Éléments de Géométrie Algébrique — English

§17. Regular rings

17.1. Definition of regular rings

Proposition (17.1.1).

Let $A$ be a Noetherian local ring of dimension $n$, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field. The following conditions are equivalent:

a) The canonical surjective homomorphism of graded $k$-modules (15.1.1.1)

$$\varphi :S_{\bullet }(\mathfrak{m}/{\mathfrak{m}}^2)\to g{r}_{\mathfrak{m}}(A)$$

(where the right-hand side is the graded $k$-module associated with $A$ equipped with the $\mathfrak{m}$-preadic filtration) is bijective.

b) One has $r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2)=n$.

c) The ideal $\mathfrak{m}$ admits a system of generators with $n$ elements.

d) The ideal $\mathfrak{m}$ admits a system of generators which is an $A$-regular sequence.

By Nakayama's lemma, $r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2)$ is the smallest number of elements in a system of generators of $\mathfrak{m}$, so b) and c) are equivalent. On the other hand, if $(x_i{)}_{1\le i\le n}$ is a system of generators of $\mathfrak{m}$ whose classes ${\bar{x}}_{i}mod{\mathfrak{m}}^2$ form a basis of the $k$-vector space $\mathfrak{m}/{\mathfrak{m}}^2$, then $S_{\bullet }(\mathfrak{m}/{\mathfrak{m}}^2)$ is isomorphic to the polynomial ring

$k[T_1,\cdots ,T_n]$; taking into account that every $A$-module of finite type is separated for the $\mathfrak{m}$-preadic filtration $(0_I,\mathrm{7.3.5})$, it follows from (15.1.9) that conditions a) and d) are equivalent; furthermore, since every $A$-regular sequence of elements of $\mathfrak{m}$ has at most $n$ elements (16.4.1), one sees that d) implies c). It remains to prove that c) implies a). For brevity put $S=S_{\bullet }(\mathfrak{m}/{\mathfrak{m}}^2)=k[T_1,\cdots ,T_n]$, $G=g{r}_{\mathfrak{m}}(A)$, and consider the exact sequence $0\to \mathfrak{J}\to S{\to }^{\varphi }G\to 0$, where the kernel $\mathfrak{J}$ of $\varphi$ is a graded ideal of $S$. For every integer $s\ge 0$ one has therefore

(17.1.1.1)               C(s+n-1, n-1) = long(S_s) = long(G_s) + long(𝔍_s).

Suppose $\mathfrak{J}\ne 0$, so that there exists a homogeneous element $u\in \mathfrak{J}$ of degree $h\ge 0$; ${\mathfrak{J}}_s$ contains $u{S}_{s-h}$ for $s\ge h$, and since $u\ne 0$ and $S$ is an integral domain, $u{S}_{s-h}$ is isomorphic to $S_{s-h}$ as a $k$-vector space; one would therefore have $long({\mathfrak{J}}_s)\ge long(S_{s-h})=C(s-h+n-1,n-1)$, whence for $s\ge h$,

(17.1.1.2)              long(G_s) ≤ C(s+n-1, n-1) − C(s-h+n-1, n-1).

Now, the right-hand side of (17.1.1.2) is a polynomial in $s$ of degree $\le n-2$; but one has $G_s={\mathfrak{m}}^s/{\mathfrak{m}}^{s+1}$, and for $s$ large enough, $long(G_s)$ is a polynomial in $s$ of degree exactly equal to n-1 whose leading coefficient is > 0 (16.2.1), which contradicts the inequality (17.1.1.2). Q.E.D.

Definition (17.1.2).

One says that a Noetherian local ring which satisfies the equivalent conditions of (17.1.1) is regular.

Corollary (17.1.3).

A regular local ring is an integral domain, integrally closed, and a Cohen-Macaulay ring.

The last assertion follows from (17.1.1, d)); on the other hand, if $A$ is regular, $g{r}_{\mathfrak{m}}(A)$ is an integral domain and completely integrally closed, being isomorphic to $k[T_1,\cdots ,T_n]$; one concludes that $A$ also possesses these two properties (Bourbaki, Alg. comm., chap. III, §2, n° 3, cor. of prop. 1 and chap. V, §1, n° 5, prop. 15).

Examples (17.1.4). — (i) A regular local ring of dimension 0, being an integral domain by (17.1.3), is necessarily a field, and conversely.

(ii) For a Noetherian local ring of dimension 1 to be regular, it is necessary and sufficient that it be a discrete valuation ring: indeed, to say that a Noetherian local ring of dimension 1 is regular means that its maximal ideal is principal, and the conclusion follows from Bourbaki, Alg. comm., chap. VI, §3, n° 6, prop. 9.

(iii) Let $k$ be a field; the ring of formal power series $A=k[[T_1,\cdots ,T_n]]$ is a regular ring of dimension $n$; indeed, it is clear that the $T_i$ $(1\le i\le n)$ generate the maximal ideal $\mathfrak{m}$ of $A$ and form an $A$-regular sequence, since $A/(T_{1}A+\cdots +T_{i}A)$ is isomorphic to $k[[T_{i+1},\cdots ,T_n]]$.

