Lectures on Unique Factorization Domains

Home , Graded ring, Integrally closed domain, Krull ring, Lattice (order), Noetherian ring, Regular local ring

Lectures On Unique Factorization Domains

By P. Samuel

Tata Institute Of Fundamental Research, Bombay 1964 Lectures On Unique Factorization Domains

By P. Samuel

Notes by M. Pavman Murthy

No part of this book may be reproduced in any form by print, microﬁlm or any other means without written permission from the Tata Institute of Fundamental Research, Colaba, Bombay 5

Tata Institute of Fundamental Research Bombay 1964 Contents

1 Krull rings and factorial rings 1 1 Divisorial ideals ...... 1 2 Divisors...... 2 3 Krullrings...... 4 4 Stability properties ...... 8 5 Two classes of Krull rings ...... 11 6 Divisor class groups ...... 14 7 Applications of the theorem of Nagata ...... 19 8 Examples of factorial rings ...... 25 9 Power series over factorial rings ...... 32

2 Regular rings 37 1 Regular local rings ...... 37 2 Regular factorial rings ...... 41 3 The ring of restricted power series ...... 42

3 Descent methods 45 1 Galoisian descent ...... 45 2 The Purely inseparable case ...... 50 3 Formulae concerning derivations ...... 51 4 Examples: Polynomial rings ...... 53 5 Examples: Power series rings ...... 56

Chapter 1

Krull rings and factorial rings

In this chapter we shall study some elementary properties of Krull rings 1 and factorial rings.

1 Divisorial ideals

Let A be an integral domain (or a domain) and K its quotient ﬁeld. A fractionary ideal U is an A-sub-module of K for which there exists an element d ∈ A(d , 0) such that dU ⊂ A i.e. U has “common denominator” d). A fractionary ideal is called a principal ideal if it is generated by one element. U is said to be integral if U ⊂ A. We say that U is divisorial if U , (0) and if U is an intersection of principal ideals. Let U be a fractionary ideal and V non-zero A-submodule of K. Then the set U : V = x ∈ K xV ⊂ U is a fractionary ideal. The following formulae are easy to verify.

(1) ( Ui : ( V j) = (Ui : V j). Ti Pj Ti, j (2) U : VV ′ = (U : V ) : V ′.

(3) If x ∈ K, x , 0, then V : Ax = x−1V .

1 2 1. Krull rings and factorial rings

Lemma 1.1. Let U be a fractionary ideal , (0) and V a divisorial ideal. Then V : U is divisorial.

Axi Proof. Let V = Axi. Then V : U = (Axi : U ) = ( ) Ti Ti Ti aT∈U a

2 Lemma 1.2. (1) Let U , (0) be a fractionary ideal. Then the smallest divisorial ideal containing U , denoted by U¯,isA : (A : U ).

(2) If U , V , (0), then U¯ = V¯ ⇔ A : U = A : V .

Proof. (1) By Lemma 1.1, A : (A : U ) is divisorial. Obviously U ⊂ A : (A : U ). Suppose now that U ⊂ Ax, x , 0, x ∈ K. Then A : U ⊃ A : Ax = Ax−1. Thus A : (A : U ) ⊂ A : Ax−1 = Ax and (1) is proved.

(2) is a trivial consequence of (1) and the proof of Lemma 1.2 is complete.

2 Divisors

Let I(A) denote the set of non-zero fractionary ideals of the integral domain A. In I(A), we introduce an equivalence relation ∼, called Artin equivalence (or quasi Gleichheit) as follows: U ∼ V ⇔ U¯ = V¯ ⇔ A : U = A : V . The quotient set I(A)/ ∼ of I(A) by the equivalence relation ∼ is called the set of divisors of A. Thus there is an l-1 correspondence between the set D(A) of divisors and the set of divisorial ideals. Let d denote the canonical mapping d : I(A) → I(A)/ ∼. Now, I(A) is partially ordered by inclusion and we have U ⊂ V ⇒ U¯ ⊂ V¯. Thus this partial order goes down to the quotient set I(A)/ ∼ by d. If U ⊂ V , we write d(V ) ≤ d(U ). On I(A) we have the structure of an ordered commutative monoid given by the composition law (U , V ) UV -with A acting as the unit element. Let U , U ′, V ∈ I(A) and U ∼ U ′i.e. A : U = A : U ′. We have A : UV = (A : U ) : V = (A : U ′) : V = A : U ′V . Hence UV ∼ 3 U ′V . Thus D(A) acquires the structure of a commutative monoid, with 2. Divisors 3 the composition law (U¯, V¯) → UV . We write the composition law in D(A) additively. Thus d(UV ) = d(U ) + d(V ) for U , V ∈ I(A). Since the order in D(A) is compatible with the compositional law in D(A), D(A) is a commutative ordered monoid with unit. We note that

d(U ∩ V ) ≥ sup(d(U ), d(V ) and d(U + V ) = inf (d(U ), d(V ). Let K∗ be the set of non-zero elements of K. For x ∈ K∗, we write d(x) = d(Ax); d(x) is called a principal divisor.

Theorem 2.1. For D(A) to be a group it is necessary and suﬃcient that A be completely integrally closed.

We recall that A is said to be completely integrally closed if when- ever, for x ∈ K there exist an a , 0, a ∈ As.t.axn ∈ A for every n, then x ∈ A. We remark that if A is completely integrally closed, then it is integrally closed. The converse also holds if A is noetherian. A valuation ring of height > 1 is an example of an integrally closed ring which is not completely integrally closed.

Proof. Suppose D(A) is a group. Let x ∈ K and a be a non-zero element of A such that axn ∈ A, for every n ≥ 0. Then a ∈ Ax−n = V which is divisorial. Set d(V ) = β and α = d(x−1). nT=0

Now β = sup(nα). But β + α = Supn≥0((n + 1)α) n≥0 = sup (qα). Thus β + α ≤ β. Since D(A) is a group −β exists, 4 q≥1 and therefore

α = (β + α) − β ≤ β − β = 0 i.e. d(x) ≥ 0.

Hence Ax ⊂ A i.e. x ∈ A. Conversely suppose that A is completely integrally closed. Let U be a divisorial ideal. Then U = xU ′, U ′ ⊂ A. Since we already know 4 1. Krull rings and factorial rings

that principal divisors are invertible, we have only to prove that integral divisorial ideals are invertible. Let V be a divisorial ideal ⊂ A. Then V .(A : V ) ⊂ A. Let V (A : V ) ⊂ Ax, for some x ∈ K. Then x−1V (A : V ) ⊂ A. Thus x−1V ⊂ A : (A : V ) = V , since V is divisorial. Thus V ⊂ V x. By induction V ⊂ V xn, for every n ≥ 0. Consider an element b , 0, b ∈ V . Then b(x−1)n ⊂ V ⊂ A for every n ≥ 0. Hence x−1 ∈ A i.e. Ax ⊃ A. Thus V (A : V ) ∼ A. Theorem 2.1 is completely proved. Notice that we have d(U ) + d(A : U ) = 0. Let us denote by F(A) the subgroup generated by the principal divisors. If D(A) is a group, the quotient group D(A)/F(A) is called the divisor class group of A and is denoted by C(A). In this chapter we shall study certain properties of the group C(A).

3 Krull rings

Let Z denote the ring of integers. Let I be a set. Consider the abelian group Z(I). We order Z(I) by means of the following relation: (I) for (αi), (βi) ∈ Z , (αi) ≥ (βi) if αi ≥ βi, for all i ∈ I. 5 The ordered group Z(I) has the following properties: (a) any two elements of Z(I) have a least upper bound and a greatest lower bound i.e. Z(I) is an ordered lattice. (b) The positive elements of Z(I) satisfy the minimum condition i.e. given a nonempty subset of positive elements of Z(I), there exists a minimal element in that set. Conversely any ordered abelian group satisfying conditions (a) and (b) is of the form Z(I) for some indexing set I (for proof see Bourbaki, Algebre, Chapter VI). Let A be an integral domain. We call A a Krull ring if D(A) ≈ Z(I), the isomorphism being order preserving. Theorem 3.1. Let A be an integral domain. Then A is Krull if and only if the following two conditions are satisﬁed. (a) A is completely integrally closed. (b) The divisorial ideals contained in A satisfy the maximum condition. In fact the above theorem is an immediate consequence of Theorem 2.1 and the characterization of the ordered group Z(I) mentioned above. An immediate consequence of Theorem 3.1 is: 3. Krull rings 5

Theorem 3.2. A noetherian integrally closed domain is a Krull ring.

We remark that the converse of Theorem 3.2 is false. For example the ring of polynomials in an inﬁnite number of variables over a ﬁeld K is a Krull ring, but is not noetherian. In fact this ring is known to be factorial and we shall show later that any factorial ring is a Krull ring. (I) Let ei = (δi j) j∈I ∈ Z , where δi j is the usual Kronecker delta. The ei are minimal among the strictly positive ele- 6 ments. Let A be a Krull ring and let ϕ be the order preserving isomor- (I) −1 phism ϕ : D(A) → Z . Let Pi = ϕ (ei). We call the divisors Pi the prime divisors. Let P(A) denote the set of prime divisors. Then any d ∈ D(A) can be written uniquely in the form

d = n P X P P∈P(A)

∗ where nP ∈ Z and nP = 0 for almost all P. Now let x ∈ K . Consider the representation

d(x) = v (x)P, v (x) ∈ Z, v (x) = 0 X P P P P∈P(A) for almost all P ∈ P(A). Since d(xy) = d(x) + d(y) we have, vP(xy) = vP(x) + vP(y) for all P. Further d(x + y) ≥ d(Ax + Ay) = inf(d(x), d(y)). This, in terms of vP, means that vP(x + y) ≥ inf(vP(x), vP(y)). We set vP(0) = +∞. Thus the vP are all discrete valuations of K. These are called the essential valuations of A. Let P be a prime divisor. Let Y be the divisorial ideal corresponding to P. As P is positive, Y is an integral ideal. We claim that Y is a prime ideal. For let x, y ∈ A, xy ∈ Y . Then d(xy) ≥ P i.e. d(x) + d(y) ≥ P i.e. vP(x) + vP(y) ≥ 1. As vP(x) ≥ 0, vp(y) ≥ 0, we have vP(x) ≥ 1 or vP(y) ≥ 1; i.e. x ∈ Y or y ∈ Y . Further the divisorial ideal corresponding to nP is x ∈ A vP(x) ≥ n , n ≥ 0. The prime ideal 7 Y is the centre of the valuation vP on A (i.e. the set of all elements x ∈ A s.t. vP(x) ≥ 0). Since the prime divisors are minimal among the set of positive divisors, the corresponding divisorial ideals, which we call prime divisorial ideals, are maximal among the integral divisorial 6 1. Krull rings and factorial rings

ideals. The following lemma shows that the divisorial ideals which are prime divisorial and this justiﬁes the terminology ’prime divisorial’.

Lemma 3.3. Let G be a prime ideal , (0). Then G contains some prime divisorial ideal.

Proof. Take an x ∈ G , x , 0. Let d(x) = niPi, (ﬁnite sum), ni > 0, Pi Y Pi ∈ P(A). Let i be the prime ideal corresponding to Pi. Let y ∈ Y ni , ≥ ≥ ⊂ i , y 0. Then vPi (y) ni. Hence d(y) d(x) i.e. Ay Ax. Thus Q Y ni G G Y G i ⊂ Ax ⊂ . As is prime, i ⊂ for some i. Q Corollary. A prime ideal is prime divisorial if and only if it is height 1.

(We recall that a prime ideal is of height one if it is minimal among the non-zero prime ideals of A).

Proof. Let Y be a prime divisorial ideal. If Y is not of height 1, then Y ⊃ G , where G is a non-zero prime ideal. By the above lemma G , contains a prime divisorial ideal Y ′. Thus Y ⊃ Y ′. This contradicts , the maximality of Y ′ among integral divisorial ideals. Conversely let Y be a prime ideal of height 1. Then, by the above lemma, Y contains a prime divisorial ideal Y ′. Hence Y = Y ′ and the proof of the Corollary is complete.

8 Lemma 3.4. Let Y be a divisorial ideal corresponding to a prime divisor P. Then the ring of quotients AY is the ring of vP. a Proof. Let ∈ AY , a ∈ A, s ∈ A − Y . Then v (s) = 0, v (a) ≥ 0, s P P a v ≥ 0. Conversely let x ∈ K∗ with v ≥ = P P(x) 0. Set d(x) n(Q)Q, s PQ and let G be the prime divisorial ideal corresponding to Q. Let V = G −n(Q). As the prime divisor Q, with n(Q) < 0 are diﬀerent from n(Q)<0 P, we have V 1 Y . Take s ∈ V , s < Y . Then vQ(sx) ≥ 0 for all Q i.e. d(sx) ≥ 0 i.e. sx ∈ A. Hence x ∈ AY . This proves the lemma. 3. Krull rings 7

Corollary . A = AY , Y running through all prime ideals of height TY one. We shall now give a characterization of Krull rings in terms of discrete valuation rings.

Theorem 3.5 (“valuation Criterion”). Let A be a domain. Then the following conditions are equivalent:

(a) A is a Krull ring.

(b) There exists a family (vi)i∈I of discrete valuations of K such that

(1) A = Rvi (i.e. x ∈ A if and only if vi(x) ≥ 0), where Rvi denotes the Ti ring of vi,

(2) For every x ∈ A, vi(x) = 0, for almost all i ∈ I.

Proof. (a) ⇒ (b). In fact A = RvP and condition (2) of (b) is P∈TP(A) obvious from the very deﬁnition of vP.

(b) ⇒ (a). Since a discrete valuation ring is completely integrally 9 closed and any intersection of completely integrally closed domains is completely integrally closed, we conclude that A is completely integrally closed. Now let x ∈ K∗. Then

Ax = y ∈ K vi(y) ≥ vi(x), for i ∈ I . n o

Thus, because of condition (2), any divisorial ideal is of the form (I) x ∈ K vi(x) ≥ ni, i ∈ I, (ni) ∈ Z , and conversely. Any integral di- n V o visorial ideal is defined by the conditions vi(x) ≥ ni, ni ≥ 0, ni = 0 for almost all i ∈ I. There are only finitely many divisorial ideals V ′ containing V (in fact the number of such ideals is (1 + ni)). Hence A satisfies the maximum condition for integral Qi divisorial ideals therefore by Theorem 3.1, A is a Krull ring.

Remark. Let G be a prime divisorial ideal of A deﬁned by vi(x) ≥ ni, ni ≥ 0, ni = 0 for almost all i. Let Yi be the centre of vi on A. Then 8 1. Krull rings and factorial rings

Y ni G Y G G i ⊂ . Hence i ⊂ for some i. But height = 1. Hence Qi Yi = G . Thus every prime divisorial ideal is the centre of some vi. Now

AG ⊂ Rvi . But AG , being a discrete valuation ring, is a maximal subring

of K. Hence AG = Rvi = ring of vQ, where Q is the prime divisor corresponding to G . Thus every essential valuation of A is equivalent to some vi. The family (vi)i∈I may be ‘bigger’, but contains all essential valuations.

4 Stability properties

In this section we shall see that Krull rings behave well under localisa- tions polynomial extensions etc.

10 Proposition 4.1. Let K be a ﬁeld and Aα a family of Krull rings. Assume that any x ∈ B = Aα, x , 0 is a unit in almost all Aα. Then B is a Tα Krull ring.

Proof. By theorem 3.5, Aα = R(vα,i), where R(vα,i) are discrete val- iT∈Iα uation rings and every x ∈ Aα is a unit in almost all R(vα,i), i ∈ I. Now B = R(vα,i). Let x , 0, x ∈ B. Then by assumption, x is a unit in Tα,i almost all Aα, i.e. vα,i(x) = 0, for all i ∈ Iα, and almost all α. Then x

is not a unit in at most a ﬁnite number of the Aβ, say Aβ1 ,... Aβt . Now

vβ ji (x) = 0 for almost all i, j = 1,..., t. Thus vα,i(x) = 0 for almost all α and i. The proposition follows immediately from Theorem 3.5.

Corollary. (a) A ﬁnite intersection of Krull rings is a Krull ring.

(b) Let A be a Krull ring, K its quotient ﬁeld. Let L be a subﬁeld of K. Then A ∩ L is a Krull ring.

Proposition 4.2. Let A be a Krull ring. Let S be any multiplicatively closed set with 0 < S . Then the ring of quotients S −1A is again a Krull −1 ring. Further the essential valuations of S A are those valuation vP for which Y ∩ S = φ, Y being the prime divisorial ideal corresponding to P. 4. Stability properties 9

−1 Proof. By Theorem 3.5 we have only to prove that S A = AY , YT∈I I, being the set of prime ideals of height one which do not intersect −1 S . Trivially, S A ⊂ AY . Conversely let x ∈ AY . We have YS∈I YS∈I vP(x) ≥ 0, for any prime divisor corresponding to Y ∈ I. For a 11 prime divisorial ideal G with G ∩ S , ∅, choose sG ∈ G ∩ S . Set −vQ(x) G s = sG , where Q is the prime divisor corresponding to . vQQ(x)<0 −1 Then sx ∈ A i.e. x ∈ S A and the proposition is proved. Proposition 4.3. Let A be a Krull ring. Then the ring A[X] of polynomials is again a Krull ring.

