ·m1 ∧ · · · ∧ mi ∧ · · · ∧mn

which map exists by the universal property. Moreover, one can check that dn−1f ◦ dnf even for n = 2.

Remark 7.6. On Λ(M) the dnf induce a map of graded algebras df : Λ(M)→ Λ(M) of degree −1 satisfyingdf ◦ df = 0. Moreover, if x ∈ ΛiM and y ∈ ΛjM such that

df (x ∧ y) = df (x) ∧ y + (−1)ix ∧ df (y)

Such that graded algebras are called differential graded algebras (in short, DGA).

Example 7.7. Let M = A so every f : A → A module homomorphism is multiplication by an elementx := f(1). Then the homology in degree zero of the above sequence is A/xA, in degree 1 it is zero if andonly if x is a non-zerodivisor.

Definition 7.8. Let C• and D• two chain complexes concentrated in nonnegative degrees. Then the tensorproduct of them is

(C ⊗D)n =⊕i+j=n

Ci ⊗Dj

gets a natural differentiation by the formula dC⊗Dn (x⊗ y) = dCi (x)⊗ y + (−1)ix⊗ dDj (y).

Example 7.9. Let f1, f2 be two A→ A module homomorphisms. Then K(f1)⊗K(f2) is the following:

A // A⊕A // A

where the second map if (x, y) 7→ f1(x) + f2(y) and the first is

A ∼= A⊗A 3 x⊗ y 7→ (f2(y)x, f1(x)y

Therefore, using the canonical isomorphism between A⊗A and Λ2(A⊕A) we get K(f1)⊗K(f2) ∼= K(f1, f2)where (f1, f2) : A⊕A→ A is an A-linear map.

More generally,

Proposition 7.10. If M1,M2 are A-modules fi : Mi → A then define

f = (f1, f2) : M1 ⊕M2 → A

Then K(f) ∼= K(f1)⊗K(f2).

Proof. Recall that Λn(M1⊕M2) ∼=⊕

i+j=n ΛiM1⊗ΛjM2 where the latter is nothing else but the n-th termof K(f). To compare df with df1 ⊗ df2 view them as degree −1 maps Λ(M1 ⊕M2) → Λ(M1 ⊕M2). Theycoincide in degree 1 and satisfy the same equation d(x ∧ y) = d(x) ∧ y + (−1)ix ∧ dy.

37


Let f = (f1, . . . , fr) : Ar → A and take fi(1) =: xi since f is the same as sum of multiplications by xi’s. Let

K(x) = K(x1, . . . , xn) := K(f)

so it is a complex of length r of free A-modules of rank(rn

). If M is an A-module then define K(x,M) :=

K(x)⊗AM .

Theorem 7.11. If x = (x1, . . . , xn) is an M -regular sequence then K(x,M) is acyclic (i.e. it has nohomology in positive degrees). In particular, for M = A, K(x) is a free resolution of A|(x1,...,xr).

Corollary 7.12. The k[x1, . . . , xn]-module k has a finite free resolution of length n, in particular, pdk[x1,...,xn](k) ≤n.

Similarly, if A is a regular local ring of dimension n then gldim(A) ≤ n.

Corollary 7.13. If I = (x1, . . . , xn) then

Tori(A/I,M) ∼= Hi(K(x)⊗AM)

Exti(A/I,M) ∼= HiHom(K(x),M)

Lemma 7.14. If C• is any complex of A-modules and x ∈ A then there exists an exact sequence of complexes

0 // C• // C• ⊗A K(x) // C•[−1] // 0

where (C•[−1])i = Ci−1. Moreover, in the corresponding long exact sequence

. . . // Hi(C•) // Hi

(C• ⊗A K(x)

)// Hi−1(C•) // Hi−1(C•) // . . .

the map Hi−1(C•)→ Hi−1(C•) is multiplication by (−1)i−1x.