Proposition (17.1.5).

For a Noetherian local ring $A$ to be regular, it is necessary and sufficient that its completion Â be so.

Indeed, the maximal ideal of Â is $\mathfrak{m}\hat{A}$, and one knows that ${\mathfrak{m}}^h/{\mathfrak{m}}^{h+1}$ and $(\mathfrak{m}\hat{A}{)}^h/(\mathfrak{m}\hat{A}{)}^{h+1}$ are isomorphic $k$-vector spaces; condition a) of (17.1.1) is therefore the same for $A$ and Â.

Definition (17.1.6).

Given a Noetherian local ring $A$ with maximal ideal $\mathfrak{m}$, one calls a regular system of parameters of $A$ a system of parameters for $A$ (16.3.5) that generates $\mathfrak{m}$.

The existence of such a system is equivalent to the fact that $A$ is regular (17.1.1); if $(x_i{)}_{1\le i\le n}$ is a regular system of parameters, it is a minimal system of generators of $\mathfrak{m}$, whose classes mod ${\mathfrak{m}}^2$ form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$ over $k$; by virtue of (17.1.1, a)) and (15.1.9), $(x_i)$ is a maximal $A$-regular sequence (whence the terminology).

One should beware, however, that in a regular ring, a system of parameters for $A$ which is also an $A$-regular sequence is not necessarily a regular system of parameters, as the example of the $k$-th powers of a regular system of parameters shows (15.1.20).

Proposition (17.1.7).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field, $(x_i{)}_{1\le i\le r}$ a sequence of elements of $\mathfrak{m}$, $\mathfrak{J}=x_{1}A+\cdots +x_{r}A$. The following conditions are equivalent:

a) $A$ is regular and the $x_i$ are part of a regular system of parameters of $A$.

a′) $A$ is regular and the classes ${\bar{x}}_i$ of the $x_i$ mod ${\mathfrak{m}}^2$ are linearly independent over $k$.

b) The $x_i$ are part of a system of parameters for $A$ and $A/\mathfrak{J}$ is a regular ring.

Moreover, when these conditions are satisfied, $\mathfrak{J}$ is a prime ideal.

The last assertion follows trivially from the fact that $A/\mathfrak{J}$, being regular, is an integral domain (17.1.3). Put $n=\dim (A)$. The equivalence of a) and a′) is immediate; indeed, the classes mod ${\mathfrak{m}}^2$ of the elements of a regular system of parameters form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$ over $k$, and conversely, if the ${\bar{x}}_i$ $(1\le i\le r)$ are linearly independent over $k$, one can find $n-r$ elements $x_i\in \mathfrak{m}$ $(r+1\le i\le n)$ such that the ${\bar{x}}_i$ $(1\le i\le n)$ form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$, hence $(x_i{)}_{1\le i\le n}$ is a regular system of parameters. To see that a) implies b), let us remark that since the $x_i$ are part of a system of parameters of $A$, one has $\dim (A/\mathfrak{J})=n-r$ (16.3.6); let $\mathfrak{n}=\mathfrak{m}/\mathfrak{J}$ be the maximal ideal of $A/\mathfrak{J}$. One has an exact sequence

(17.1.7.1)              0 → (𝔪² + 𝔍)/𝔪² → 𝔪/𝔪² → 𝔫/𝔫² → 0

since ${\mathfrak{n}}^2=({\mathfrak{m}}^2+\mathfrak{J})/\mathfrak{J}$; hypothesis a) implies that $r{g}_k(({\mathfrak{m}}^2+\mathfrak{J})/{\mathfrak{m}}^2)=r$, hence $r{g}_k(\mathfrak{n}/{\mathfrak{n}}^2)=n-r=\dim (A/\mathfrak{J})$, which proves b) by virtue of (17.1.1, b)).

Conversely, to prove that b) implies a), note that the fact that the $x_i$ are part of a system of parameters implies that $\dim (A/\mathfrak{J})=n-r$ (16.3.6); on the other hand, the hypothesis that $A/\mathfrak{J}$ is regular implies $r{g}_k(\mathfrak{n}/{\mathfrak{n}}^2)=n-r$ (17.1.1); moreover one obviously has $r{g}_k(({\mathfrak{m}}^2+\mathfrak{J})/{\mathfrak{m}}^2)\le r$, so one deduces from (17.1.7.1) that $r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2)\le r+(n-r)=n$; hence $A$ is regular by virtue of (16.2.6) and (17.1.1). Furthermore, the relation $r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2)=n$ then implies $r{g}_k(({\mathfrak{m}}^2+\mathfrak{J})/{\mathfrak{m}}^2)=r$, and consequently the ${\bar{x}}_i$ are linearly independent over $k$. Q.E.D.

Corollary (17.1.8).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $t$ an element of $\mathfrak{m}$. The following conditions are equivalent:

a) $A/tA$ is regular and $t$ is not a zero-divisor in $A$.

b) $A$ is regular and $t\notin {\mathfrak{m}}^2$.