Proof. Let A be defined by the discrete valuation rings vi i∈I. If v is a discrete valuation of A, then v can be extended to A[X], by putting q v¯(ao + a1 x + ··· + aq x ) = min(v(a j)) and then to the quotient field K(X) j of A[X] (K is the quotient field of A) by puttingv ¯( f /g) = v¯( f )−v¯(g). Let

Φ= v¯i i∈I. On the other hand, let Ψ denote the set p(X)-adic valuation of K(X), where p(X), runs through all irreducible polynomials K[X]. We now prove that A[X] is a Krull ring with Φ Ψ as a set of valuations deﬁning it. Let f ∈ K(X). If ω( f ) ≥ 0 for allSω′ ∈ Ψ, then f ∈ K[X], q say f = ao + a1X + ··· + aqX : ai ∈ K. If further v( f ) ≥ 0, for all v ∈ Φ, then v(ai) ≥ 0, v ∈ Φ, i = 1,..., q. Since A is a Krull ring, ai ∈ A, i = 1,..., q. Hence f ∈ A[X]. To prove that A[X] is a Krull ring, it only remains to verify that for f ∈ A[X], v( f ) = 0 = ω( f ) for almost all v ∈ Φ, ω ∈ Ψ. Since KX is a principal ideal domain, ω( f ) = 0, for almost all ω ∈ Ψ. Furtherv ¯i( f ) = min(vi(a j)) = 0 for almost all i ∈ I, j since A is a Krull ring.

Corollary. Let A be a Krull ring. Then A X1,..., Xn is a Krull ring. 12 Remark . Since, in a polynomial ring in a an infinite number of variables, a given polynomial depends only on a finite number of variables the above proof shows that a polynomial ring in an infinite number of variables over A is also a Krull ring, Proposition 4.4. Let A be a Krull ring. Then the ring of formal power series A[[X]] is again a Krull ring. 10 1. Krull rings and factorial rings

Proof. We note that A [X] = Vα[[X]]S K [X] , where S is the multiplicatively closed set 1, x, xT2,... , K, theT quotient ﬁeld of A and Vα the essential valuation rings of A. Now Vα [X] S and K [X] , being noetherian and integrally closed, are Krull rings, Now the proposition is an immediate consequence of Proposition 4.1.

Proposition 4.5. Let A be a Krull ring and K its quotient ﬁeld. Let K′ be a ﬁnite algebraic extension of K and A′, the integral closure of A in K′. Then A′ is also a Krull ring.

Proof. Let K′′ be the least normal extension of K containing K′ and let A′′ be the integral closure of A in K′′. Since A′ = A′′ K′, by Corollary (b), Prop. 4.1, it is suﬃcient to prove that A′′ is a KrullT ring. Let Φ be the family of essential valuations of A. Let Φ′′ be the family of all discrete valuations of K′′ whose restriction to K are in Φ. It is well known (see Zariski-Samuel. Commutative algebra Vol. 2) that every discrete valuation v of K extends to a discrete valuation of K′′ and 13 that such extensions are ﬁnitely many in number. We shall show that A′′ is a Krull ring by using the valuation criterion with Φ′′ as family of valuations. We prove (i) x ∈ A′′ if and only if ω′(x) ≥ 0, for all ω ∈ Φ′′

Proof. Let x ∈ A′′. Then x satisﬁes a monic polynomial, say xn + n−1 an−1 x + ···+ ao = 0, ai ∈ A. If possible let ω(x) < 0, for some ω ∈ Φ. n−1 n−1 n Then ω(−an−1 x −···− ao) ≥ inf(ω(an−1 x ),...,ω(ao) > ω(x ), ′′ since ω(ai) ≥ 0. Contradiction. Conversely let x , 0, x ∈ K with ω(x) ≥ 0, for all ω ∈ Φ′′. Let σ be any K-automorphism of K′′. Then ωoσ ∈ Φ′′. Hence ω(σ(x)) ≥ 0 for all K-automorphisms σ of K′′. Let us now consider the minimal polynomial f (X) of x over K; say r r−1 f (X) = X + αr−1X + ··· + αo, αi ∈ K. Since the αi are symmetric polynomials in the σ(x), we have v(αi) ≥ 0, for all v ∈ Φ. Since A is a ′′ Krull ring, αi ∈ A and (i) is proved. (ii) For x , 0, x ∈ A , ω(x) = 0, for almost all ω ∈ Φ′′.

n n−1 Proof. Let x + an−1 x + ··· + ao = 0 be an equation satisﬁed by x (which expresses the integral dependence of x). We may assume ao , 0. n n−1 If ω(x) > 0, then ω(ao) = ω(x +an−1 x +···+a1 x) ≥ ω(x) > 0. Since 5. Two classes of Krull rings 11

A is a Krull ring there are only a ﬁnitely many v ∈ Φ such that v(ao) > 0 and since every v ∈ Φ admits only a ﬁnite number of extensions, (ii) is proved and with it, the proposition.

5 Two classes of Krull rings

Theorem 5.1. Let A be a domain. The following conditions are equivalent.

(a) Every fractionary ideal V , (0) of A is invertible. (i.e. there exists 14 a fractionary ideal V −1 such that VV −1 = A).

(b) A is a Krull ring and every non-zero ideal is divisorial.

(d) A is a noetherian, integrally closed domain such that every prime ideal , (0) is maximal.

Proof. (a) ⇒ (b). Let V −1 exist. Then V −1 = A : V . For VV −1 ⊂ A ⇒ V −1 ⊂ A : V and VV −1 = A and V (A : V ) ⊂ A together imply A : V ⊂ V −1. Further V = A : (A : V ); the condition that U and V are Artin equivalent becomes “U = V ′′. Thus D(A) is a group. The following lemma now proves the assertion (a) ⇒ (b) because of Theorem 2.1.

Lemma. V ⊂ A invertible ⇒ V is ﬁnitely generated. (and thus (a) implies that A is noetherian).

−1 −1 Proof. Since VV = A, we have 1 = xiyi, xi ∈ V , yi ∈ V . For P1 x ∈ V , we have x = xi(yi x) i.o. V = Axi. Pi Pi (b) ⇒ (c), Since A is a Krull ring and every non-zero ideal is divisorial, every non-zero prime ideals is of height 1. (c) ⇒ (a). Let V be any fractionary ideal. Then V (A : V ) ⊂ A and since D(A) is a group, V (A : V ) is Artin-equivalent to A. Hence V (A : V ) is not contained in any prime divisorial ideal. But by (c). Since every 12 1. Krull rings and factorial rings

prime ideal , (0) is prime divisorial, V (A : V ) is not contained in any maximal ideal, and hence V (A : V ) = A. 15 (a) ⇒ (d). That A is noetherian is a consequence of the lemma used in the proof of the implication (a) ⇒ (b). That A is integrally closed and every prime ideal , (0) is maximal is a consequence of the fact that (a) ⇒ (c). (d) ⇒ (c). This is an immediate consequence of the fact that a noetherian integrally closed domain is a Krull ring. The proof of Theorem 5.1 is now complete.

Deﬁnition 5.2. A ring A is called a Dedekind ring if A is a domain satisfying any one of the equivalent conditions of Theorem 5.1.

Remark. (1) The condition (c) can be restated as: A is a Dedekind ring if A is a Krull ring and its Krull dimension is atmost 1.

(2) Let A be a Dedekind ring and K its quotient ﬁeld. Let K′ be a ﬁnite algebraic extension of K and A′, the integral closure of A in K′. Then A′ is again a Dedekind ring.

Proof. By Proposition 4.5 it follows that A′ is a Krull ring. For a non- zero prime ideal Y ′ of A′, we have, by the Cohen-Seidenberg theorem, height (Y ′) = height (Y ′ ∩A) ≤ 1. Now (2) is a consequence of Remark (1).

(3) Let A be a domain and U , a fractionary ideal. Then U is invertible if and only if U is a projective A-module. If further, A is noetherian, then U is projective if and only if U is locally principal (i.e UM is principal for all maximal ideals M of △).

We shall say that a ring A satisﬁes the condition (M) if A satisﬁes the maximum condition for principal ideals. For instance Krull rings satisfy (M).

16 Theorem 5.3. For a domain A, the following conditions are equivalent.

a) A is a Krull ring and every prime divisorial ideal is principal. 5. Two classes of Krull rings 13 b) A is a Krull ring and every divisorial ideal is principal. c) A is a Krull ring and the intersection of any two principal ideals is principal. d) A satisfies (M) and any elements of A have a least common multiple (1.c.m.) (i.e. Aa ∩ Ab is principal for a, b ∈ A). e) A satisfies (M) and any two elements of A have greatest common divisor (g.c.d.). f) A satisfies (M) and every irreducible element of A is prime. (We recall that a ∈ A is irreducible if Aa is maximal among principal ideals, an element p ∈ A is prime if A p is a prime ideal). g) A has the unique factorization property. More precisely, there exists a subset P ⊂ A, 0 < P such that every x , 0, x ∈ A can be written in on and only way as x = u. pn(p), n(p) ≥ 0, n(p) = 0 for almost pQ∈P all p, u being a unit. Proof. (a) ⇒ (b). Since prime divisors generate D(A), we have D(A) = F(A). (b) ⇒ (c). Trivial. (c) ⇒ (b). Let V be any divisorial ideal , (0).

We shall show that V is principal. We may assume that V is integral. Let V = Acλ, cλ ∈ K. Consider the set of all divisorial ideals VJ = 17 λT∈I Acλ A, where J runs over all finite subsets I. We have V ⊂ VJ ⊂ λT∈J T A, for all finite subsets J ⊂ I. Since A is a Krull ring, any integral divisorial ideal is defined by the inequalities v(x) ≥ nv, nv > 0, nv = 0 for almost all v, v running through all essential valuations of A. Hence there are only finitely many divisorial ideals between V and A. Hence V = λ∈J Ac, for some finite set J ⊂ I. By choosing a suitable common denominatorT for cλ, λ ∈ J, we may assume that cλ ∈ A. Now (c) ⇒ (b) is immediate. 14 1. Krull rings and factorial rings

(b) ⇒ (a). Trivial. (c) ⇒ (d). Trivial. (d) ⇒ (e). This is an immediate consequence of the following elementary property of ordered abelian groups, namely: in an ordered abelian group G, the existence of sup (a, b) is equivalent to the existence of inf(a, b), for a, b ∈ G. (Apply this, for instance to F(A)) (e) ⇔ ( f ) ⇔ (g). This follows from divisibility arguments used in elementary number theory.

v (x) (g) ⇒ (c). For K ∈ A, x , 0, we write x = ux p p , u being a pQ∈P unit. The set of all {vp}p∈P, defines a set Φ of discrete valuations of the quotient field X of A. It is clear that A satisfies the valuation criterion for Krull rings with Φ as the set of valuations. Further, max(v (x)v (g)) for a, b ∈ A, Aa ∩ Ab = Ac, where c = uxuy p P P . pQ∈P Hence (g) ⇒ (c) and the proof of Theorem 5.3 is complete. Definition 5.4. A is said to be factorial if A is a domain satisfying any one of the conditions of Theorem 5.3.

18 Remark 5.5. Let A be a noetherian domain with the property that every prime ideal of height 1 is principal. Then A is factorial.

Proof. We shall prove the condition ( f ). Set b ∈ A be an irreducible element. By Krull’s Principal Ideal Theorem, Ab ⊂ Y , Y , a prime ideal of height 1. By hypothesis, Y is principal. Since Ab is maximal among principal ideals, Ab = Y . Of course A satisﬁes (M).

6 Divisor class groups

Let A be a Krull ring. We recall that the divisor class group c(A) of A is D(A) , where D(A) is the group of divisors of A and F(A) the subgroup F(A) of D(A) consisting of principal divisors of A. If A is a Dedekind ring, C(A) is called the group of ideal classes. By Theorem 5.3, it is clear that a Krull ring A is factorial if and only if C(A) = 0. 6. Divisor class groups 15

Let A and B be Krull rings, with A ⊂ B. From now on we shall use the same notation for a prime divisor and the prime divisorial ideal corresponding to it. Let P, p be prime divisors of B and A respectively.

We write P p if P lies above p i.e. P∩A = p. If P p, the restriction of vP to the quotient field of A is equivalent to vp, and we denote by e(P, p), the ramification index of vp in vP. For a p ∈ P(A), we define

j(p) = e(P, p)P, P ∈ P(B). X P p

The above sum is ﬁnite since x ∈ p, x , 0 is contained in only ﬁnite many P, P ∈ P(B). Extending j by linearity we get a homomorphism of D(A) into D(B) which we also denote by j. We are interested in 19 the case in which j induces a homomorphism of ¯j : C(A) → C(B) i.e. j(F(A)) ⊂ F(B). For x ∈ A, we write dA(x) = d(Ax) ∈ D(A) and dB(x) = d(Bx) ∈ D(B).

Theorem 6.1. Let A and B be Krull rings with A ⊂ B. Then we have j(dA(x)) = dB(x) if and only if the following condition is satisﬁed. (NBU). For every prime divisor P of B, height (P ∩ A) ≤ 1.

Proof.

j(d (x) = j( v (x)p) A X p p∈P(A) = v (x) e(P, p)P = v (x)P X p X X P p∈P(A) P|p p,P|p = v (x)P. X P P,P∩A∈P(A)

If P ∩ A = (0), then vP(x) = 0. Therefore,

j(d (x)) = v (x)P. (1) A X P height(P∩A)≤1 16 1. Krull rings and factorial rings

Now, if (NBU) is true, then j(dA(x)) = vP(x)P = dB(x). On the P∈PP(B) other hand let j(dA(x)) = dB(x) for every x ∈ A. Let P ∈ P(B) with P ∩ A , (0). Choose x ∈ P ∩ A, x , 0. We have by (1) above

j(d (x)) = v (x)P = d (x) = v (x)P. A X P B X P height (P∩A)≤1 P∈P(B)

Since vP(x) > 0, we have height (P ∩ A) = 1 and the theorem is proved. When (NBU) is true we have j(F(A)) ⊂ F(B) and therefore j induces a canonical homomorphism ¯j : C(A) → C(B). 20 We now give two sufficient conditions in order that (NBU) be true. Theorem 6.2. Let A and B be as in Theorem 6.1. Then (NBU) is satisfied if any one of the following two conditions are satisfied. (1) B is integral over A.

(2) B is a flat A-module (i.e the functor ⊗B is exact). A Further, if (2) is satisfied we have j(U ) = U B, for every divisorial ideal U of A. Proof. If (1) is satisfied, (NBU) is an immediate consequence of the Cohen-Seidenberg theorem.

Suppose now that (2) is satisﬁed. Let P ∈ P(B) with U = P ∩ A , (0). Suppose U is not divisorial. Choose a non-zero element x ∈ U . n U 1 Let d(x) = vpi (x)p p ∈ P(A). Since height σ > 1 we have p iP=1 i i i for i = 1,..., n. By an easy reasoning on prime ideals there exists n a y ∈ U , y < p . Then dA(x) and dA(y) do not have any component fS=1 i in common and therefore

dA(xy) = dA(x) + dA(y) = Sup(dA(x), dA(y)).

This, in terms of divisorial ideals, means that Ax ∩ Ay = Axy. Since B is A-ﬂat, we have Bxy = Bx ∩ By; that is dB(x) and dB(y) do not have any component in common. But x, y ∈ P ∩ A. Contradiction. 6. Divisor class groups 17

We shall now prove that for any divisorial ideal U ⊂ A, j(U ) = U B. Since A is a Krull ring U is the intersection of ﬁnitely many n principal ideals, say U = Axi, so that dA(U ) = sup(dA(xi)). Since 21 iT=1 n B is A-ﬂat, we have BU = Bxi. Thus BU is again divisorial. On iT=1 the other hand, dB(BU ) = sup(dB(xi)) = sup( j(dA(x)). Noting that j is i i order preserving and that any order preserving homomorphism of Z(I) into Z(J) is compatible with the formation of sup and inf (to prove this we have only to check it component - wise), we have

BU = sup( j(dA(xi)) = j(sup(dA(xi))) = j(dA(U )). i i Theorem 6.3 (Nagata). Let A be a Krull ring and S , a multiplicatively closed set in A(0 < S ). Consider the ring of quotients S −1A (which is A-ﬂat). We have (a) ¯j : C(A) → C(S −1A) is surjective.