Corollary 7.15. There exists an exact sequence

0 // Hi(C•)/xHi(C•) // Hi

(C• ⊗A K(x)

)// Ker

(Hi−1(C•)

x·→ Hi−1(C•))

// 0

Proof of Lemma 7.14. : We know that K(x) = Ax·→ A. The Corresponding exact sequence is

0 // A // K(x) // A[−1] // 0

what we can tensor with C•, where the following term appears:(C• ⊗K(x)

)= (Ci ⊗A A)⊕ (Ci−1 ⊗A A) ∼= Ci ⊕ Ci−1

with the differential Ci ⊕ Ci−1 → Ci−1 ⊕ Ci−2 given by[∂ (−1)i−1x0 ∂

]where ∂ is the differential of C•. Then

0 // Ci //

��

Ci ⊕ Ci−1 //

��

Ci−1 //

��

0

0 // Ci−1 // Ci−1 ⊕ Ci−2 // Ci−2 // 0

To compute the connecting homomorphism in the long exact sequence can be computed by applying theSnake lemma on the above diagram. In details, take α ∈ Ker(Ci−1 → Ci−2), lift it to (0, α) ∈ Ci ⊕Ci−1 andmap it into ((−1)i−1xα, 0) ∈ Ci−1 ⊕ Ci−2 by applying the matrix above.

38


Proof of Theorem 7.11. : We proceed by induction. Recall thatK(x,M) = [M x·→M ]. ThenH1(K(x),M

)=

Ker(M x·→M) and H0(K(x),M

)= M/xM so the induction starts.

Now, assume that r > 1 and apply the above argument and the Corollary 7.15 of the Lemma 7.14 toHi−1(C•) and Hi(C•). We get that

0 // H0(K(x)⊗Hi(C)

)// Hi

(K(x)⊗ C•) // H1

(K(x)⊗Hi−1(C•)

)// 0

For the inductive step, let C• = K(x1, . . . , xr)⊗AM and x = xr hence

Hi

(K(x),M) ∼= Hi

(C• ⊗A K(x)

)since K(x) = K(x1)⊗ · · · ⊗K(xr) by Proposition 7.10. The induction hypothesis says that Hi(C•) = 0 forall i > 0 hence – by the above short exact sequence – we get Hi

(K(x,M)

)= 0 for all i > 1.

Let’s determine the terms in the short exact sequence: for i = 1

H1(K(x,M)

) ∼= H1(K(x)⊗H0(C•)

) ∼= H1(K(x)⊗M/(x1, . . . , xr−1)M

)= Ker

(Nr−1

xr→Mr−1

)= 0

which is zero since x is a regular sequence.

Theorem 7.16. Let A be a Noetherian local ring, M be a finitely generated A-module and x1, . . . , xr ∈ Pwhere P is the maximal ideal of A. Then the following are equivalent:

1. (x1, . . . , xr) is an M -regular sequence

2. Hi

(K(x,M)

)= 0 for i > 0

3. H1(K(x),M)

)= 0

Remark 7.17. Under the assumption of the theorem, this implies that every permutation of an M -regularsequence is M -regular.

Proof. 1) ⇒ 2) is the previous theorem, 2) ⇒ 3) is trivial and 3) ⇒ 1): In the case r = 1, the assumptionsays that Ker

(M

x·→M)

= 0 hence it is a non-zerodivisor.For r > 1: set C• := K(x1, . . . , xr−1)⊗AM . By the Lemma

. . . // H1(C•) // H1(C•) // H1(K(xr)⊗ C

)H1(K(x),M) = 0

It means that H1(C•)xr·→ H1(C•) is surjective. But xr ∈ P hence H1(C•) = 0 by Nakayama. Therefore, the

induction hypothesis says that x1, . . . , xr−1 is M -regular but

0 = H1(K(x,M) ∼= Ker(M/(x1, . . . , xr−1)M xr·→ M/(x1, . . . , xr−1)M

)proving that x is a regular sequence.

Definition 7.18. (Grothendieck) A Noetherian local ring A is a complete intersection ring if there exists aregular complete local ring R such that A ∼= R/I where I is generated by a regular sequence.

Fact: Regular ⇒ complete intersection ⇒ Cohen-Macauley

Theorem 7.19. (J. Tate, Assumus) Let A be a Noetherian local ring and x1, . . . , xr is a minimal system ofgenerators of the maximal ideal. Then A is a complete intersection ring if and only if the following gradedA-algebras are isomorphic, i.e.

∞⊕i=0

Hi

(K(x)

) ∼= Λ(H1(K(x))

)

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