This is the special case $r=1$ of (17.1.7) (taking into account that an element of $\mathfrak{m}$ which is not a zero-divisor is part of a system of parameters (16.4.1)).

Corollary (17.1.9).

Let $A$ be a regular local ring, $\mathfrak{m}$ its maximal ideal, $\mathfrak{J}$ an ideal of $A$ contained in $\mathfrak{m}$. The following conditions are equivalent:

a) The ring $A/\mathfrak{J}$ is regular.

b) $\mathfrak{J}$ is generated by a sequence $(x_i{)}_{1\le i\le r}$ which is part of a regular system of parameters of $A$.

We have already seen (17.1.7) that b) implies a). Conversely, suppose $A/\mathfrak{J}$ is regular (which implies that $\mathfrak{J}$ is prime) and let $n=\dim (A)$, $n-r=\dim (A/\mathfrak{J})$; with the notations of (17.1.7), the exact sequence (17.1.7.1) gives $r{g}_k(({\mathfrak{m}}^2+\mathfrak{J})/{\mathfrak{m}}^2)=r$, hence there exists a sequence $(x_i{)}_{1\le i\le r}$ of elements of $\mathfrak{J}$ that is part of a regular system of parameters of $A$. Put ${\mathfrak{J}}^{\prime }=x_{1}A+\cdots +x_{r}A$; it follows from (17.1.7) that ${\mathfrak{J}}^{\prime }$ is a prime ideal of $A$ and that $A/{\mathfrak{J}}^{\prime }$ is regular and of dimension $n-r$; but since $A/\mathfrak{J}=(A/{\mathfrak{J}}^{\prime })/(\mathfrak{J}/{\mathfrak{J}}^{\prime })$ and $\mathfrak{J}/{\mathfrak{J}}^{\prime }$ is prime in the integral domain $A/{\mathfrak{J}}^{\prime }$, the dimensions of $A/\mathfrak{J}$ and $A/{\mathfrak{J}}^{\prime }$ can be equal only if $\mathfrak{J}={\mathfrak{J}}^{\prime }$ (16.1.2.2).

17.2. Recollections on the projective dimension and the injective dimension of modules

(17.2.1) Let $A$ be a ring, $M$ an $A$-module. Recall (M, VI, 2) that one calls the projective dimension (resp. injective dimension) of $M$, and denotes by $\dim .proj(M)$ or $\dim .pro{j}_A(M)$ (resp. $\dim .inj(M)$ or $\dim .in{j}_A(M)$), the smallest $n$ (an integer or equal to $+\infty$) such that there exists a left projective resolution (resp. a right injective resolution) of $M$ of length $n$. It comes to the same thing (loc. cit.) to say that $n$ is the smallest number such that for every $A$-module $N$, one has $Ex{t}_A^i(M,N)=0$ (resp. $Ex{t}_A^i(N,M)=0$) for every $i>n$, or only for $i=n+1$. For every direct factor $M^{\prime }$ of $M$, one therefore has dim. proj(M') ≤ dim. proj(M) since $Ex{t}_A^i(M^{\prime },N)$ is a direct factor of $Ex{t}_A^i(M,N)$; similarly dim. inj(M') ≤ dim. inj(M). An equivalent condition to $\dim .proj(M)=n$ (resp. $\dim .inj(M)=n$) is that $n$ is the smallest number such that, for every exact sequence

                        0 → R → P_{n-1} → ⋯ → P_0 → M → 0

(resp. $0\to M\to Q_0\to \cdots \to Q_{n-1}\to C\to 0$), where all the $P_i$ for $0\le i<n$ are projective (resp. all the $Q_i$ for $0\le i<n$ are injective), $R$ is projective (resp. $C$ is injective).

Remarks (17.2.2). — (i) To say that $M$ is a projective (resp. injective) $A$-module is therefore equivalent to saying that $\dim .proj(M)=0$ (resp. $\dim .inj(M)=0$).

(ii) Let $T$ be an additive covariant functor from the category of $A$-modules to an abelian category. It follows at once from the definitions of (17.2.1) and from that of

derived functors that if $\dim .proj(M)\le n$ (resp. $\dim .inj(M)\le n$), one has $L_{i}T(M)=0$ (resp. $R^{i}T(M)=0$) for every $i>n$.

(iii) If one assumes $A$ Noetherian and $M$ of finite type, the last interpretation of the projective dimension given in (17.2.1) shows that if $\dim .proj(M)=n$, then $M$ admits a left resolution of length $n$ formed of projective modules of finite type. If moreover $A$ is a Noetherian local ring, $M$ admits a resolution of length $n$ by free modules of finite type, a projective $A$-module of finite type being then free (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 5).

Lemma (17.2.3).

Let $A$ be a ring, $M$ an $A$-module. For $\dim .inj(M)\le n$, it is necessary and sufficient that for every monogenic $A$-module $N$, one have $Ex{t}_A^{n+1}(N,M)=0$.