(b) If S is generated by prime elements then ¯j is bijective.

Proof. (a) Since P(S −1A) = pS −1A p ∈ P(A), p∩S = φ , ¯j is surjective by Theorem 6.2, (2). Let us look at the kernel of ¯j. Let H be the subgroup of D(A) generated by prime divisors p with p ∩ S , φ. Then it is clear that

(H + F(A)) H Ker(¯j) = ≈ (6.4) F(A) (H ∩ F(A)) Suppose that S is generated by prime elements. Let p ∈ P(A), with p ∩ S , φ, say s1 ··· sn ∈ p, where si are prime elements. Then since p is minimal p = Asi for some si. Thus H ⊂ F(A) and hence ¯j is a bijection.

′Theorem 6.3 (Nagata). Let A be a noetherian domain and S a multi- −1 plicatively closed set of A generated by prime elements p i∈I. If S A i is a Krull ring then A is a Krull ring and ¯j is bijective. 22 18 1. Krull rings and factorial rings

Proof. By virtue of Theorem 6.3, we have only to prove that A is integrally closed (then it will be a Krull ring, since it is noetherian). Now

AApi is a local noetherian domain whose maximal ideal is principal

and hence AApi is a discrete valuation ring. It suﬃces to show that −1 , , A = S A ∩ ( AApi ). We may assume Api Ap j for i j. Let iT∈I ∈ −1 ∩ ∈ ∈ = n(i) ≥ a/s S A ( AApi ), a A, s S , s i pi . We have vpi (a/s) 0, Ti Q where v is the valuation corresponding to A . By our assumption, pi Api , vpi (p j) = 0 for j i. Hence vpi (a) ≥ vpi (s) = n(i). Hence S divides a i.e. a/s ∈ A.

Corollary. Let A be a domain and S a multiplicatively closed set generated by a set of prime elements. Let S −1A be factorial. If A is noetherian or a Krull ring, then A is factorial.

Proof. By Theorems 6.3 and 6.3, ¯j : C(A) → C(S −1A) is bijective.

Theorem 6.4 (Gauss). Let R be a Krull ring. Then ¯j : C(R) → C(R[X]) is bijective. In particular, R is factorial if and only if R[X] is factorial. (Since R[X] is R-ﬂat, ¯j is deﬁned).

Proof. Set A = R[X], S = R∗, the set of non-zero elements of R. Then S −1A = K[X], where K is the quotient ﬁeld of R. Thus C(S −1A) = 0 i.e. (H + F(A)) C(A) = Ker(¯j : C(A) → C(S −1A)) = F(A)

23 where H is the subgroup of D(A) generated by P ∈ P(A), with P ∩ R , (0) (see formula 6.4). Hence D(A) = H + F(A). Since R[X] is R-ﬂat, by Theorem 6.2, (2) we have

j(P ∩ R) = (P ∩ R)R[X] = P, for P ∈ P(A), P ∩ R , (0).

Hence j(D(R)) = H and therefore ¯j is surjective, since D(A) = H+F(A). Now an ideal U of R is principal if and only if U R[X] is principal. Therefore ¯j is injective. Thus ¯j is bijective. Let A be a noetherian ring and M an ideal contained in the radical of A (i.e. the intersection of all maximal ideals of A). If we put on A the 7. Applications of the theorem of Nagata 19

M -adic topology, then (A, M ) is called a Zariski ring. The completion Aˆ of A is again a Zariski ring and it is well known that Aˆ is A-flat and A ⊂ Aˆ. Theorem 6.5 (Mori). Let (A, M ) be a Zariski ring. Then if Aˆ is a Krull ring, then so is A. Further j : C(A) → C(Aˆ) is injective. In particular if Aˆ is factorial, so is A. Proof. Let K and L be the quotient fields of A and Aˆ respectively; K ⊂ L. To prove that A is a Krull ring we observe that A = Aˆ ∩ K. For if a a ∈ Aˆ ∩ K, a, b ∈ A, then a ∈ Abˆ ∩ A = Ab, i.e. ∈ A. Hence b b A is a Krull ring. By virtue of Theorem 6.2 (2), to prove that ¯j is an injection it is enough to show that an ideal U of A, is principal if AÛ is U AÛ principal. Let AÛ = Aˆα, α ∈ Aˆ. Now ≈ is generated by MU AˆMU A a single element as an -module say by x( mod MU ), x ∈ U . Then M U = Ax + MU . By Nakayama’s-lemma U = Ax and the theorem is 24 proved.

7 Applications of the theorem of Nagata

We recall that a ring A is called graded if A = An, An being abelian nP∈Z groups such that Ap Aq ⊂ Ap+q, for p, q ∈ Z, and the sum being direct. An ideal of A is graded if it generated by homogeneous elements. Proposition 7.1. Let Λ be a graded Krull ring. Let DH(A) denote the subgroup of D(A) generated by graded prime divisorial ideals and let FH(A) denote the subgroup of DH(A) generated by principal ideals. DH(A) Then the canonical mapping → C(A), induced by the inclusion FH(A) i : DH(A) → D(A), is an isomorphism.

Proof. If A = Ao, there is nothing to prove. Hence we may assume A , Ao. Let S be the set of non-zero homogeneous elements of A. −1 −1 −1 Then S A is again a graded ring; infact S A = (S A) j, where jP∈Z −1 a o o (S A) j = a, b ∈ A, a, b homogeneous, d a − d b = j . b

20 1. Krull rings and factorial rings

Here do x denotes the degree of a homogeneous element x ∈ A. We −1 −1 −1 note that (S A)o = K is a ﬁeld and that S A ≈ K t, t , where t is a homogeneous element of smallest strictly positive degree . Now t is transcendental over K and therefore S −1A is factorial. Hence C(A) ≈ Ker(¯j), where ¯j is the canonical homomorphism ¯j : C(A) → C(S −1A), and C(S −1A) = 0. H Hence C(A) ≈ , where H is the subgroup of D(A), gen- (H ∩ F(A)) 25 erated by prime ideals Y of height 1 with Y ∩ S , φ. Since DH(A) ∩ F(A) = FH(A), and since prime divisorial ideals are of height 1, the proposition is a consequence of the following

Lemma 7.2. Let A = An be a graded ring and Y a prime ideal in A nP∈Z and let U be the ideal generated by homogeneous elements of Y . Then U is a prime ideal.

U < U < U Proof. Let xy ∈ , x = xi, y = yi, x , y . Let xio y jo P P < U < U be the lowest components of x, y such that xio , y jo . Then U Y Y Y Y xio y jo ∈ ⊂ . Since is prime, xio or y jo ∈ , say xio ∈ . Then U xio ∈ , a contradiction.

Corollary . Let A = An be a graded ring and Y a prime ideal of nP∈Z height 1. Then Y is graded if and only if Y ∩ S , φ.

Remark. If U is a graded ideal of A, then the least divisorial ideal A : (A : U ) containing U is also graded (straight forward proof). Thus the divisors corresponding to graded divisorial ideals of A form a subgroup of D(A); this subgroup obviously contains DH(A); furthermore, since, given a graded integral divisorial ideal U , all the prime divisorial ideals containing U are graded (by the corollary), we see that this subgroup is in fact DH(A).

The above proposition can be applied for instance to the homogeneous coordinate ring of a projective variety. The following proposition connects the divisor class group of a projective variety V with the divisor class group of a suitable aﬃne open subset of V. 7. Applications of the theorem of Nagata 21

26 Proposition 7.3. Let A be a graded Krull ring and p a prime homogeneous element , 0 with do p = 1. Let A′ be the subring of K (quotient a ﬁeld of A) generated by A and , where a runs over the non-zero o pdoa homogeneous elements of A. Then C(A′) ≈ C(A).

′ −1 ′ Proof. We note that A = (S A)o and that p is transcendental over A . The inclusions A′ → A′[p] → A′ p, p−1 induce isomorphisms C(A′) ≈ C(A′[p]), C(A′[p]) ≈ C(A′[p, p−1]), the ﬁrst isomorphism follows from Theorem 6.4 and the second from Theorem 6.3. Now A p−1 = A′ p, p−1 . But again by Theorem 6.3, C(A) ≈ C(A p−1 ) and the proof of the proposition is complete.

Let V be an arithmetically normal projective variety. We prove that the homogeneous coordinate ring of V is factorial if and only if the local ring of the vertex of the projecting cone is factorial; in fact we have the following

Proposition 7.4. Let A = Ao + A1 + A2 + ··· be a graded Krull ring and suppose that Ao is a ﬁeld. Let M be the maximal ideal A1 +A2 +··· . Then C(A) ≈ C(AM ).

Proof. We have only to prove that ¯j : C(A) → C(AM ) is injective. Because of Proposition 7.1 and of the remark following it is suﬃcient to prove that if V is a graded divisorial ideal such that V AM is principal, then so is V . Suppose that V is a graded divisorial ideal with V AM principal. Since AM is a local ring, there is a homogeneous element u ∈ V such that V AM = uAM . Let x ∈ V be any homogeneous y element. Then x = u, y ∈ A, z ∈ A − M . Let z

y = yq + yq+1 + ··· , z = zo + z1 + z2 + ··· , yi ∈ Ai, i ≥ q, yq , 0,

z j ∈ A j, zo , 0. Thus x(zo + z1 + z2 + ··· ) = (yq + yq+1 + ··· )u.

27 Hence xzo = yqu. Since zo is invertible, we conclude that V = An and the proposition is proved. 22 1. Krull rings and factorial rings

Adjunction of indeterminates. Let A be a local ring and M its maximal ideal. We set A(X)loc = A[X]M A[X] and by induction A(X1,..., Xn)loc = A(X1,..., Xn−1)loc (Xn)loc. We remark that A(X)loc is a local ring and that if A is noetherian, properties of A like its dimension, multiplicity, regularity and so on are preserved in passing from A to A(X)loc. Further A(X)loc is A-ﬂat (in fact it is faithfully ﬂat). Proposition 7.5. Let A be a local Krull ring. Then

¯j : C(A) → C(A(X1,..., Xn)loc) is an isomorphism. Proof. It is suﬃcient to prove this when n = 1. Since by Theorem 6.5 C(A) ≈ C(A[X]), we see that ¯j is surjective. Let V be a divisorial ideal of A for which V A(X)loc is principal. Since A(X)loc is a local ring, we may assume that V A(X)loc = A(X)locα, α ∈ V . Let y ∈ V . f (X) Then y = .α, where f (X), g(X) ∈ A[X] and atleast one of the g(X) coeﬃcients of g(X) is invertible in A. Looking at a suitable power of X in y.g(X) = α f (X) we see that y ∈ Aα i.e. V = Aα. Hence ¯j is injective. This proves the proposition.

28 Proposition 7.6. Let A be a domain and a, b ∈ A with Aa ∩ Ab = Aab. The following results hold. (a) The element aX − b is prime in A[X]. (b) If further, we assume that A is a noetherian integrally closed domain and that Aa and Aa + Ab are prime ideals, then the ring A′ = A[X] is again integrally closed and the groups C(A) and C(A′) (aX − b) are canonically isomorphic. b Proof. (a) Consider the A-homomorphism ϕ : A[X] → A given a b by ϕ(X) = . It is clear that the ideal (aX − b) ⊂ Ker(ϕ). Con- a versely we show by induction on the degree that if a polynomial P(X) ∈ Ker(ϕ), then P(X) ∈ (aX − b). This is evident if 7. Applications of the theorem of Nagata 23

o n n−1 d (P) = 0. If P(X) = cnX + cn−1X + ··· + co(n > 0), the b relation P( ) = 0, shows that bnc ∈ Aa. Since Aa ∩ Ab = Aab, it a n follows that cn ∈ Aa, say cn = dna, dn ∈ A. Then the polynomial n−1 P1(X) = P(X) − dn(aX − b)X ∈ Ker(ϕ) and has degree ≤ n − 1. By induction we have P1(X) ∈ (aX−b) and hence P(X) ∈ (aX−b). Thus (aX − b) = Ker(ϕ) and (a) is proved. b 1 1 1 1. We note that A′ ≈ A ⊂ A and A ≈ A′ . By Theo- a a a a rem 6.3, the proof of (b) will be complete if we show that a is a prime element in A′. But A′ A[X] A[X] A ′ ≈ = ≈ X . A a (a, aX − b) (a, b) (a, b)

By assumption (a, b) is a prime ideal and therefore a is a prime element in A′. 29

Remark 1. In (b), if a, b are contained in the radical of A, and if the ideal Aa+ Ab is prime, then a and b are prime elements (for proof see P. Samuel: Sur les anneaux factoriels, Bull. Soc. math. France, 89 (1961), 155-173).

Remark 2. Let A be a noetherian integrally closed local domain and let the elements a, b ∈ A satisfy the hypothesis of the above proposition. Set A′′ =A(X) loc /(aX − b). Then it follows from (a) that A′′ is a Krull ring. We have a commutative diagram

β α / / ′′ A I / A(X)log 9/ A II tt9 II tt II tt II tt I$ tt A[X]/(aX−b)

′′ Since A is a ring of quotients of A[X]/(aX−b), it follows from (b) that βoα induces a surjective mapping ϕ : C(A) → C(A′′). We do not know if ϕ is an isomorphism. If ϕ is an isomorphism we can get another proof 24 1. Krull rings and factorial rings

of the fact that a regular local ring is factorial (see P. Samuel : Sur les anneaux factoriels, Bull. Soc. math. France, t.89, 1961).

Proposition 7.7 (C.P. Ramanujam). Let A be a noetherian analytically normal local ring and let M be its maximal ideal. Let B = A X1,..., h Xn . Then the canonical mapping j : C(B) → C(MM B) is an isomor- i phism. Proof. By Theorem 6.3 (a), ¯j is surjective and Ker( j) = (H + F(B))/ F(B), where H is the subgroup of D(B) generated by prime ideals Y of height one in B with Y 1 M B and F(B) is the group of principal 30 ideals. Thus we have only to prove that if Y is a prime ideal of height one of B with Y 1 M B, the then Y is principal. This, we prove in two steps. (i) Assume that A is complete. Let Y be a prime ideal of height one with Y 1 M B. It is clear that Y is generated by a finite number of elements f j ∈ Y such that f j < M B. Set R = A X1,..., Xn−1 and let M (R) denote the maximal ideal of R. h i We claim that any f ∈ B − M B is an associate of a polynomial q q−1 g(Xn) = Xn +aq−1Xn +···+ao, ai ∈ M (R). To prove this we first remark that by applying an A-automorphism of B given by Xi t(i) Xi + Xn , t = 1,..., n − 1, Xn Xn with t(i) suitably chosen, we may assume that the series f j are regular in Xn, (for details apply Zariski and P.Samuel : Commutative algebra p.147, Lemma 3 ′ to the product of the f j s). Now since the Weierstrass Preparation Theorem is valid for the ring of formal power series over a com- q q−1 plete local ring, it follows that f = u(Xn + aq−1Xn + ··· + ao), u invertible in B = R Xn , ai ∈ M (R) and q being the order of h i f mod M (R). Thus Y is generated by σ = Y ∩ R[Xn]. Now, since B is R[Xn] - flat it follows by Theorem 6.2, (2), that σ is divisorial. Now by Proposition 7.5 C(R[Xn]) → C(R(Xn)loc) is an isomorphism. Since σ 1 M (R) R[Xn], it follows that σ is principal. Hence Y is principal. (ii) Now we shall deal with the case in which A is not complete. The completion Bˆ of B is the ring Aˆ Y1,..., Yn . Let Y be a minimal h i 8. Examples of factorial rings 25

prime ideal of B, with Y 1 MB. Since Bˆ is B-ﬂat, the ideal Y Bˆ is divisorial. Furthermore, since Y 1 M B, all the components k of Y Bˆ are such that k 1 M Bˆ, and therefore principal by (i). Thus Y Bˆ is principal. Hence Y is principal by Theorem 6.6. 31

8 Examples of factorial rings

Theorem 8.1. Let A be a factorial ring. Let A X1,..., Xn be graded by assigning weights ωi to xi (ωi > 0). LetF(X1,..., Xn) be an irreducible isobaric polynomial. Let c be a positive integer prime to ω, the weight c of F. Set B = A X1,... Xn, Z /(Z − F(X1,..., Xn)) = A x1,..., xn, z , c z = F(x1,..., xn). Then B is factorial in the following two cases. (a) c ≡ 1( mod ω)

(b) Every ﬁnitely generated projective A-module is free.

c Proof. (a) Since B/zB ≈ A X1,..., Xn, Z]/(Z − F, Z) ≈ A X1,..., Xn / dωi ′ (F), it follows that z is prime in B. Now, set xi = z xi , where c c−1 ′ ′ c = 1 + dω. Then z = F(x1,..., xn) = z F(x1,..., xn), i.e. ′ ′ −1 ′ ′ ′ ′ −1 z = F(x1,..., xn) so that B z = A x1,..., xn, F(x1,..., xn) . ′ ′ Since x1,..., xn are algebraically independent over A, we see that −1 −1 B z is factorial. Now B = B z ∩ K x1,..., xn, z , where K is y ∈ ∈ the quotient ﬁeld of A j for, let r K x1,..., xn, z with y B. z Then since (zr) is a primary ideal not intersecting A, we have y ∈ r r Bz = B ∩ K x1,..., xn, z z . Hence B is a Krull ring and therefore factorial by Theorem 6.3.