With the notations of (17.2.1), it suffices to prove that $C$ is injective; for every $A$-module $N$, $Ex{t}_A^{n+1}(N,M)$ is isomorphic to $Ex{t}_A^1(N,C)$ (M, V, 7), so one has $Ex{t}_A^1(N,C)=0$ for every $A$-module $N$ of the form $A/\mathfrak{J}$, where $\mathfrak{J}$ is an arbitrary ideal of $A$. The exact sequence of Ext then shows that the canonical homomorphism $\mathrm{Hom}(A,C)\to \mathrm{Hom}(\mathfrak{J},C)$ is surjective for every ideal $\mathfrak{J}$ of $A$, which implies that $C$ is an injective $A$-module (M, I, 3.2).

Lemma (17.2.4).

Let $A$ be a Noetherian ring, $M$ an $A$-module of finite type. For $\dim .proj(M)\le n$, it is necessary and sufficient that for every monogenic $A$-module $N$, one have $Ex{t}_A^{n+1}(M,N)=0$.

One knows indeed (M, VI, 2.5) that the condition $\dim .proj(M)\le n$ is equivalent to $Ex{t}_A^{n+1}(M,N)=0$ for every $A$-module $N$ of finite type. To see that the condition of the statement is also sufficient, one argues by induction on the number of generators $m$ of $N$: there is a submodule N_1 of $N$ generated by $m-1$ elements and such that $N_2=N/N_1$ is monogenic; from the exact sequence $0\to N_1\to N\to N_2\to 0$, one then deduces the exact sequence Ext^{n+1}_A(M, N_1) → Ext^{n+1}_A(M, N) → Ext^{n+1}_A(M, N_2), and the induction hypothesis shows that the condition of the statement does indeed imply $Ex{t}_A^{n+1}(M,N)=0$.

Corollary (17.2.5).

Let $A$ be a Noetherian ring, $M$ an $A$-module. One has

(17.2.5.1)              dim. inj_A(M) = sup_𝔪(dim. inj_{A_𝔪}(M_𝔪))

and if $M$ is of finite type

(17.2.5.2)              dim. proj_A(M) = sup_𝔪(dim. proj_{A_𝔪}(M_𝔪))

where $\mathfrak{m}$ runs through the set of prime ideals (or the set of maximal ideals) of $A$.

Indeed, if $N$ is an $A$-module of finite type (hence of finite presentation), one has, for every multiplicative subset $S$ of $A$ and every $i\ge 0$,

                        S^{-1} Ext^i_A(N, M) ≅ Ext^i_{S^{-1} A}(S^{-1} N, S^{-1} M)

by flatness, considering a free resolution of $M$ and using the fact that the preceding relation is true for $i=0$ (Bourbaki, Alg. comm., chap. II, §2, n° 7, prop. 19). In particular $Ex{t}_{A_{\mathfrak{m}}}^i(A_{\mathfrak{m}}/\mathfrak{J}{A}_{\mathfrak{m}},M_{\mathfrak{m}})=(Ex{t}_A^i(A/\mathfrak{J},M){)}_{\mathfrak{m}}$ for every prime ideal $\mathfrak{m}$ and every ideal $\mathfrak{J}$ of $A$; taking (17.2.3) into account, and the fact that every

ideal of $A_{\mathfrak{m}}$ is of the form $\mathfrak{J}{A}_{\mathfrak{m}}$ for a suitable ideal $\mathfrak{J}$ of $A$, one deduces formula (17.2.5.1) from what precedes and from Bourbaki, Alg. comm., chap. II, §3, n° 3, th. 1. One proceeds similarly for (17.2.5.2), this time using (17.2.4) and exchanging the roles of $M$ and $N$.

For Noetherian rings and modules of finite type, the study of projective dimension or injective dimension is therefore reduced by (17.2.5) to the case of local rings. One then has the following:

Lemma (17.2.6).

Let $A$ be a Noetherian local ring, $k$ its residue field, $M$ an $A$-module of finite type. For $\dim .proj(M)\le n$, it is necessary that $To{r}_i^A(M,k)=0$ for $i>n$, and sufficient that $To{r}_{n+1}^A(M,k)=0$.

Necessity is a special case of remark (17.2.2, (ii)), applied to the covariant functor $M\mapsto k{\otimes }_{A}M$. To prove that the condition is sufficient, one must, with the notations of (17.2.1), establish that $R$ is projective when the $P_i$ are assumed of finite type; now $To{r}_{n+1}^A(M,k)$ is isomorphic to $To{r}_1^A(R,k)$ (M, V, 7); and one knows that, since $R$ is of finite type, the condition $To{r}_1^A(R,k)=0$ implies that $R$ is free (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 5).

Corollary (17.2.7).

Under the hypotheses of (17.2.6), let $x$ be an $M$-regular element of $A$ belonging to the maximal ideal of $A$; then one has

(17.2.7.1)              dim. proj(M/xM) = dim. proj(M) + 1.

Indeed, from the exact sequence $0\to M{\to }^{x}M\to M/xM\to 0$ defined by multiplication by $x$ in $M$, one deduces the exact sequence of Tor:

        Tor^A_i(M, k) →^x Tor^A_i(M, k) → Tor^A_i(M/xM, k) → Tor^A_{i-1}(M, k) →^x Tor^A_{i-1}(M, k)

for every $i\ge 1$. Now, for every $A$-module $N$, the homothety of ratio $x$ in $M{\otimes }_{A}N$ comes both from the homothety of ratio $x$ in $M$ and from the homothety of ratio $x$ in $N$; one concludes at once, since the homothety of ratio $x$ in $k$ is zero by definition, that for every $i$, the homomorphism $To{r}_i^A(M,k){\to }^{x}To{r}_i^A(M,k)$ is zero; in other words, one has the exact sequence

                        0 → Tor^A_i(M, k) → Tor^A_i(M/xM, k) → Tor^A_{i-1}(M, k) → 0.

If $\dim .proj(M)=n$, then $To{r}_n^A(M,k)\ne 0$ and $To{r}_{n+1}^A(M,k)=To{r}_{n+2}^A(M,k)=0$ by virtue of (17.2.6). It follows from what precedes that one has $To{r}_{n+1}^A(M/xM,k)\ne 0$ and $To{r}_{n+2}^A(M/xM,k)=0$, hence $\dim .proj(M/xM)=n+1$ by (17.2.6).

Proposition (17.2.8) (M. Auslander).

Let $A$ be a ring. The following conditions are equivalent:

a) Every $A$-module is of projective dimension $\le n$.

a′) Every $A$-module of finite type is of projective dimension $\le n$.

b) Every $A$-module is of injective dimension $\le n$.

c) For every pair of $A$-modules M, N one has $Ex{t}_A^{n+1}(M,N)=0$.

c′) For every pair of $A$-modules M, N such that $M$ is of finite type (or only monogenic), one has $Ex{t}_A^{n+1}(M,N)=0$.

This follows at once from (17.2.1) and (17.2.3).

The smallest number $n$ (an integer or $+\infty$) for which the equivalent conditions of (17.2.8) are satisfied is called the global cohomological dimension (or simply cohomological dimension) of $A$ and denoted $\dim .coh(A)$.

Proposition (17.2.9).

Let $A$ be a Noetherian ring. The following conditions are equivalent:

a) $\dim .coh(A)\le n$.

b) Every $A$-module of finite type is of injective dimension $\le n$.

c) For every pair of $A$-modules of finite type M, N, one has $Ex{t}_A^{n+1}(M,N)=0$.

This follows at once from the definition and from (17.2.4).

Corollary (17.2.10).

If $A$ is a Noetherian ring, one has

(17.2.10.1)             dim. coh(A) = sup_𝔪(dim. coh(A_𝔪))

where $\mathfrak{m}$ runs through the spectrum of $A$ (or the set of maximal ideals of $A$).

This follows from (17.2.9) and (17.2.5).

Proposition (17.2.11).

Let $A$ be a Noetherian local ring, $k$ its residue field. For $\dim .coh(A)\le n$, it is necessary that $To{r}_i^A(k,k)=0$ for $i>n$, and sufficient that $To{r}_{n+1}^A(k,k)=0$.

Taking (17.2.6) into account, it suffices to prove that the relations $\dim .coh(A)\le n$ and $\dim .pro{j}_A(k)\le n$ are equivalent. It is clear that the first implies the second by definition. Conversely, if $\dim .pro{j}_A(k)\le n$, one has $To{r}_i^A(M,k)=0$ for $i>n$ and every $A$-module $M$ by virtue of (17.2.2, (ii)); hence $\dim .proj(M)\le n$, which proves the proposition by virtue of (17.2.8).

Corollary (17.2.12).

Let $A$ be a Noetherian ring. For $\dim .coh(A)\le n$, it is necessary that, for every maximal ideal $\mathfrak{m}$ of $A$, one have $To{r}_i^{A_{\mathfrak{m}}}(A/\mathfrak{m},A/\mathfrak{m})=0$ for $i>n$, and sufficient that these relations be satisfied for $i=n+1$.

This follows at once from (17.2.11) and (17.2.10).

Proposition (17.2.13).

Let $A$, $B$ be two Noetherian local rings, $\varphi :A\to B$ a local homomorphism making $B$ a flat $A$-module. Then one has

(17.2.13.1)             dim. coh(A) ≤ dim. coh(B).

Suppose that $\dim .coh(B)=n$ is finite; it suffices to prove that for every pair $(M,N)$ of $A$-modules of finite type, one has $To{r}_i^A(M,N)=0$ for $i>n$ (17.2.11). Since $B$ is a faithfully flat $A$-module $(0_I,\mathrm{6.6.2})$, it comes to the same thing $(0_I,\mathrm{6.4.1})$ to show that one has

(17.2.13.2)             Tor^A_i(M, N) ⊗_A B = 0.