(b) Since c is prime to ω, there exists a positive integer e such that ′ ce c e ≡ 1( mod ω). Now by (a) B = A x1,..., xn, u , with u = ′ e ′ F(x1,..., xn) is factorial. Further B = B[u], u = z and B is a free B-module with 1, u, u2,..., ue−1 as a basis. It follows that B is the intersection of B′ and of the quotient ﬁeld of B, and is therefore a 32 Krull ring. Now B can be graded by attaching a suitable weight to Z. Let U be a graded divisorial ideal. Since B′ is factorial, U B′ is 26 1. Krull rings and factorial rings

principal. As B′ is free over B, U is a projective B-module. Now by Nakayama’s lemma for graded rings it follows that U is free and therefore principal. The proof of (b) is complete.

Examples. (1) Let a, b, c be positive integers which are pairwise relatively prime. Let A be a factorial ring. Then the ring B = A x, y, z , with zc = xa + yb, is factorial.

(2) Let R denote the field of real numbers. Then the ring B = R x, y, z with z3 = x2 + y2 is factorial. Theorem 8.2 (Klein-Nagata). Let K be a field of characteristic , 2 and A = K x1,..., xn with F(x1,..., xn) = 0, where F is a non-degenerate quadratic form and n ≥ 5. Then A is factorial. Proof. Extending the ground field K to a suitable quadratic extension ′ K if necessary, the quadratic form F(X1,..., Xn) can be transformed ′ ′ ′ into X1 X2 − G(X3,..., Xn). Let A = K ⊗ A = K x1,..., xn , x1 x2 = K G(x3,..., xn). Since F is non-degenerate and n ≥ 5, G(X3,..., Xn) is ′ ′ 1 irreducible and therefore xl is a prime element in A . Now A = x1 ′ 1 K x1, x3,..., xn, . x1

Since x1, x3,..., xn are algebraically independent, it follows from Theo- rem 6.3 that A′ is factorial. Now as A′ is A-free, for any graded divisorial ideal U of A, U A′ is divisorial and hence principal. Therefore U is a projective ideal. Since a ﬁnitely generated graded projective module is 33 free over A, we conclude that U is principal. Thus A is factorial.

Remark 1. The above theorem is not true for n ≤ 4. For instance, A = K x1, x2, x3, x4 with x1 x2 = x3 x4 is evidently not factorial. Remark 2. We have proved that if A is a homogeneous coordinate ring over a field K such that K′ A is factorial for some ground field ex- NK tension K′ of K, then A is factorial. This is not true for affine coordinate rings (see the study of plane conics later in this section). 8. Examples of factorial rings 27

Remark 3. The above theorem is a particular case of theorems of Sev- eri, Lefshetz and Andreotti, which in turn are particular cases of the following general theorem proved by Grothendieck.

Theorem . (Grothendieck). Let R be a local domain which is a complete intersection such that RY is factorial for every prime ideal Y with height Y ≤ 3. Then R is factorial. (We recall that R is a complete intersection if R = A/U , where A is a regular local ring and U an ideal generated by an A-sequence). (For proof of the above theorem see Grothendieck: Seminaire de Geometric algebrique, exposéXI, IHES (Paris), 1961-62). Study of plane conics. Let C be a projective non singular curve over a ground field K. Let A be the homogeneous coordinate ring of C. The geometric divisors of C can be identified with elements of DH(A). Then FH(A) = G1(C) + Zh, where Gl(C) denotes set of divisors of C linearly equivalent to zero and h denotes a hyper plane section. Let now C be a conic in the projective plane P2. Since the genus of C is zero we have Gl(C) = Go(C) where Go(C) is the set of divisors of degree zero of C. 34 (i.e. its Jacobian variety is zero). Let d denote the homomorphism of DH(A) into Z given by d(U ) = degree of U , for U ∈ DH(A). Then −1 d (2Z) = Go(C) + Zh = Gl(C) + Zh = FH(A). Hence C(A) ≈ Imd/2Z. Thus A is factorial if and only if Imd = 2Z. Suppose C does not carry any K-rational points. Then A is factorial. For if not, C(A) ≈ Z/(2) and there exists a divisor U ∈ DH(A) with d(U ) = 1. By the Riemann-Roch Theorem, we have l(U ) ≥ d(U ) − g + 1 = 2, where l(U ) denotes the dimension of the vector space of functions f on C with ( f ) + U ≥ 0. Thus there exists a function f on C with ( f ) + U ≥ 0 and thus we obtain a positive divisor of degree 1, i.e. C carries a rational point: Contradiction. Conversely if C carries a rational point P, then P is a divisor of degree 1 and C(A) ≈ Z/(2) i.e. A is not factorial. Thus we have proved (a) The homogeneous coordinate ring A of C is factorial if and only if C does not have rational points over K. Let now C′ be a conic in the affine place over K. Let A′ be its coordinate ring. Let C be its projective closure in P2. Let I be the subgroup 28 1. Krull rings and factorial rings

of DH(A) generated by the divisors at inﬁnity (A is the homogeneous coordinate ring of C). Then C(A′) ≈ DH(A)/(FH(A) + I). Thus (i) if C has no rational points over K, then by (a)DH(A) = FH(A) and therefore C(A′) = 0, so that A′ is factorial;

(ii) if C has rational points over K at inﬁnity, then DH(A) = FH(A)+I and A′ is factorial;

35 (iii) if C has rational points, but not at inﬁnity, then I ⊂ FH(A) and ′ C(A ) ≈ C(A) ≈Z/(2) ; in this case A is not factorial. Examples. (i) C′ ≡ x2 + 2y2 + 1 = 0 over the rationals. Then A′ is factorial. However the coordinate ring of C′ over Q(i) is not factorial.

(ii) C′ ≡ x2 + y2 − 1 = 0. The coordinate ring of C′ over Q is not factorial. But the coordinate ring of C′ over Q(i) is factorial. The above examples show that unique factorization is preserved nei- ther by ground field extension nor by ground field restriction. Study of the real sphere. Let R denote the field of real numbers and C, the field of complex numbers. We shall consider the coordinate ring of the sphere X2 + Y2 + Z2 = 1 over R and C. Proposition 8.3. (a) The ring A = R x, y, z , x2 +y2 +z2 = 1 is factorial (b) The ring A = C x, y, z , x2 + y2 + z2 = 1 is not factorial. Proof. (a) We have A/(z − 1) ≈ R X, Y, Z (Z − 1, X2 + Y2 + Z2 − 1) R X, Y, Z (X3 + Y2, Z − 1) ≈R X, Y /(X2 + Y2). 1 1 Hence Z − 1 is prime in A. Set t = , so that z = 1 + . Now, z − 1 t 1 2 since x2 + y2 + z2 − 1 = 0, we have x2 + y2 + 1 + + − 1 = 0 i.e. t2 t 1 (tx)2 + (ty)2 = −2t − 1 i.e. t ∈ R[tx, ty]. Now A[t] = R[tx, ty, ] is t factorial. Hence by Theorem 6.3, A is factorial. 8. Examples of factorial rings 29

(b) Since (x+iy)(x−iy) = (z+1)(z−1), we conclude that A = C[x, y, z], x2 + y2 + z2 = 1 is not factorial.

Let K denote the ﬁeld of complex numbers or the ﬁeld of reals, and 36 A = K x, y, z, x2+y2+z2 = 1. Let M be the module M = Adx+Ady+Adz, with the relation xdx + ydy + zdz = 0.

Proposition 8.4. The A-module M is projective

(a) If K = R, then M is not free

(b) If K = C, then M is free.

Proof. Since the elements v1 = (0, z, −y), v2 = (−z, 0, x), v3 = (y, −x, 0) 3 of A satisfy the relation xv1 + yv2 + zv3 = 0, we have a homomorphism 3 u : M → A , given by u(dx) = v1, u(dy) = v2, u(dz) = v3. Let ν be the homomorphism ν : A3 → M given by ν(a, b, c) = a(ydz − zdy) + b(zdx − xdz) + c(xdy − ydx). It is easy to verify that νou is the identity on M. Hence M can be identiﬁed with a direct summand of A3. Hence M is projective. Now the linear form ϕ : A3 → A given by ϕ(a, b, c) = ax + by + cz is zero on M. But A3/M is a tossion-free module of rank 1. Hence M = ker ϕ. On the other hand we have ϕ(A3) = A, since x2 + y2 + z2 = 1. Hence M ⊕ A ≈ A3. Thus M is equivalent to a free module.

(a) If K = R, then M is not free. We remark that M is the A-module of sections of the dual bundle of the targent bundle to the sphere S 2. Since there are no non-degenerate continuous vector ﬁelds on S 2, the tangent bundle is not trivial, nor is its dual.

(b) If K = C, then M is free. For, the tangent bundle to the complex- iﬁcation of S 2 is trivial (this complexiﬁcation being the product of two complex projective lines).

Remark. R.Swan (Trans, Amer. Math. Soc. 105(1962), 264-277(1962) 37 5 2 has proved the following. The ring A = R x1, x2,..., x5 , xi = 1 is iP=0 30 1. Krull rings and factorial rings

factorial. Now S 7 can be fibred by S 3, the base being S 4. Let V be the bundle of tangent vectors along the fibres for this fibration and M, the corresponding module. Then M is not free, where as M C is free NR over A C. Further A C is factorial. Moreover M is not equivalent NR NR to a free-module.

Grassmann varieties. Let E be a vector space of dimension n over a field K. Let G = Gn,q be the set of all q dimensional subspaces of E(q ≤ n). Then set G can be provided with a structure of a projective variety as given below. q We call an element x ∈ ∧E a decomposed multi-vector if x is of the form x1∧· · ·∧xq, xi ∈ E. We have x1∧· · ·∧xq = 0 if and only if x1,..., xq ∗ are linearly dependent. Further x1 ∧···∧xq = λy1 ∧···∧yq, λ ∈ K if and only if x1,..., xq and y1,..., yq generate the same subspace. In the set of all decomposed multivectors we introduce the equivalence relation ∗ x1 ∧··· xq ∼ y1 ∧···∧ yq if x1 ∧···∧ xq = λy1 ··· yq for some λ ∈ K . Then the set Gn,q can be identified with the quotient set which is a subset q n of P(∧E) the (q) − 1 dimensional projective space defined by the vector q space ∧E. It can be shown that with this identification, Gn,q is a closed q subset of P(∧E) in the Zariski topology). The projective variety Gn,q is known as the Grassmann variety. As GL(n, K) acts transitively on Gn,q, it is non-singular. 38 Let L be a generic q-dimensional subspace of E with a basis x1,... xq, n say xi = λi je j, 1 ≤ i ≤ q, λi j ∈ K. Then jP=1

x ∧···∧ x = d (λ)e ∧···∧ e , 1 q X i1,...,iq i1 iq i1···iq

where di1,...,iq = det(λki j ). Let xi j, 1 ≤ i ≤ q, 1 ≤ j ≤ n be algebraically independent elements over K. Let B = K xi j 1≤i≤q the polynomial ring 1≤ j≤n in nq variables. For any subset H = i1,..., iq , i1 < i2 < ··· < iq of cardinality q, we denote by dH(x) the q by q determinant det(xki j ). It is 8. Examples of factorial rings 31

′ clear that A = K dH(x) H∈J where J is the set of all subsets of cardinality q of i,..., n , is the homogeneous coordinate ring of Gn,q. Proposition 8.5. The ring A is factorial.

Proof. It is known that the ring A is normal (See J. Igusa: On the arith- metic normality of Grassmann variety, Proc. Nat. Acad. Sci. U.S.A. Vol. 40, 309 - 313). Consider the element d = d{1,...,q}(x) ∈ A. We first prove that d is prime in A. Consider the subvariety S of Gn,q de- ′ ′′ fined by d = 0. Let E be the subspace generated by e1,..., eq and E the subspace generated by eq+1,..., en. (We recall that e1,..., en is a basis of E). Now α ∈ S if and only if dim(prE′ (α)) < q, i.e. if and ′′ only if α ∩ E , (0). Let Z = (0,..., 0, Zq+1,..., Zn), where the Zi are algebraically independent. Let x1,..., xq−1 be independent generic points of E, independent over k(z). Then Z ∧ x1 ∧···∧ xq−1 is a generic 39 point of S , and therefore S is irreducible. Let Y be the prime ideal defining S . Then A.d = Y (s) for some s. We now look at the zeros of A.d which are singular points. These zeros are given by the equations ∂ (d) = 0, 1 ≤ i ≤ q, 1 ≤ t ≤ n, or equivalently, by equating to 0 xit the sub-determinants of d of order q − 1. Hence α is a singular zero ′ of A.d if and only if prE(α) has codimension ≥ 2 i.e. if and only if dim(α ∩ E′′) ≥ 2. Hence A.d has at least one simple zero. That is, s = 1 and A.d is a prime.

The co-ordinate ring of the aﬃne open set U deﬁned by d , 0 is the ′ a ring A = a ∈ A, a homogeneous = A/ . We shall describe ddo(a)/q (1−d) ′ ′′ the ring A in another way. Let α ∈ Gn,q. Then α ∈ U ⇔ α ∩ E = (0). Let y1 = (1, 0,..., 0, y1q+1,... y1n),..., yq = (0,..., 1, yq,q+1,..., yqn), where the yi j are algebraically independent over K. Then y1 ∧···∧ yq is a generic point of U. Set y = (yi j)1≤i≤q where yi j = δi j, i ≤ q, 1≤ j≤n ′ j ≤ q. Then A = K dH(y) H∈J. But d1,...,i,...,q, j(y) = ±yi j, 1 ≤ i ≤ q, ′ ′ q + 1 ≤ j ≤ n. HenceA = K yi j 1≤i≤q, . Hence A is factorial. Hence, q+1≤ j≤n by Proposition 7.3, the ring A is factorial. 32 1. Krull rings and factorial rings

Remark 1. The ring A provides an example of a factorial ring which is not a complete intersection.

40 Remark 2. We do not know any example of a factorial ring which is not a Cohen-Macaulay ring.

Remark 3. We do not know any example of a factorial ring which is not a Gorenstein ring.

A local ring A is said to be a Gorenstein ring if A is Cohen-macaulay and every ideal generated by a system of parameters is irreducible.

9 Power series over factorial rings

Theorem 9.1. Let A be a noetherian domain containing elements x, y, z satisfying

(i) y is prime, Ax ∩ Ay = Axy;

(ii) zi−1 < Ax + Ay, zi ∈ Ax j + Ayk, where i, j, k are integers such i jk − i j − jk − ki ≥ 0.

Then A [T] is not factorial. We ﬁrst list here certain interesting corollaries of the above theorem.

Corollary 1. There exist factorial rings A (also local factorial ones) such that A [T] is not factorial. Let k be a ﬁeld and let A′ = k x, y, z with zi = x j+ yk, (i, j, k) = 1, i jk − i j − jk − ki ≥ 0 (for instance i = 2, j = 5, k = 7). Then by Theorem 8.1 the ring A′ is factorial, and so is ′ the local ring A = A(x,y,z). But x, y, z satisfy the hypothesis of the above theorem. Therefore A′ [T] and A [T] are not factorial. 41 Corollary 2. There exists a local factorial ring B such that its comple- ˆ ′ ′ tion B is not factorial. Set A = A(x,y,z),B = A[T](M ,T), where A is as in the proof of Corollary 1 and M is the maximal ideal of A. Then Bˆ is factorial. Now Bˆ = Aˆ [T] . Further Bˆ is also the completion of the local ring A [T] . Thus ifBˆ is factorial, so is A [T] by Mori’s Theorem 9. Power series over factorial rings 33

(see for instance, Sur les anneaux factorials, Bull. Soc. Math. France, 89, (1961), 155 - 173). Contradiction.

Corollary 3. There exists a local non-factorial ring B such that its associated graded ring G(B) is factorial.

7 5 2 4 3 We set A1 = k[u, v, x, y, z], z = u x + v y . We observe that z is prime 1 ′ ′ 1 3 ′ 2 ′ 5 ′2 in A1 and that A1 = k x , y , u, v, , x = z x , y = z y , z = u x + z z 4 3 1 v y . Hence A1 is factorial and therefore is A1. Take A = A1(u,v,x,y,z) , z B = A [T] . Since x, y, zA satisfy the hypothesis of the above theorem with i = 7, j = 2, k = 3, the ring B = A [T] is not factorial. But G(B) = G(A)[T] = A1[T] is factorial.