Now, if $L_{\bullet }=(L_j)$ is a right resolution of $M$ by free $A$-modules, it follows from the fact that $B$ is a flat $A$-module that $L_{\bullet }{\otimes }_{A}B=(L_j{\otimes }_{A}B)$ is a right resolution of the $B$-module $M{\otimes }_{A}B$ by free $B$-modules; moreover, one has $(L_j{\otimes }_{A}B){\otimes }_B(N{\otimes }_{A}B)=(L_j{\otimes }_{A}N){\otimes }_{A}B$, whence one concludes at once that the left-hand side

of (17.2.13.2) equals $To{r}_i^B(M{\otimes }_{A}B,N{\otimes }_{A}B)$; the hypothesis on $B$ implies that this $B$-module is zero for $i>n$ (17.2.2, (ii)), whence the conclusion.

(17.2.14) Let $(X,{\mathcal{O}}_X)$ be a ringed space, $\mathcal{F}$ an ${\mathcal{O}}_X$-Module; one calls the pointwise projective (resp. injective) dimension of $\mathcal{F}$ and denotes by $\dim .proj(\mathcal{F})$ (resp. $\dim .inj(\mathcal{F})$) the number sup_{x ∈ X}(dim. proj(ℱ_x)) (resp. sup_{x ∈ X}(dim. inj(ℱ_x))). One calls the pointwise cohomological dimension of $X$ and denotes by $\dim .coh(X)$ the number sup_{x ∈ X}(dim. coh(𝒪_x)). It follows from (17.2.5) and (17.2.10) that when $A$ is a Noetherian ring, $M$ an $A$-module of finite type and $X=\mathrm{Spec}(A)$, one has dim. proj(M̃) = dim. proj(M), dim. inj(M̃) = dim. inj(M) and dim. coh(X) = dim. coh(A). One calls the projective (resp. injective) dimension of $\mathcal{F}$ at a point $x\in X$ the projective (resp. injective) dimension of ${\mathcal{F}}_x$, the cohomological dimension of $X$ at a point $x$ the cohomological dimension of ${\mathcal{O}}_x$.

Proposition (17.2.15).

Let $X$, $Y$ be two ringed spaces with Noetherian local rings, $f:X\to Y$ a flat morphism. If $\dim .coh(X)\le n$, then $Y$ is of cohomological dimension $\le n$ at every point of $f(X)$.

This follows at once from (17.2.13).

17.3. Cohomological theory of regular rings

Theorem (17.3.1) (Hilbert-Serre).

Let $A$ be a Noetherian local ring. For $A$ to be of finite cohomological dimension, it is necessary and sufficient that $A$ be regular; one has then

(17.3.1.1)              dim. coh(A) = dim(A).

Suppose $A$ regular; let $\mathfrak{m}$ be its maximal ideal, $\mathbf{x}=(x_i{)}_{1\le i\le n}$ a regular system of parameters of $A$ (17.1.6); consider the complex of the exterior algebra $K_{\bullet }(\mathbf{x})$ (III, 1.1.1), which is formed of free $A$-modules, with $K_i(\mathbf{x})=0$ for $i>n$; since $(x_i)$ is a regular sequence, one has $H_i(K_{\bullet }(\mathbf{x}))=0$ for $i>0$ (III, 1.1.4 and 1.1.3.3) and $H_0(K_{\bullet }(\mathbf{x}))=A/(x_{1}A+\cdots +x_{n}A)=A/\mathfrak{m}=k$; the $K_i(\mathbf{x})$ therefore form a free resolution of $k$ of length $n$. Now, the fact that the $x_i$ belong to $\mathfrak{m}$ implies at once that in the complex $K_{\bullet }(\mathbf{x},k)=K_{\bullet }(\mathbf{x}){\otimes }_{A}k$, the boundary operator is zero in all dimensions, so that one has, by definition, $To{r}_i^A(k,k)={\Lambda }^i(k^n)$; equality (17.3.1.1) therefore follows at once from (17.2.11) (this result is essentially Hilbert's "syzygy theorem").

Let us now show that if $A$ is a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, and if $\dim .pro{j}_A(\mathfrak{m})$ is finite, then $A$ is regular, which will complete the proof of (17.3.1). We proceed by induction on $n=r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2)$. For $n=0$, one has $\mathfrak{m}=0$ and the assertion is trivial.

Lemma (17.3.1.2) (Nagata).

If every element of $\mathfrak{m}-{\mathfrak{m}}^2$ is a zero-divisor in $A$, there exists $c\ne 0$ in $A$ such that $c\mathfrak{m}=0$ (in other words, one has $\mathfrak{m}\in Ass(A)$).

One can restrict to the case where $\mathfrak{m}\ne 0$, hence $\mathfrak{m}\ne {\mathfrak{m}}^2$. The hypothesis implies that $\mathfrak{m}-{\mathfrak{m}}^2$ is contained in the union of the ideals ${\mathfrak{p}}_i$ of $Ass(A)$ (Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 3 of prop. 2); hence $\mathfrak{m}$ is contained in the union of ${\mathfrak{m}}^2$

and the ${\mathfrak{p}}_i$, hence in one of the ${\mathfrak{p}}_i$ since $\mathfrak{m}\ne {\mathfrak{m}}^2$ (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2); $\mathfrak{m}$ being maximal, this proves the lemma.

Lemma (17.3.1.3).

If $a\in \mathfrak{m}-{\mathfrak{m}}^2$, $\mathfrak{m}/Aa$ is isomorphic to a direct factor of $\mathfrak{m}/a\mathfrak{m}$.

There exists a minimal system of generators of $\mathfrak{m}$ containing $a$ (whose images in $\mathfrak{m}/{\mathfrak{m}}^2$ form a basis of this $k$-vector space); let $\mathfrak{b}$ be the ideal generated by the elements of this system other than $a$. Since the relation $xa\in \mathfrak{b}$ implies $x\in \mathfrak{m}$ (by considering the image of xa in $\mathfrak{m}/{\mathfrak{m}}^2$), one has $\mathfrak{b}\cap Aa\subset a\mathfrak{m}$, whence a homomorphism $\mathfrak{b}/(\mathfrak{b}\cap Aa)\to \mathfrak{m}/a\mathfrak{m}$ deduced from the injection $\mathfrak{b}\to \mathfrak{m}$ by passage to the quotients, and which is injective. Furthermore, $\mathfrak{b}+Aa=\mathfrak{m}$, and the composite homomorphism

                𝔪/Aa = (𝔟 + Aa)/Aa ≅ 𝔟/(𝔟 ∩ Aa) → 𝔪/a𝔪 → 𝔪/Aa

is the identity; whence the lemma.

Lemma (17.3.1.4).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $E$ an $A$-module of finite type and of finite projective dimension. If $a\in \mathfrak{m}$ is $A$-regular and $E$-regular, then $E/aE$ is an $(A/aA)$-module of finite projective dimension, at most equal to $\dim .pro{j}_A(E)$.

We argue by induction on $h=\dim .pro{j}_A(E)$, the case $h=0$ being trivial since $E$ is then a projective $A$-module, hence $E/aE$ is a projective $(A/aA)$-module. There exists an exact sequence

$$0\to N\to L\to E\to 0$$

where $L$ is free and $\dim .pro{j}_A(N)=h-1$ (17.2.2, (iii)), with $N$ of finite type. Moreover, the sequence

$$0\to N/aN\to L/aL\to E/aE\to 0$$

is exact (15.1.18). Since $a$ is $A$-regular, it is also $N$-regular since $L$ is free; the induction hypothesis implies that $N/aN$ is an $(A/aA)$-module of projective dimension $\le h-1$, and since $L/aL$ is a free $(A/aA)$-module, $E/aE$ is an $(A/aA)$-module of projective dimension $\le h$.

Let us now examine two cases:

I. — Suppose first that every element of $\mathfrak{m}-{\mathfrak{m}}^2$ is a zero-divisor in $A$, in which case (17.3.1.2) there exists $c\ne 0$ in $A$ such that $c\mathfrak{m}=0$. Let us show that then $\mathfrak{m}=0$. Were this not so, let us first note that $\mathfrak{m}$ could not be a projective $A$-module, for it would be free (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 5), which contradicts the relation $c\mathfrak{m}=0$. One would therefore have $n=\dim .coh(A)\ge 1$. Since $\mathfrak{m}\in Ass(A)$, there would exist an exact sequence of $A$-homomorphisms

$$0\to k\to A\to E\to 0.$$

But this is absurd, for by virtue of the relation $n\ge 1$, the exact sequence of Tor would give the exact sequence $0\to To{r}_{n+1}^A(E,k)\to To{r}_n^A(k,k)\to 0$; now one has $To{r}_{n+1}^A(E,k)=0$ (17.2.2, (ii)) and $To{r}_n^A(k,k)\ne 0$ (17.2.11), and we have reached a contradiction.

II. — One can therefore restrict to the case where there exists $a\in \mathfrak{m}-{\mathfrak{m}}^2$ which is an $A$-regular element, and consequently also $\mathfrak{m}$-regular. Consider the ring $A^{\prime }=A/aA$ and its

maximal ideal ${\mathfrak{m}}^{\prime }=\mathfrak{m}/aA$; it is clear that $r{g}_k({\mathfrak{m}}^{\prime }/{\mathfrak{m}}^{\prime 2})=n-1$. By virtue of (17.3.1.4), $\mathfrak{m}/a\mathfrak{m}$ is an $A^{\prime }$-module of finite projective dimension, hence so is ${\mathfrak{m}}^{\prime }=\mathfrak{m}/aA$, which is a direct factor of it (17.3.1.3 and 17.2.1). The induction hypothesis therefore implies that $A^{\prime }=A/aA$ is regular, which proves by (17.1.8) that $A$ is regular.

Corollary (17.3.2).

If $A$ is a regular local ring, $A_{\mathfrak{p}}$ is regular for every prime ideal $\mathfrak{p}$ of $A$.

Indeed, one has seen (17.2.10) that dim. coh(A_𝔭) ≤ dim. coh(A), hence the conclusion follows at once from (17.3.1).

Proposition (17.3.3).