Remark. 1. If A is a regular factorial ring, then so is A [T] . (see Chap- ter 2, Theorem 2.1).

2. If A is a noetherian factorial ring such that AM [T] is factorial for every maximal ideal M of A, then A [T] is factorial. 3. Suppose that A is a factorial Macaulay ring such that AY [T] is factorial for all prime ideals Y with height Y = 2. Then A[T] is factorial (for proofs of (2) and (3), see P. Samuol, on uniquefactor- ization domains, Illinois J.Math. 5(1961) 1-17).

4. Open question. Let A be a complete local ring which is factorial. 42 Then is A [T] factorial? In Chapter 3 we shall see that at least in characteristic 2, the completion Aˆ of A of Corrolary 2 is not factorial. We shall also give examples to show that C(A) → C(A [T] ) is not surjective. Finally it may be of interest to note that J. Geiser has proved that there do not exist complete factorial rings satisfying the hypothesis of Theorem 9.1.

Proof of Theorem 9.1. Let S denote the multiplicatively closed set 1, x, x2,.... Set A′ = S −1A, B = A [T] . Then S −1B ⊂ A′ [T] ; in fact A′ [T] is the T-adic completion of S −1B. But, however,S −1B is 34 1. Krull rings and factorial rings

not a Zariski ring with the T-adic topology. Let S ′ denote the set of elements of B, whose constant coeﬃcients are in S . Then S ′−1B is a Zariski ring with the T-adic topology and its completion is A′ [T] . Consider the element v = xy − zi−1T ∈ B.

2 i−1 (a) No power series y+a1T +a2T +··· B is an associate of v = xy−z T in A′ T (nor, a fortiori in S ′−1B). 1 c Proof. If possible, suppose that (xy − zi−1T) ( + T + ··· ) ∈ B, with x xα 1 c cy zi−1 + +···∈ ′ − ∈ − i−1 α−2 ∈ α−1 α T A [T] . Then − A i.e cy z x Ax . x x xα 1 x Since zi−1 < Ax + Ay we have α ≥ 2. Further since Ax ∩ Ay = Axy we have c ∈ Axα−2, say c = c′ xα−2. Then zi−1 − c′y ∈ Ax. Contradiction yt b (b) There exists an integer t and an element v′ = + 1 T + ··· + x x2 b n T n+1 + ··· such that u = vv′ ∈ B, where v = xy − zi−1T. xn+1

Proof. Take t ≥ i j. We have to ﬁnd elements b1, b2,... bn,... of A such that bn bn−1 i−1 i−1 n xy − z ∈ A i.e. b y − b − z ∈ Ax xn+1 xn n n 1 t 43 for n ≥ 1. We set b0 = y . Assume that the bl for l ≤ ni j have been t(n) j k determined and that bni j = y Fn(x , y ), where t(n) ≥ i j and Fn(X, Y) is a form of degree ni. This is trivially veriﬁed for n = 0.

i−1 ni j+1 The congruence bni j+1y − bni jz ∈ Ax may be solved by taking t(n)−1 j k i−1 bni j+1 = y Fn(x , y )z . Similarly t(n)−r j k r(i−1) bni j+r = y Fn(x , y )z , 0 ≤ r < i j. Further the relation

i−1 (n+1)i j b(n+1)i jy − bni j+i j−1z ∈ Ax

implies that

t(n)−i j+1 j k i j(i−1) (n+1)i j b(n+1)i jy − y Fn(x , y )Z ∈ Ax . 9. Power series over factorial rings 35

But zi ∈ Ax j + Ayk, say zi = cx j + dyk. Now we have to solve the t(n)−i j+1 (n+1)i j j congruence, b(n+1)i j ≡ y G( mod Ax ) where G = (cx + k j(i−1) j k j(i−1) dy ) Fn(x , y ). The form G(X, Y) = (cX + dY) Fn(X, Y) is of degree ni + (i − 1) j. The monomials in G(x j, yk) are of the form x jαykβ, α + β = ni + (i − 1) j. By reading modulo Ax(n+1)i j we can ‘neglect’ the terms for which jα ≥ (n + 1)i j i.e. α ≥ (n + 1)i. For the remaining terms we have β > ni + (i − 1) j − (n + 1)i = i j − j − i. j k (i j− j−i)k j k (n+1)i j Thus G(x , y ) ≡ y Fn+1(x , y )( mod Ax ), where Fn+1 is a form of degree ni + (i − 1) j − (i j − j − i) = (n + 1)i. Now b(n+1)i jy ≡ t(n)+(i jk− jk−ki−i j)+1 j k (n+1)i j y Fn+1(x , y ) ( mod Ax ). We may solve this by t(n)+i jk− jk−ki−i j j k taking b(n+1)i j = y Fn+1(x , y ), i.e. we may take t(n + 1) = t(n) + i jk − jk − ki − i j and (b) is proved.

′ ′ Set u = vv , with v, v as in (b). Let u = u1,... us be the decomposition of u into prime factors in B; since the constant term of u is a power of y and since y is prime, the constant term of each ul is a power of y. Consider R = S ′−1B; R is factorial. Now v′ ∈ Rˆ and therefore uRRv = Rv. Further v is prime in R (since the constant term of v is y times an invertible element in S −1A). Now unique factorization in R implies that v is an associate of some u j in R. This contradicts (a).

Chapter 2

Regular rings

Let A be a noetherian local ring and M its maximal ideal. We say 45 that A is regular if M is generated by an A-sequence. We recall that x1,..., xr ∈ A is an A-sequence if, for i = 0,..., r − 1, xi+1 is not a zero divisor in A/(x1,..., xi). It can be proved that a regular local ring is a normal domain. Let A be a noetherian domain. We say that A is regular if AM is regular for every maximal ideal M of A.

1 Regular local rings

Let A be a noetherian local ring and M its maximal ideal. Let E be a ﬁnitely generated module over A. Let x1,..., xn ∈ E be such that the elements xi mod M E form a basis for E/M E; then the xi generate E (by Nakayama’s lemma). Such a system of generators is called a minimal system of generators. Let x1,..., xn be a minimal system of n generators of E. Let F = Aei be a free module of rank n. Then the iP=1 ϕ sequence 0 → E1 → Fo −→ E → 0 is exact where ϕ(ei) = xi. Now E1 is ﬁnitely generated. Choosing a minimal set of generators for E1, we can express E1 as a quotient of a free module F2. Continuing in this fashion we get an exact sequence of modules.

···→ Fn → Fn−1 →···→ Fo → E → 0;

37 38 2. Regular rings

we call this a minimal resolution of E. We say that the homological dimension of E (notation hd E) is < n if Fn+1 = 0 in a minimal resolution of E. If Fi , 0 for every i, then we put hdE = ∞. It can be proved that 46 hd E does not depend upon the minimal resolution (since two minimal systems of generators of E diﬀer by an automorphism of E). We recall that E is free ⇔ the canonical mapping M ⊗ E → E is injective ⇔ Tor A 1 (E, A/M ) = 0. From this it easily follows that for a ﬁnitely generated A-module E A we have hdE < n if and only if Tor n+1(E, A/M ) = 0.

Theorem 1.1 (Syzygies). Let A be a regular local ring and d the number of elements in an A-sequence generating the maximal ideal M of A. Then for any ﬁnitely generated module E. We have hdE ≤ d.

We state a lemma which is not diﬃcult to prove.

Lemma 1.2. Let A be a noetherian local ring and G a ﬁnitely generated A-module with hdG < ∞. Let a be a non-zero divisor for G. Then G hd = hd G + 1. aG

Now if M = (x1,..., xd), where x1,..., xd is an A-sequence, then by means of an immediate induction and a use of the above lemma we A ′ get hd(A/M ) = d. Hence Tord+1 (E, A/M ) = 0 for any module E . Hence hdE ≤ d.

Theorem 1.3 (Serre). Let A be a local ring with maximal ideal M such that hdM < ∞. Then A is regular.

We ﬁrst observe that the hypothesis of the theorem implies that for an A-module E, we have hdE ≤ hd(A/M ) = hd m + 1. We prove the theorem by induction on the dimension d of the A/m vector space M /M 2 . If d = 0, then M = 0; A is a ﬁeld and therefore 47 regular. Suppose d > 0. Then we claim that under the hypothesis of the theorem there exist an element b ∈ M −M 2 which is not a zero divisor. This follows from the following lemma. 1. Regular local rings 39

Lemma 1.4. Let A be local ring. Suppose that every x ∈ M − M 2 is a zero divisor. Then any finitely generated module of finite homological dimension is free. Proof. Let G be a module with hdG < ∞. If hdG > 0, we find, by resolving G, an A-module E such that hdE = 1. Let 0 → F1 → Fo → E → 0 be a minimal resolution of E. Then F1 is free and F1 ⊂ M Fo. Now, since every element of M − M 2 is a zero divisor, it follows that every element of M is a zero divisor and M is associated to (0). Hence there exists an a , 0, such that aM = 0. Hence a F1 = 0. This contradicts the fact that F1 is free. Hence hdG = 0, and G is free. Applying this lemma to ‘our’ A we see that if every element of M − M 2 is a zero divisor, then M is free. Since M consists of zero divisors we have M = 0 i.e. A is a field. Thus if d > 0 there is a b ∈ M − M 2 ′ ′ such that b is not a zero divisor. Set A = A/Ab, M = M /Ab. We claim that M /Ab is a direct summand of M /M b. Let ψ denote the M M canonical surjection /M b → /Ab . Let b, q1,..., qd−1 be a minimal d−1 set of generators of M . Set σ = Aqi. Let ϕ be the canonical mapping iP=1 b σ → M /M b . Then Ker(ϕ) = σ ∩ M ⊂ σ ∩ Ab. On the other hand if d−1 λb ∈ σ, then λb = λiqi, λi ∈ A. But b, q1,..., qd is a minimal set of iP=1 generators of M . Hence λ, λi ∈ M . Thus σ ∩ M b = σ ∩ Ab. σ + Ab θ M Thus we have a canonical injection M / = −→ , since 48 Ab Ab M b σ + Ab/Ab ≈ σ/σ∩Ab = σ/σ∩M b. It is easy to see that ψoθ = IM . Hence M /Ab is a direct summand of M /M b . We now have the following lemma easily proved by induction on hd(E)). Lemma 1.5. Let A be a commutative ring and E an A-module with hdE < ∞. Let b ∈ A be a non-zero divisor for A and E. Then hdA/AbE/bE < ∞.

From the above lemma, it follows that hdA/bAM /M b < ∞. Since M /Ab is a direct summand of M /M b we have hdA/Ab M /Ab < ∞. M M Since dimA/M /Ab M 2Ab/Ab = dimA/M /M 2+Ab = d − 1, A/Ab is regular by induction hypothesis. Hence M /Ab is generated by an 40 2. Regular rings

A/Ab-sequence, say x1,... xd−1 modulo Ab. Then b, x1,..., xd−1 is an A-sequence and generates M . Thus A is regular. For any local ring A we deﬁne the global dimension of A to be hdA/M , where M is the maximal ideal; notation : gl dim A = δ(A). For any A-module E we have hdAE ≤ δ(A). Corollary. Let A be a regular local ring and p a prime ideal with p , M , where M is the maximal ideal of A. Then Ap is regular and δ(Ap) < δ(A).

Proof. We have δ(Ap) = hdAp Ap/pAp = hdAp A/p ⊗ Ap ≤ hdAA/p, the last inequality, being a consequence of the fact that Ap is A-flat. M 49 Choose x ∈ − p. Since x is not a zero divisor for A/p we have by p + x Lemma 1.2, hd A/ / (A/ ) = hd = 1 + hd A/ ≤ δ(A) i.e. A p x p A p A p hdAA/p ≤ δ(A) − 1. Hence δ(Ap) < δ(A). Theorem 1.6 (Auslander-Buchsbaum). Any regular local ring is factorial. Proof. Let A be a regular local ring. We prove the theorem by induction on the global dimension δ(A) of A. If δ(A) = 0, then A is a field and therefore factorial. Suppose δ(A) > 0. Let x be an element of an A-sequence generating M . Then x is a prime element. By Nagata’s 1 theorem, we have only to prove that B = A is factorial. For any max- x imal ideal M of B, we have BM = Ap, where p is a prime ideal with x < p. By the corollary to Theorem 1.3, we see that Ap is regular and δ(Ap) < δ(A). Hence by the induction hypothesis, Ap is factorial. Thus BM is factorial for every maximal ideal M of B (i.e. B is locally factorial). Let σ be a prime ideal of height 1 in B. Then σ is locally principal i.e. σ is a projective ideal. Now B being a ring of quotients of the regular local ring A, the ideal admits a finite free resolution. By making an induction on the length of the free resolution of σ we conclude that there exist free modules F, L such that σ ⊕ L ≈ F. By comparing the ranks we see that L ≈ Bn, F ≈ Bn+1 for some n. Taking the (n + 1)th n j n+1− j exterior power we have ∧(σ) ⊗ ∧ (L) ≈ B. Since σ is a modulo Lj=1 2. Regular factorial rings 41

j of rank 1 ∧ (σ), j ≥ 2 is a torsion-module. Hence σ ≈ B that is, σ is principal and therefore B is factorial. Hence A is factorial.

2 Regular factorial rings

We recall that a regular ring A is a noetherian domain such that AM is 50 regular for any maximal ideal M of A. We say that domain A is locally factorial if AM is factorial for every maximal ideal M of A.

Theorem 2.1. If A is a regular factorial ring then the rings A[X] and A [X] are regular factorial rings. Proof. We ﬁrst prove that A[X] and A [X] are regular. Let B = A[X]. Let M be a maximal ideal of B. Set p = A ∩ M . Then BM =

(Ap[X])M Ap[X]. Since a localisation of a regular ring is again regular, we see that Ap is regular. Thus to prove that B is regular we may assume that A is a local ring with maximal ideal m(A) and that M ∩ A = m(A). Since A is regular, m(A) is generated by an A-sequence, say a1,..., ar. Now B/m(A)B ≈ A/m(A)[X]. Thus M /m(A)B = (F¯(X)), where F(X) ∈ M is such that the class F¯(X) of F(X) ( mod m(A)) is irreducible in A [X]. Now a ,..., a , F(X) is BM -sequence and generates M /BM m(A) 1 r (in fact M = (a1,..., arF(X))). Hence BM is regular for every maximal ideal M i.e. B is regular. We shall now prove that C = A [X] is regular. Let M be a maximal ideal of C. Since X ∈ Rad(C), M =M + XC, where M is a maximal ideal of A. Now AM ⊂ CM. Since AM is regular, M AM is generated by an AM - sequence, say m1,..., md. Then m1,..., md, X is a CM-sequence which generates MCM. Thus CM is regular i.e. C is regular.

We now prove that B = A[X], C = A [X] are factorial. That B 51 is factorial has already been proved (see Chapter I, Theorem 6.5). To prove that C is factorial, we note that K is in the radical Rad (C) of C C and ≈ A is factorial. Now the following lemma completes the proof XC of the theorem. 42 2. Regular rings

Lemma 2.2. Let B be a locally factorial noetherian ring (for instance a regular domain). Let x ∈ Rad (B). Assume that B/xB is factorial. Then B is factorial.

Proof. Let σ be a prime ideal of height 1 in B. Then σ is locally principal i.e. σ is projective. If x ∈ σ, then σ = Bx. If x < σ, then σ ∩ Bx = σx. Thus σ/σx = σ/(σ∩Bx) ≈ (σ + Bx)/Bx i.e. σ/σx is a projective ideal in B/Bx. Since B/Bx is factorial and σ/σx divisorial, we see that σ/σx is principal in B/Bx. Hence, by Nakayama’s lemma, σ is principal.

Corollary. Let A be a principal ideal domain. Then A [X1,..., Xn] is factorial.

In particular if K is a ﬁeld, then K [X1,..., Xn] is factorial.

3 The ring of restricted power series

Let A be a commutative ring and let M be an ideal of A. We provide α A with the M - adic topology. Let f = aαX ∈ A [X1,..., Xd] , = α = α1 ··· αd P α (α1,...,αd), X X1 Xd . We say that f is a restricted power series if aα → 0 as |α|→∞, |α| = α1 + ··· + αd. It is clear that the set of all restricted power series is a subring of A [X1,..., Xd] which 52 we denote by A x1,..., Xd ; we have the inclusions A X1,..., Xd ⊂ A{X1,... Xd}A [X1,..., Xd] . In fact A X1,..., Xd is the M (X1,..., Xd)- adic completion of AX1,..., Xd . In particular, if A is noetherian so is A X1,..., Xd . Further A [X1,..., Xd] is the completion of A X1,..., Xd for the (X1,..., Xd)-adic topology. But this is not of interest, since A X1,..., Xd is not a Zariski ring with respect to the (X1,..., Xd)-adic topology.

Lemma 3.1. Let A be a commutative ring and M an ideal of A with M ⊂ Rad (A). Let A X1,..., Xd denote the ring of restricted power series, A being provided with the M -adic topology. Then M ⊂ Rad (A{X1,..., Xd}). 3. The ring of restricted power series 43

Proof. Let m ∈ M . Consider 1 + ms(X), where s(X) ∈ A X1,..., Xd . Set s(X) = ao + t(X), t(X) being without constant term. Since 1 + mao 1 is invertible, we have 1 + ms(X) = (1 − mu(X)), where u(X) is a 1 + mao 1 restricted series without constant term. Now, = 1 + mu(X) + 1 − mu(X) m2u(X)2 + ··· is clearly a restricted power series. Thus 1 + ms(X) is invertible in A X1,..., Xd i.e. m ∈ Rad (A X1,..., Xd ). Theorem 3.2. [P. Salmon]Let A be a regular local ring and let M denote its maximal ideal. Then R = A{X1,..., Xd} is regular and factorial, the power series being restricted with respected to the maximal ideal.

Proof. Let p be a maximal ideal of R. By Lemma 3.1, M R ⊂ Rad

(R) ⊂ p. Now p/M R is a maximal ideal of R/M R = (A/M ) X1,..., Xd . It is well known that any maximal ideal of (A/M ) X1,..., Xd is generated by d elements which form an (A/M ) X1,...,Xd -sequence. Thus 53 p/M R is generated by an R/M R-sequence. But, A being regular, M R is generated by an R-sequence. Therefore p is generated by an R-sequence. By passing to the localisation, we see that pRp is generated by a Rp- sequence. Hence R is regular.

We now prove that R is factorial. The proof is by induction on gl. dim A = δ(A). If δ(A) = 0, then A is a ﬁeld and R = A X1,..., Xd hence R is factorial. Let δ = δ(A) > 0 and m1,..., mδ generate M . Now R is regular and therefore locally factorial. By Lemma 3.1 m1 ∈ Rad

(R). Further R/m1R ≈ (A/mA) X1,..., Xd , δ(A/m1A) = δ − 1; hence, by induction hypothesis, R/m1R is factorial. Using Lemma 2.2 we see that R is factorial.

Remark 1. Let A be a local ring which is factorial. Then it does not 2 3 7 imply that A{T} is factorial. Take A = k x, y, z (x,y,z), z = x + y . As in the proof of Theorem 9.1, Chapter 1, we can prove that there exist y b b b b , b ,... ∈ A such that (xy−zT) ( + 1 T+ 2 T 2+···+ n T n+1+··· ) = 1 2 x x2 x3 xn+1 u ∈ B = A{T}. In fact it can be checked that we can take the elements bi 2 2 8 2 14 n 2+6n such that u = y −xT −xyT −3xy T ··· , −αn xy T ..., where αn is 44 2. Regular rings

3n an integer such that 0 ≤ αn ≤ 2 . By providing A with the (x, y, z)-adic topology, we see that the power series u is restricted. Now the proof of Theorem 9.1 verbatum carries over and we conclude that the restricted power series ring A{T} is not factorial.

Remark 2. In the above example, if we take k = R or C, the real number 54 ﬁeld or the complex number ﬁeld respectively, then we can speak of the convergent power series ring over A. Now the above power series u is 3n convergent since 0 ≤ αn ≤ 2 . Hence the convergent power series ring over A is also not factorial. Chapter 3

Descent methods

1 Galoisian descent 55 Let A be a Krull ring and let K be its quotient field. Let G be a finite group of automorphisms of A. Let A′ denote the ring of invariants of A with respect to G and let K′ be the quotient field of A′. Then A′ = A∩K′, so that A′ is a Krull ring. Since (x − s(x)) = 0, x ∈ A, we see that sQ∈G A is integral over A′. Thus we have the homomorphism j : D(A′) → D(A) and ¯j : C(A′) → C(A) (see Chapter 1 §6). We are interested in −1 computing Ker (¯j). Let D1 = j (F(A)). Then Ker (¯j) = D1/F(A). Let S be a system of generators of G. Let d ∈ D1, with j(d) = (a), a ∈ K. The divisor j(d) is invariant under G, i.e. (s(a)) = (a), s ∈ G. Hence s(a)/a ∈ U, the group of units of A. Let h denote the homomorphism ∗ ∗ S ∗ S h : K → (K ) given by x (s(x)/x)s∈S . Then h(a) ∈ h(K ) ∩ U , Now if a = a′u, a′ ∈ K, u ∈ U, then s(a)/a = s(a′)/a′ · s(u)/u′. Thus h(a) is determined uniquely modulo h(U), and we therefore have a ho- ∗ S momorphism ϕ : D1 → (h(K ) ∩ U )h(U) with d h(a) ( mod h(U)), where a(d) = (a), a ∈ K.

Theorem 1.1. The mapping ϕ induces a monomorphism θ : Ker (¯j) → (h(K∗) ∩ US ) . Furthermore, if no prime divisor of A is ramiﬁed over h(U) A′, then θ is an isomorphism.

45 46 3. Descent methods

56 Proof. Let d ∈ D1. Then ϕ(d) = 0 ⇔ h(a) = h(u), u ∈ U ⇔ s(a)/a = s(u)/u, for all s ∈ S . a a a ⇔ s( ) = for all s ∈ G ⇔ = a′ ∈ K′ u u u a ′ ′ ⇔ j(d) = (a) = ( ) = (a ) = j((a ) ′ ). A u A A A

′ ′ But, since j is injective, we have d = (a )A′ i.e. Ker(ϕ) = F(A ). Hence θ is a monomorphism. Now assume that no prime divisor of A is ramiﬁed over A′. Let α ∈ (h(K∗) ∩ US ) , α = h(a) ( mod h(U)). Since h(K∗) = h(A∗), .h(U) we may assume that a ∈ A. Since s(a)/a ∈ U, for s ∈ S , the divisor (a) is invariant under G. Now, by hypothesis for any prime divisor ′ ′ ′ Y ∈ D(A ), we have j(Y ) = Y1 + ··· + Yg, where the Yi form ′ a complete set of prime divisor lying over Y . Further the Yi are conjugate to each other. Since the divisor (a) is invariant under G, the prime divisors which are conjugate to each other occur with the same coeﬃcient in (a) so that (a) is the sum of divisors of form j(Y ′), Y ′ ∈ P(A′). Hence θ is surjective and therefore an isomorphism.

Remark 1. For S = G the group (h(K∗)∩(U)G) h(U) is the cohomology s(x) group H1(G, U): in fact a system ( ) for x ∈ K∗ is the most gen- x s∈G ∗ 1 ∗ eral cocycle of G in K (since H (G1K ) = 0, as is well known), whence h(K∗)∩(U)G = Z1(G, U); on the other hand h(U) is obviously the group B1(G, U) of coboundaries. The preceding theorem may also be proved by the following cohomological argument. As usual, if G operates on a 57 set E, we denote by EG the set of invariant elements of E; we recall that EG = Ho(G, E). Now, since H1(G, K∗) = 0, the exact sequence

0 → U → K∗ → F(A) → 0 gives the exact cohomology sequence

0 → UG → (K∗)G → F(A)G → H1(G, U) → 0. 1. Galoisian descent 47

On the other hand, since UG is the group of units in A′ = AG, we have

0 → UG → (K∗)G → F(A′) → 0 and therefore,

0 → F(A′) → F(A)G → H1(G, U) → 0

In other words, H1(G, U) = (invariant principal divisors of A) / (divisors of A induced by principal divisors of A′). This gives immediately a monomorphism θ : ker(¯j) → H1(G, U). If A is divisorially unramiﬁed over A′, one sees, as in the theorem, that every invariant divisor of A comes from A′, thus θ is surjective in this case.

Remark 2. Suppose G is a ﬁnite cyclic group generated by an element say s. Then we may take S = {s}. By Hilbert’s Theorem 90, the group h(K∗) is precisely the group of elements of norm 1. Thus (h(K∗) ∩ U)/h(U) is the group of units of norm 1 modulo h(U).

Remark 3. The hypothesis of ramiﬁcation is essential in the above theorem. For instance let A = Z[i], i2 = −1, G = {1,σ}, σ(i) = −i. Then A′ = Z, C(A′) = C(A) = 0. Hence Ker(¯j) = 0. However, U ∩ h(K∗) = ∗ {1, −1, i, −i}, h(U) = {1, −1}. Thus (h(K ) ∩ U) h(U) ≈ Z/(2). We note that the prime number 2 is ramiﬁed in A. 58 Examples: Polynomial rings.

1. Let k be a fied and A = k x1,..., xd , the ring of polynomials in d variables, d ≥ 2. Let n be an integer with (n, p) = 1, p being the characteristic of k and let k contain a primitive nth root of unity w. Consider the automorphism s : A → A with xi wxi, 1 ≤ i ≤ d and let G be the cyclic group of order n generated by s. Then the ring ′ of invariants A is generated by the monomials of degree n in the xi; geometrically this is the n-tuple model of the projective space. Set n n Fi(X) = X − xi . Now any ramified prime divisor of A must contain ′ n−1 Fi (xi) = nxi . Thus there is no divisorial ramification in A. Here U = k∗ and the group of units of norm 1 is the group of nth roots of 48 3. Descent methods

Z unity. Further h(U) = {1}, (h(K∗) ∩ U) h(U) ≈ , by Remark 2. (n) Since A is factorial, by Theorem 3.1, we have C(A′) ≈ Z/(n).

2. Let k, w, n, be as in (1) and A = k[x, y]. Let s be the k-automorphism of A defined by x wx, y w−1y. The ring of G-invariants A′ = k xn, yn, xy , i.e. A′ is the affince coordinate ring of the surface Zn = XY . Again as in (1) there is no divisorial ramification, U = k∗ and C(A′) ≈ Z/(n).

3. Let k be a ﬁeld and A = k X1,... Xn . Let An denote the alternating group. Now An acts on A. If the characteristic of k is , 2, then the ′ ring of An-invariants is A = k s1,..., sn, △ where s1 ··· sn denote 59 the elementary symmetric functions and △ = (xi − x j). If char- 1Q< j) ′ acteristic k = 2, then △ is also symmetric and A = k s1,..., sn,α , 1 where α = ( (xi − x j) + (xi + x j)). 2 iQ< j iQ< j

As the coefficients of (xi − x j) + (xi + x j) are divisible by 2, the iQ< j iQ< j element α has a meaning in characteristic 2. Further there is no divisorial ramification in A over A′. For the only divisorial ramifications of A over ′ k s1,..., sn are those prime divisors which contain F (xi) = (x j − xi), Qj,i ′ where F(X) = (X − x j). Since △ = (xi − x j) ∈ A (in characteristic Q iQ< j 2, △ is in fact in k s1,..., sn ), there is no divisorial ramification in A ′ ′ 1 over A . Hence C(A ) ≈ H (An, U), by the remark following Theorem ∗ ∗ ′ 1 1.1. But U = k and An acts trivially on k . Hence C(A ) ≈ H (An, U) is ∗ the group of homomorphisms of An into k . Thus if n ≥ 5, An is simple and therefore C(A′) = 0 i.e. A′ is factorial. The only non-trivial cases we have to consider are, n = 3, 4. For n = 3, An is the cyclic group of order 3. Hence C(A′) = 0 if k does not contain cube roots of unity, otherwise C(A′) ≈ Z/(3). We now consider the case n = 4. We have [A4, A4] = 1, (2 2)(3 4), (1 4)(2 3), (1 3)(2 4) and A4 A4, A4 ≈ Z/(3). ∗ Now the group of homomorphisms of A4 into k is isomorphic to the ∗ group of homomorphisms of A4 A4, A4 into k . Hence, is in the case n = 3, C(A′) = 0 if k does not contain cube roots of unity; otherwise 1. Galoisian descent 49

′ C(A )≈Z/(2).

Example. Power series rings. We ﬁrst prove the following lemma.

Lemma 1.2. Let A be a local domain, M its maximal ideal. Let s be 60 an automorphism of A of order n, (n, char (A/M )) = 1. Let U denote the group of units of A and h the mapping K∗ → K∗ with x s(x)/x, K being the quotient ﬁeld of A. Then (1 + M ) ∩ h(K∗) ⊂ h(U) (i.e. im(H′(G, I + M ) → H′(G, U)) = 0, where G is the group generated by s).

∗ Proof. Let u ∈ 1 + M , u = s(x)/x, x ∈ K . Then u has norm 1, i.e. − − N(u) = u1+s+···+sn 1 = 1. Set v = 1 + u + u1+s + ··· + u1+···+sn 2 . Then v ≡ n.1( mod M ). Since n is prime to the characteristic of A/M , it follows − that v is a unit. Further we have s(v) = 1 + us + us+s2 + ··· us+1···+sn 1 and us(v) = v i.e. u = s(v−1) v−1 ∈ h(U) and the lemma is proved. In the examples (1) and (2) of polynomial rings we replace the rings A = K x1,..., xd and A = K[x, y] respectively by A = K [x1,..., xd] and A = K [x, y] . Since in A we have U (1 + M ) ≈ k∗, we obtain the same results as in the case of ring of polynomials, in view of the above lemma.

Proposition 1.3. Let A be a local ring, M its maximal ideal. Let G be a ﬁnite group of automorphisms of A, acting trivially on k = A/M . Further, assume that there are no non-trivial homomorphisms of G into k∗ and that (Card(G), Char (k)) = 1. Then H1 (G, U) = 0, U being the group of units of A. In particular, if A is factorial, so is A′.

Proof. Let (us)s∈G, us ∈ U, be a 1-cocycle of G with values in U. Then uss′ = s(us′ ).us. Reducing modulo M , we getu ¯ ss′ = u¯ s′ ·u¯ s, since G acts trivially on k. We have made the hypothesis that there are no non-trivial ∗ homomorphisms of G into k . Hence us ∈ 1 + M , s ∈ G. Set y = ut. 61 tP∈G Then y = Card (G) · 1 ( mod M ). Thus y ∈ U. Now s(y) = s(ut) = tP∈G ust 1 −1 −1 1 = y i.e. us = s(y ) y . Hence H (G, U) = 0. Pt us us 50 3. Descent methods

Corollary . Let A = k [x1,..., xn] . Let k be of characteristic p > n or 0. Let An be the alternating group on n symbols. Then for n ≥ 5, H1(G, U) = 0, i.e. the ring of invariants A′ is factorial.

For further information about the invariants of the alternating group we refer to Appendix 1.

2 The Purely inseparable case

Let A be a Krull ring of characteristic p , 0, and let K be its quotient field. Let △ be a derivation of K such that △(A) ⊂ A. Set K′ = Ker(△) and A′ = A∩K′. Then A′ is again a Krull ring and Ap ⊂ A′, K p ⊂ K′. In particular, A is integral over A′. Hence the mapping j : D(A′) → D(A) of the group of divisors goes down to a mapping ¯j : C(A′) → C(A) of the corresponding divisor class groups. We are interested in computing −1 ′ Ker(¯j). Set D1 = j (F(A)), so that Ker(¯j) = D1/F(A ). Let d ∈ D1, ∗ and j(d) = (a), a ∈ K . From the definition of j it follows that ep divides vp(a), where p is a prime divisor of A, vp the corresponding valuation and ep the ramification index of vp. Hence there exists an ′ ′∗ ′ ′ a ∈ K such that vp(a) = vp(a ), i.e. a = a .u, u being a unit in Ap. △u Thus △a/a = △a′/a′ + △u/u = . Since △(A ) ⊂ A , it follows that u p p 62 △a/a ∈ Ap, for all prime divisors p of A, i.e. △a/a ∈ A. We shall call ax ∈ K, a logarithmic derivative if x = △t/t for some t ∈ K∗. The set of all logarithmic derivative is an additive subgroup of K. Set L = △t/t △t/t ∈ A, t ∈ K∗ . Let U denote, as before, the group of units ′ ′ of A and set L = △u/u u ∈ U . Now L ⊂ L . For ad ∈ Di with ∗ L L ′ j(d) = (a), a ∈ K , △a/a ∈ is uniquely determined modulo . Let ′ ′ ϕ denote the homomorphism : D1 → L /L , d △a/a( mod L ) if j(d) = (a). Now ϕ(d) = 0 ⇔ △a/a = △u/u, for u ∈ U ⇔ △(a/u) = 0 ′ ′ ′ ′ ′ i.e. a/u = a ∈ K ⇔ (a)A = (a )A. But j((a )A′ ) = (a )A′ and j is ′ ′ injective. Hence d = (a )A′ . Thus Ker(ϕ) = F(A ). We have proved the first assertion of the following theorem.

Theorem 2.1. (a) We have a canonical monomorphism ϕ : Ker(¯j) → L L ′. 3. Formulae concerning derivations 51

(b) If [K : K′] = p and if △(A) is not contained in any prime ideal of height 1 of A, then ϕ¯ is an isomorphism.

Proof. To complete the proof of the theorem, we have only to show thatϕ ¯ is surjective under the hypothesis of (b). As K/K′ is a purely inseparable extension, every prime divisor of A′ uniquely extends to a prime divisor of A. Thus a divisor d = np p ∈ Im( j) if and only p∈PP(A) p ′ if ep np where ep denotes the ramification index of p. Since A ⊂ A , ∗ it follows that for any prime division p of A, ep = 1 or p. Let a ∈ K be such that △a/a ∈ A. It is sufficient to prove that for a prime divisor p of A, if n = vp(a) is not a multiple of p, then ep = 1. Let t be a n uniformising parameter of vp. Let a = ut , u being a unit in Ap. Then △u/u + n △t/t = △a/a ∈ Ap. Hence, △t/t ∈ Ap, i.e. △ induces a 63 derivation △¯ on the residue class field k = Ap tAp. By hypothesis, since △A 1 p, we have △¯ , 0. Let k′ be the residue class field of p ∩ A′. Then k′ ⊆ Ker(△¯ ) ⊂ k. Thus f = [k : k′] , 1. Since [K : K′] = p, and , p ′ ′ k ⊂ k , we have f = p. Now the inequality ep f ≤ [K : K ] = p gives ep = 1. The proof of Theorem 2.1 is complete.

3 Formulae concerning derivations

Let K be a ﬁeld of characteristic p , 0 and let D : K → K be a derivation. Let tD denote the derivation x t. D(x). We note that Dp, the pth iterate of D, is again a derivation.

Proposition 3.1. Let D : K → K be a derivation of K(char K = p , 0). Assume that [K : K′] = p. Then

(a) Dp = aD, a ∈ K′ = Ker D.

(b) If A is a Krull ring with quotient ﬁeld K such that D(A) ⊂ A, that a ∈ A′ = K′ ∩ A.

Proof. (a) By hypothesis K = K′(z), zp ∈ K′. For any K′- derivation △z △ of K, △ = D. In particular Dp = aD for a ∈ K. Hence Dz 52 3. Descent methods

p+1 p 2 2 Da = D(D a) = D(a · Da) = (Da) + aD a. On the other hand, Dap+1 = Dp(Da) = aD2a. Hence Da = 0 i.e. a ∈ K′.

(b) Since A = Ap and D(Ap) ⊂ Ap, we have only to deal with p∈TP(A) the case when A is a discrete valuation ring. Let v denote the corresponding valuation. Let t ∈ A, with v(t) = 1. We have Dpt = a · Dt. 1 If v(Dt) = 0, then a = , Dpt ∈ A. Assume v(Dt) > 0. Then for Dt x ∈ A, we have v(Dx) ≥ v(x). For is x = utn with v(u) = 0. In particular, v(Dpt) ≥ v(Dt) i.e. a ∈ A.

64 Proposition 3.2. Let D : K → K be a derivation of, K(char K = p , 0). Let K′ = Ker D and [K : K′] = p. An element t ∈ K is a logarthmic derivative (i.e. there exists an x ∈ K such that t = Dx/x) if and only if

Dp−1(t) − at + tp = 0,

where Dp = aD.

Proof. We state ﬁrst the following formula of Hochschild (Trans. A.M.S. 79(1955), 477-489).

Let K be a ﬁeld of characteristic p , 0 and D a derivation of K. Then

(tD)p = tpDp + (tD)p−1(t).D

(t ∈ K, tD denotes the derivation x t.Dx). We have to prove the proposition only in the case when t , 0. 1 Let now t be a logarthmic derivative, say t = Dx/x. Set △ = D. t Then by Hochschild’s formula, we have,

Dp = (t△)p = tp△p + (t△)p−1(t)△. = tp△p + Dp−1(t) ·△ = aD.

But △n x = x, for n ≥ 1. Hence a.Dx = tp x + Dp−1(t)x, i.e.

tp − at + Dtp−1(t) = 0. 4. Examples: Polynomial rings 53

Conversely assume that t is such that Dp−1t − at + tp = 0. Set △ = 1 .D. Again by Hoschschild’s formula, we have Dp = (t△)p = tp△p + t Dp−1(t).△ = aD = a.t△, i.e. a.t△ = tp△p + (at − tp)△ i.e. tp(△p −△) = 0, i.e. △p −△ = 0, i.e. (△− (p − 1)I) ··· (△− 2I)(△− I) = 0, where I is the identity mapping of K into K. Choose y ∈ K with y1 = △y , 0 and set 65 y2 = (△− I)y1,..., yp = (△− (p − 1)I)yp−1(= 0). Then, there exists a j such that y j−1 , 0 and y j = (△− jI)y j−1 = 0. Hence ∗ △y j−1 = jy j−1, i.e. △y j−1 y j−1 = j ∈ Fp, Fp being the prime ﬁeld of characteristic p. Let n be the inverse of j modulo n △x y j−1 p. Set x = y j−1. Then = n = n j = 1, i.e. △x = x i.e. t = Dx/x. x y j−1

4 Examples: Polynomial rings

Let k be a factorial ring of characteristic p , 0. Set A = k[x, y]. Let D be a k-derivation of A and A′ = Ker(D). The group of units U of A is the group of units of k. Hence here L ′ = 0. Since A is factorial, we by Theorem 2.1, an injection of C(A′) = Ker(¯j) into L . (We recall that L is the group of logarthmic derivatives contained in A and that L ′ is the group of logarthmic derivatives of units.) We shall now consider certain special k-derivations of A.

(a) The Surface Z p = XY. Consider the derivation D of A k[x, y] with Dx = x and Dy = −y. Then k[xp, yp, xy] ⊂ A′ = Ker(D). Let L, K, K′ denote the quotient fields of k, A, A′ respectively. Now L[xp, yp, xy] is the coordinate ring of the affine surface Z p = XY. Since the surface Z p = XY has only an isolated singularity (at the origin), it is normal. But k[xp, yp, xy] = L[xpyp, xy] ∩ k[x, y]. Hence k[xp, yp, xy] is normal. Since A′ is integral over k[xp, yp, xy] and has the same quotient field as k[xp, yp, xy], we have A′ = k[xp, yp, xy]. We note that the hypothesis of Theorem 2.1 (b) is satisfied here. 66 Hence C(A′) = L . Now L = DP/P P ∈ K, DP/P ∈ A . For o o L P ∈ A, we have d (DP) ≤ d (P). Hence DP/P ∈ if and only if 54 3. Descent methods

a b a b DP/P ∈ k. The formula D(x y ) = (a − b)x y shows that L = Fp, the prime ﬁeld of characteristic p. Hence C(A′) ≈ Z/(p).

(b) The surface Z p = Xi + Y j. Again we take A = k[x, y], k a factorial ring of characteristic p , 0. Let D be the k-derivation of A given by Dx = jy j−1, Dy = −ixi−1, where i, j are positive integers prime to p. Let K, K′, L denote the quotient fields of A, A′ = Ker(D) and k respectively. We have k[xp, yp, xi + y j] ⊂ A′ and [L(xp, yp, xi + y j) : K] = p. Hence K′ = L(xp, yp, xi + y j). Now L[xp, yp, xi + y j] is the coordinate ring of the affine surface Z p = Xi + Y j which is normal since it has only an isolated singularity (at the origin). Hence L[xp, yp, xi + y j] is integrally closed. But k[xp, yp, xi + y j] = L[xp, yp, xi +y j]∩k[x, y]. Hence k[xp, yp, xi +y j] is integrally closed. Since A′ is integral over k[xp, yp, xi +y j], we have A′ = k[xp, yp, xi + y j]. We remark that our D satisfies the hypothesis of Theorem 2.1 (b). Hence C(A′) = DP/P P ∈ K, DP/P ∈ A . We shall now L compute . We attach weights j and i to x and y respectively. By Proposition 3.1, we have Dp = aD with a ∈ A′. It is easily checked that if G is an isobaric polynomial of weight w, then DG is isobaric weight w+i j−i− j and therefore DpG is isobaric of weight w + p(i j − i − j). Now Dp x = aDx. Comparing the weights we see that a is isobaric of weight (p − 1)(i j − i − j).

67 Let F be a polynomial which is a logarthmic derivative. Let Fα of weight α (respectively Fβ of weight β) be the component of smallest (respectively largest) weight of F. By Proposition 3.2, F is a logarthmic derivative if anly only if Dp−1F − aF = −F p. Comparing the weights of the components with smallest and largest weights on both sides, we get p−1 p p−1 weight (D Fα − aFα) ≤ weight (Fα) and weight (D Fβ − aFβ) ≥ p weight (Fβ ). That is pα ≥ α+(p−1)(i j−i− j), pβ ≤ β+(p−1)(i j−i− j). Hence i j − i − j ≤ α ≤ β ≤ i j − i − j i.e. α = β = i j − i − j. Hence F must be isobaric of weight i j − i − j. Set d = (i, j), i = dr, j = ds. Thus, the monomials that can occur in F are of the form xλyµ, λ j + µi = i j − i − j i.e. λs + µr = drs − r − s, i.e. (λ + 1)s = (ds − µ − 1)r. Since (r, s) = 1, λ + 1 is a multiple of r. Thus the smallest value of λ admissible is r − 1, the corresponding µ being (d −1)s−1. Thus F is necessarily of the form 4. Examples: Polynomial rings 55

d−1 nr−1 (d−n)s−1 ′ F = bn x y . If d = 1, then L = 0 and A is factorial. If nP=1 p−1 d > 1, the coefficients of D F −aF will be linear forms in b1,..., bd−1 p th and those of −F are p powers of b1,..., bd−1. Thus F is a logarth- p ′ mic derivative if and only if bn = Ln(b), Ln′ (b) = 0, 1 ≤ n ≤ d − 1, ′ ′ 1 ≤ n ≤ t, where Ln(b), Ln′ (b) are linear forms occuring as the coeffi- cient of Dp−1F − aF. ′ p Ln′ (b) indicates the ones which do not occur in −F . The hypersur- p faces bn = Ln(b) intersect at a finite number of points in the profective space Pd−1 and, by Bezout’s theorem, the number of such points in the algebraic closure of L is atmost pd−1. As L is an additive subgroup of A, L is a p-group of type (p,..., p) of order p f , f ≤ d − 1. Hence we 68 have proved the Theorem 4.1. Let k be a factorial ring of characteristic p , 0, and let i, j be two positive integers prime to p and d = (i, j). Then the group C(A′) of divisor classes of A′ = k[X, Y, Z] with Z p = Xi + Y j is a finite group of type (p,... p) of order p f with f ≤ d − 1. In particular A′ is factorial if i and j are coprime. We can say more about C(A′) in the case p = 2. Let k be of characteristic 2. Then D2 = 0, i.e. a = 0. The equation for the logarthmic derivative then becomes DF = F2. As above F is of the form d−1 nr−1 (d−n)s−1 F = bn x y . Here i, j, r, s, d are all odd integers. If nP=1 nr−1 (d−n)s−1 nr−1+dr−1 (d−n)s−2 n is odd, then D(bn x y = bn x y . The corre- 2 2 2mr−2 2(d−m)s−2 sponding term in D F is bm x y , where 2m = n + d = 2 2q + 1 + d(n = 2q + 1), bn = bm. Set d = 2c − 1. Then m = q + c. 2 Thus b2q+1 = bq+c. On the other hand let n be even, say n = 2q. The nr−1 (d−n)s−1 nr−2 (d−n)s−1+ds−1 D(bn x y ) = bn x y . The corresponding term in 2 2 2mr−2 2(d−m)s−2 2 D F is bm x y , where bn = bm and nr − 2 = 2mr − 2 i.e. 2 2m = n = 2q. Hence b2q = bq. Thus F is a logarthmic derivative if and 2 2 only if the equations b2q+1 = bq+c and b2q = bq, d +1 = 2c, are satisfied. Consider the permutation of (1, 2,..., d−1) given by, (2q) = q, 1 ≤ q ≤ c − 1, (2q − 1) = qQ+ c, 0 ≤ q ≤ c − 2. Now the equationsQ for Q 2 the logarthmic derivative can be written as bq = bΠ(q), 1 ≤ q ≤ 2c − 2. Let U1,..., Ul be the orbits of the group generated by and let Card Q 56 3. Descent methods

69 (Ui) = u(i). Then u(1)+···+u(l) = 2c−2. If Ue = (q1,... qu(e)), then the equations b = b2 ,..., b = b2 are equivalent to b2u(e) = b . q1 Π(q1) qu(e) Π(qu(e)) q1 q1 2u(e) 2 Thus the solutions of b = b give rise to solution of bm = bΠ(m), where 2u(e) u(e) m ∈ Ul. But the solutions of b = b in k is the group F(2 )∩k, where F(2u(e)) is the ﬁeld consisting of 2u(e) elements. Hence the group L of 1 logarthmic derivatives is isomorphic to (F(2u(e)) ∩ k). Hence we have eQ=1 proved the following Theorem 4.2. Let k be a factorial ring of characteristic 2 and let i, j be odd integers and d = (i, j). Let A′ = k[X, Y, Z],Z2 = Xi + Y j. The group C(A′) is of the type (2,..., 2) and of order 2u with u ≤ d − 1. If k contains the algebraic closure of the prime ﬁeld, then the order of C(A′) is 2d−1. Remark . It would be interesting to know if the above theorem is true for arbitrary non-zero characteristics. We remark that for p = 3 and for the surfaces Z3 = X2 + Y4, Z3 = X4 + Y8, the analogue of the above result can be checked.

5 Examples: Power series rings

Let A be a Krull ring and D : A → A, a derivation of A. Let L denote the group of logarthmic derivatives contained in A and L ′ the group of logarthmic derivatives of units of A. Set q = A · D(A). We have, L ′ ⊂ q ∩ L . We prove the other inclusion in a particular case.

70 Lemma 5.1. Let A be a factorial ring of characteristic 2 and D : A → A be a derivation of A satisfying D2 = aD, with a ∈ Ker(D). Assume that there exist x, y ∈ Rad (A) such that q = (Dx, Dy). Then L ′ = L ∩ q. Proof. Let t ∈ L ∩ q, say t = cDx + dDy. If r = (Dx, Dy). By consid- 1 ering the derivation D, we may assume that Dx and Dy are relatively r prime. Since t ∈ L , by Proposition 3.2, we have Dt + at + t2 = 0. Substituting t = cDx + dDy in this equation, we get Dx(Dc + c2Dx) = Dy(Dd + d2Dy). 5. Examples: Power series rings 57

Since Dx and Dy are relatively prime, there is an α ∈ A such that

Dc + c2Dx = αDy, Dd + d2Dy = αDx.

Set u = 1 + cx + dy + (cd + α)xy. The element u is a unit in A. A straight forward computation shows that Du = tu. The proof of the lemma is complete.

(a) The surface Z2 = XY in characteristic 2. Let k be a regular factorial ring. Then, by Theorem 2.1, A = k [x, y] is factorial. Let D be the k-derivation of A given by Dx = x, Dy = y. Then as in §4, A′ = ker(D) = k [x2, y2, xy] = k [X, YZ] , Z2 = XY. Here, q = (x, y) and [K : K′] = 2. Hence, by Theorem 2.1, C(A′) ≈ L /L ′. By Lemma 5.1, we have L /L ′ = L (L ∩ q = (L + q q. This, a b a b and the formula D(x y ) = (a − b)x y show that (L + q)/q ≈ F2 = Z/(2).

+ + (b) The Surface Z2 = X2i 1 + Y2 j 1 in characteristic 2. Let k be a 71 regular factorial ring and A = k [x, y] . Let D be the k-derivation deﬁned by Dx = y2 j, Dy = x2i. Then A′ = k [x2, y2, x2i+1+y2 j+1] = k [X, Y, Z] , Z2 = X2i+1 +Y2 j+1. We have q = AD(A) = (x2i, y2 j) and [K : K′] = 2. Hence C(A′) ≈ L /L ′. Since D2 = 0, an element F ∈ A is a logarthmic derivative if and only if DF = F2. We assign the weights 2 j + 1 and 2i + 1 to x and y respectively. For an F ∈ A with F = Fl, where Fl is an isobaric polynomial of weight l, lP≥q Fq , 0, we call q the order of F, 0(F) = q. As in Theorem 4.1, D elevates the weight of an isobaric polynomial by 4i j − 1, Hence, if F ∈ L and 0(F) = q, then 0(F) = 2q = 0(DF) = q + 4i j − 1. Hence q ≥ 4i j − 1.

L L L L Let q = F F ∈ , 0(F) ≥ q . Now q q≥4i j−1 filters and L ′ L L′ L ′ ′ L L ′ q = q ∩ filters . Hence C(A ) = / is filtered by Cq = ′ ′ L L ′ L ′ L L ( q + )/L ≈ q/L q. In view of Lemma 5.1, we have q = q for q large, ie.e Cq = 0, for q large. Since the Cq are vector spaces over F2, 58 3. Descent methods

′ the extension problem here is trivial. Hence C(A ) ≈ Cq/Cq+1. q≥4Pi j−1 Since 0(x2i) = 2i(2 j + 1) > 4i j and 0(y2 j) = 2 j(2i + 1) > 4i j, we ′ have L = L ∩ q ⊂ L4i j. Therefore C4i j−1 C4i j = L4i j−1 L4i j. By f Theorem 4.2, L4i j−1 L4i j is a ﬁnite group of type (2,..., 2) of order 2 , with f ≤ d − 1, d =(2i + 1, 2 j + 1). We now determine Cq Cq+1, for 72 q ≥ 4i j. Let A(q) denote the k-free module generated by monomials of (q) weight q. Let ϕq : Lq → A be the homomorphism given by ϕq(F) = component of F of weight q, F ∈ Lq. Then ker(ϕq) = Lq+1. We shall now prove

(q) ′ ϕq(Lq) = A ∩ A , q ≥ 4i j, (*) L ′ (q) ′ ϕq( q ) = A ∩ A ∩ q, q ≥ 4i j. (**)

Note that (∗∗) is a consequence of (∗) and the fact that

′ Lq = Lq ∩ q.

Proof if (*). Let F = Fq + Fq+1 + ···∈ Lq, Fq being of weight q. Since 2 DF = F , and weight DFq = q + 4i j − 1 < 2q, we have DF = 0, (q) ′ (q) ′ i.e. ϕq(F) = Fq ∈ A ∩ A . Conversely, let Fq ∈ A ∩ A . We have to find Fn, n ≥ q, Fn isobaric polynomial weight n, such that F = 2 Fn ∈ Lq, i.e. DF = F . Hence we have to determine Fn such nP≥q 2 that DFn = 0, if n is even or n + 4i j − 1 < 2q and DFn = F m, if 2m = n + 4i j − 1(m < n). Thus Fn have to determined by ‘integrating’ 2 the equation DFn = G , where G = 0 or an isobaric polynomial of weight q. Because of the additivity of the derivation, we have only to handle the case G = xαyβ,α(2 j + 1) + β(2i + 1) ≥ q ≥ 4i j. In this 2(α−i) 2β+1 case, either α ≥ i or β ≥ j. If α ≥ i, we take Fn = x y and 2α+i 2(β− j) if β ≥ j, we take Fn = x y . Thus proves (∗) and hence also (q) ′ (q) ′ (∗∗). This gives Cq Cq+1 ≈ (A ∩ A ) (A ∩ A ∩ q), q ≥ 4i j. Hence Cq Cq+1, q ≥ 4i j is a k-free module of finite rank, say n(q). Hence ′ 73 C(A ) ≈ C4i j−1 C4i j ⊕ C4i j, where C4i j is a k-free module of finite rank N(i, j) = n(q). We now determine the integer N(i, j). We observe q≥P4i j that in A, the ideal q admits a supplement generated by the monomials 5. Examples: Power series rings 59 xqyb such that a < 2i, b < 2 j. Since x2i+1 + y2i+1 ∈ q, in A′, the ideal q ∩ A′ admits a supplement generated by the monomials x2ay2b such that 2a < 2i, 2b < 2 j. Thus N(i, j) is equal to the number of monomials x2ay2b, with 0 ≤ 2α < 2i, 0 ≤ 2β < 2 j and weight of x2ay2b ≥ 4i j. Hence we have the

Theorem 5.2. Let k be a factorial ring of characteristic 2, and i, j two integers with (2i + 1, 2 j + 1) = d. Let A′ = k [X, Y, Z] , where Z2 = X2i+1 + Y2 j+1. Then the divisor class group C(A′)≈ H ⊕ G, where, H is a group of type (2,..., 2) of order 2 f , f ≤ d − 1; (if k contains the d−1 algebraic closure of the prime ﬁeld F2, then H is of order 2 ); further G ≈ kN(i, j), where N(i, j) is the number of pairs of integers (a, b) with 0 ≤ a < i, 0 ≤ b < j and (2 j + 1)a + (2i + 1)b ≥ 2i j.

Remarks. 1) The function N(i, j) ∼ i j/2

2) N(i, j) = 0 if and only if the pair (a, b) = (i − 1, j − 1) does not satisfy the inequality (2 j + 1)a + (2i + 1)b ≥ 2i j, i.e. if (i, j) satisﬁes the inequality 2i j − i − j < 2. This is satisﬁed only by the pairs (1, 1), (1, 2) and (1, 3), barring the trivial cases i = 0 or j = 0. Hence, upto a permutation the only factorial ring we obtain is, except for the trivial cases, k [X, Y, Z] , Z2 = X3 + Y5. In view of Theorem 4.1 and Theorem 5.2, the pairs (2 i + 1, 2 j + 1) , (3,5), (5, 3), for which 2i + 1 74 and 2 j + 1 are relatively prime, provide examples of factorial rings whose completions are not factorial.

(c) Power series ring. Let k be a regular factorial ring of characteristic 2. Let A = k[x, y] (resp. k [x, y] ) and R = A [T] . We deﬁne a k-derivation D : R → R by Dx = y2 j, Dy = x2i, DT = 0. Then Ker D = A′ [T] , where A′ = k[x2, y2, x2i+1 +y2 j+1] (Resp. k [x2, y2, x2i+1 +y2i] ). For a Krull ring B, let L (B) and L ′(B) denote the group of logarthmic derivatives in B and the group of logarthmic derivatives of the units of B, respectively. We will compute C(R) = L (R) L ′(R). An F ∈ R is 2 2 n in L (R) if and only if DF = F (since D = 0). Let F = anT . Pn L 2 2 Then F ∈ (R) if and only if Dao = ao, Da2n+1 = 0, Da2n = an. Since by Lemma 5.1, L ′(R) = L (R) ∩ q, where q = (Dx, Dy), we 60 3. Descent methods

′ 2 2 have F ∈ L (R) if and only if Dao = ao, Da2n+1 = 0, Da2n = an ′ and an ∈ (Dx, Dy). Thus F ∈ L (R) (resp. .L (R)) implies ao ∈ ′ ′ ′ L (A) resp. .ao ∈ L (A)). Further, L (R) L (R) ≈ L (A) L (A) ⊕ L (R) ∩ TR . As before we assign weights 2 j + 1, 2i + 1 to x and y re- L ′(R) ∩ TR 2 spectively. Let q(n) = 0(an). Now if F ∈ L (R), then Dan = an. Hence q(2n)+q ≤ 2q(n), where q = 4i j−1. That is , q(2n)−q ≤ 2(q(n)−q). By induction, we get q(2rn) − q ≤ 2r(q(n) − q) for r ≥ 1. Since q(2rn) ≥ 0, we conclude that q(n) ≥ q. A computation similar to that in Theorem 2 5.2 shows that the ‘integration’ of Da2n = an is possible. Further if an ∈ q = (Dx, Dy), then a2n can be chosen in q. 75 Let A(q) be the set of elements of order ≥ q. In computing F ∈ L (R), each integration introduces an ‘arbitrary element’ of A′ ∩ A(q). In computing F ∈ L ′(R), each integration introduces an arbitrary constant of A′∩A(q) ∩q. Hence (L (R)∩TR) L ′(R)∩TR is the product of countably many copies of V = (A′ ∩ A(q) (A′ ∩ A(q) ∩ q). As in the last example, V is a k-free module of rank equal to the number N(i, j) of pairs (a, b) with 0 ≤ a < i, 0 ≤ b < j, (2 j + 1)2a + (2i + 1)2b ≥ q = 4i j − 1 and this inequality is equivalent to (2 j + 1)a + (2i + 1)b ≥ 2i j. Hence we have the

Theorem 5.3. Let k be a factorial ring of characteristic 2, and i, j two integers. Let A′ = k[X, Y, Z] (or k [X, Y, Z] ) with Z2 = X2i+1 + Y2 j+1. Then C(A′ [T] ) C(A′) ≈ (k [T] )N(i, j) where N(i, j) is the number of pairs (a, b)with 0 ≤ a < i, 0 ≤ b < j and (2 j + 1)a + (2i + 1)b ≥ 2i j.

Remarks. (1) Take A′ = k x2, y2, x2i+1 + y2 j+1 with (2i + 1, 2 j + 1) = 1 and N(i, j) > 0. Then A′ is factorial, but A ′ [T] is not. (We have thus to exclude only Z2 = X3 + Y5 and trivial cases.)

(2) Let A′ be the complete local ring A′ = k [X, Y, Z] , Z2 = X2i+1 + Y2 j+1. Then A′ and A′ [T] are simultaneously factorial or simultaneously non-factorial.

(3) In general, the mapping C(A′) → C(A′ [T] ) is not surjective. Appendix

The alternating group operating on a power series ring 76 We have seen (Chap. 3, §1) that the ring A′ of invariants of the alternating group An operating on the polynomial ring k[x1,..., xn] is factorial for n ≥ 5. Let us study the analogous question for the power series ring A = k [x1,..., xn] ; let U be the group of units in A, m the ′ maximal ideal ofA, and A the ring of invariants of An (operating by permutations of the variables). We recall (Chap. 3, §1) that C(A′) ≈ 1 ′ H (An, U) since A is divisorially unramiﬁed over A . We have already 1 seen (Chap. 3, §1, Corollary to Proposition 1.3) that H (An, U) = 0 if the characteristic p of k is prime to the order of An, i.e. if p > n. Thus what we are going to do concerns only ﬁelds of “small” characteristic. Theorem. Suppose that p , 2, 3. Then with the notation as above, A′ is factorial for n ≥ 5. For n = 3, 4,C(A′) is isomorphic to the group of cubic roots of unity contained in k.

′ 1 ∗ ∗ Our statement means that C(A ) ≈ H (An, k ) = Hom(An, k ). In view of the exact sequence 0 → 1 + M → U → k∗ → 0, we have only 1 to prove that H (An, 1 + M ) = 0. For this it is suﬃcient to prove that

1 H (An, (1 +M s) (1 +M s + 1) = 0 for every j ≥ 1. (1) In fact, given a cocycle (xs) in 1 + M (s ∈ An, xs ∈ 1 + M ), it is a 77 2 coboundary modulo 1 + M , i.e. there exists y1 ∈ 1 + M such that xs ≡ −1 M 2 −1 s(y1)y1 mod 1 + . We set x2,s = xsy1 s(y1) ; now x2,s is a cocycle

61 62 3. Descent methods

in 1 + M 2, and therefore a coboundary modulo 1 + M 3. By induction j j we ﬁnd elements y1,..., y j1 ··· (y j ∈ 1 + M ) and x js ∈ 1 + M such ∞ −1 that x1,s = xs and x j+1,s = x j,sy j s(y j ). The product y j converges Qj=1 −1 since A is complete; calling y its value, we have xsys(y ) = 1 for every s ∈ An, which proves that (xs) is a coboundary. In order to prove (1), we notice that the multiplicative group (1 + M j) (1 + M j+1) is isomorphic to the additive group M j M j+1, i.e. to the vector space W j of homogeneous polynomials of degree j. Our theorem is thus a consequence of the following lemma:

Lemma 1. Let S n(resp. .An) operate on k x1,..., xn by permutations of the variables, and let W j be the vector space of homogeneous polynomials of degree j. Then

1 a) H (S n, W j) = 0 if the characteristic p is , 2;

1 b) H (An, W j) = 0 if p , 2, 3

j(1) j(n) We consider a monomial x = x1 ··· x1 of degree j and its trans- forms by S n(resp. .An). These monomials span a stable subspace V of W j, and W j is a direct sum of such stable subspaces V. We need only 1 1 prove that H (S n, V) = 0(resp. .H (An, V) = 0). Now the distict trans- 78 forms xθ of the monomial x are indexed by G/H(G = S n or An), where H is the stability group of x; we have s(xθ) = xsθ for a ∈ G. We are going to prove, in a moment, that

1 H (G, V) = Hom (H, k) (G = S n or An) (2)

Let us ﬁrst see how (2) implies Lemma 1. The stability subgroup H j(i) j(i) is the set of all s in S n (or An) such that xi = x = s(x) = xs(i) = Qi Qi − x j(s 1(i)), i.e. such that j(s−1(i)) = j(i) for every i. Thus H is the set Qi of all s in S n or An which, for every exponent r, leave the set of indices s−1({r}) globally invariant. Denote by n(r) the cardinality of s−1({r}) (i.e. the number of variables xi having exponent r in the monomial x). In the case of S n, H is the direct product of the groups S n(r); since a 5. Examples: Power series rings 63

nontrivial factor group of S n(r) is necessarily cyclic of order 2, we have Hom(H, K) = 0 in characteristic , 2; hence we get a) in Lemma 1. In the case of An, H is the subgroup of S n(r) consisting of the elements Qr (sr) such that the number of indices for which sr ∈ S n(r) − An(r) is even; 1 1 thus H contains H = An(r) as an invariant subgroup, and H/H is a Qr commutative group of type (2, 2,..., 2); on the other hand a nontrivial commutative factor group of An(r) is necessarily cyclic of order 3 (this happens only for n(r) = 3, 4); thus, if p , 2 and 3, who have Hom (H, k) = 0, and this proves b). We are now going to prove (2). More precisely we have the following lemma (probably well known to specialists in homological algebra; probably, also, high-powered cohomological methods could make the proof less computational).

Lemma 2. Let G be a ﬁnite group, H a subgroup of G, k a ring, V a 79 free k-module with a basis (eθ) indexed by G/H. Let G operate on V by 1 s(eθ) = esθ. Then H (G, V) ≈ Hom (H, k).

A system (vs = as,θeθ) (s ∈ G, as,θ ∈ k) is a cocycle if and only θ∈PG/H if vss′ = vs + s(vs′ ) i.e. if and only if

ass′,θ = as,θ + as′,s−1θ. (3)

It is a coboundary if and only if there exists y = bθ eθ such that θ∈PG/H vs = s(y) − y, i.e. if and only if there exist elements bθ of k such that

as,θ = bs−1θ − bθ. (4)

Let ε denote the unit class H in G/H and, given a cocycle (vs) as above, set ϕv(h) = ah,ε for h in H. Since hε = ε(h ∈ H), (3) shows that ϕv is a homomorphism of H into k. We obviously have ϕv+v′ = ϕv + ϕv′ , whence a homomorphism

ϕ : Z1(G, V) (“cocycles′′) → Hom (H, k).

By (4), we see that ϕ is zero on the coboundaries. Conversely if ϕv = 0, we prove that (vs) is a coboundary. In fact, for θ ∈ G/H, choose 64 3. Descent methods

−1 t ∈ G such that θ = t ε, and set bθ = at,ε; this element does not depend on the choice of t since, if t−1ε = u−1ε, then ut−1 ∈ H and u = ht with h ∈ H; by (3), we have au,ε = ahv,ε = ah,ε + at,h−1ε = at,ε (since ϕa = 0). −1 −1 −1 80 Now, if θ = tε and if s ∈ G, we have s θ = (ts) θ, whence bθ = at,ε and bs−1θ = ats,ε. From (3) we get bs−1θ − bθ = ats,ε − at,ε = as,t−1ε = as,θ, thus proving that (vs) is a coboundary. Thus the proof of lemma 2 will be complete if we show that ϕ is surjective. Let c be a homomorphism of H into k. For every θ in G/H, we choose t(θ) in G such that θ = t(θ)−1ε. Then every s ∈ G may be written uniquely as s = h.t(µ) (h ∈ H,µ = s−1H). We set

as,θ = c(h), (5)

where h is the unique element of H such that t(θ) · s = h.t(s−1θ) (notice that t(θ).s.t(s−1θ)−1.ε = t(θ)s.s−1θ = t(θ).θ = ε, whence t(θ).s.t (s−1θ)−1 ∈ H). Let us verify the “cocycle condition” (3). We have ′ ass′,θ = c(h), as,θ = c(h1) and as′,s−1θ = c(h2), with t(θ)ss = h.t ′−1 −1 −1 −1 ′ ′−1 −1 (s s θ), t(θ)s = t( )s = h1t(s θ) and t(s θ).s = h2t(s s θ). From this we immediately deduce that h = h1h2. Since c is a homomorphism, we have c(h) = c(h1) + c(h2), i.e. ass′ ,θ = as,θ + as′ s−1θ. Thus θ vs = asθeθ is a cocycle. For this cocycle, we have (for h ∈ H)ϕv(h) = Pθ −1 ah,ε = c(h1), where, by (5), h1 is such that t(ε).h = h1t(h ε) = h1t(ε); since the additive group of k is commutative, we have c(h) = c(h1), whence ϕv(h) = c(h) for every h ∈ H. Q.E.D