Let $A$, $B$ be two Noetherian local rings, $\rho :A\to B$ a local homomorphism, $\mathfrak{m}$ (resp. $\mathfrak{n}$) the maximal ideal of $A$ (resp. $B$), $k=A/\mathfrak{m}$ (resp. $k^{\prime }=B/\mathfrak{n}$) the residue field of $A$ (resp. $B$). One has therefore a canonical homomorphism of $k^{\prime }$-vector spaces

(17.3.3.1)              ψ : (𝔪/𝔪²) ⊗_k k′ → 𝔫/𝔫².

(i) If $B$ is regular and if $B$ is a flat $A$-module, $A$ is regular.

(ii) The following conditions are equivalent:

a) $B$ is regular and the homomorphism $\psi$ (17.3.3.1) is injective.

b) $A$ and $B$ are regular, and for a regular system of parameters $(x_i{)}_{1\le i\le m}$ of $A$, the $\varphi (x_i)$ are part of a regular system of parameters of $B$ (in which case this property holds for every regular system of parameters of $A$).

c) $B$ and $B{\otimes }_{A}k$ are regular, and $B$ is a flat $A$-module.

d) $A$ and $B{\otimes }_{A}k$ are regular, and $B$ is a flat $A$-module.

e) $A$ and $B{\otimes }_{A}k$ are regular, and one has

(17.3.3.2)              dim(B) = dim(A) + dim(B ⊗_A k).

(i) One has dim. coh(A) ≤ dim. coh(B) by (17.2.13), so it suffices to apply (17.3.1).

(ii) When $A$ is assumed regular and $(x_i)$ is a regular system of parameters of $A$, to say that $B$ is a flat $A$-module is equivalent, by virtue of (15.1.21), to saying that the sequence of $y_i=\varphi (x_i)$ is $B$-regular (since $B{\otimes }_{A}k$ is a $k$-flat module). On the other hand, since $\dim (A)=m$, and the sequence $(y_i)$ generates the ideal $\mathfrak{m}B$, the relation (17.3.3.2), which is also written $\dim (B/\mathfrak{m}B)=\dim (B)-m$, is equivalent to saying that the sequence $(y_i{)}_{1\le i\le m}$ is part of a system of parameters of $B$ (16.3.7). One therefore sees that conditions b), d) and e) are equivalent respectively to the following:

b′) $A$ and $B$ are regular and $(y_i{)}_{1\le i\le m}$ is part of a regular system of parameters of $B$.

d′) $A$ is regular, $B/({\sum }_{i=1}^m{y}_{i}B)$ is regular and the sequence $(y_i)$ is $B$-regular.

e′) $A$ is regular, $B/({\sum }_{i=1}^m{y}_{i}B)$ is regular and the sequence $(y_i)$ is part of a system of parameters of $B$.

Now, b′) and e′) are equivalent by virtue of (17.1.7), and since d′) implies e′) (16.4.1) and is implied by b′) (17.1.7), it is equivalent to them. The conjunction

of b) and d) trivially implies c), and by virtue of (i), c) implies d); we have therefore proved the equivalence of b), c), d) and e). Furthermore, it is clear that b) implies a), the classes of the $x_i$ in $\mathfrak{m}/{\mathfrak{m}}^2$ forming then a basis of this $k$-vector space and the classes of the $y_i$ in $\mathfrak{n}/{\mathfrak{n}}^2$ a linearly independent system of this $k^{\prime }$-vector space. It remains therefore to prove that a) implies e). Put $V=\mathfrak{m}/{\mathfrak{m}}^2$, $W=\mathfrak{n}/{\mathfrak{n}}^2$; it is immediate that one has $\psi (V{\otimes }_k{k}^{\prime })=({\mathfrak{m}}^2+\mathfrak{m}B)/{\mathfrak{n}}^2$. One has on the one hand, by virtue of (16.3.9),

(17.3.3.3)              dim(B) ≤ dim(A) + dim(B ⊗_A k);

in the second place, by (16.2.6), one has

(17.3.3.4)              dim(A) ≤ rg_k(V) ⊗_k k′.

Finally, in the local ring $B{\otimes }_{A}k=B/\mathfrak{m}B$, ${\mathfrak{n}}^{\prime }=\mathfrak{n}/\mathfrak{m}B$ is the maximal ideal, $k^{\prime }$ the residue field and ${\mathfrak{n}}^{\prime }/{\mathfrak{n}}^{\prime 2}$ is isomorphic to $\mathfrak{n}/({\mathfrak{n}}^2+\mathfrak{m}B)=W/\psi (V{\otimes }_k{k}^{\prime })$; one has therefore, by (16.2.6),

(17.3.3.5)              dim(B ⊗_A k) ≤ rg_{k′} W - rg_{k′} ψ(V ⊗_k k′).

Finally, since $B$ is assumed regular, one has $\dim (B)=r{g}_{k^{\prime }}W$ (17.1.1); one therefore concludes from (17.3.3.3), (17.3.3.4) and (17.3.3.5) that one has

        rg_{k′} W ≤ dim(A) + dim(B ⊗_A k) ≤ rg_{k′} W + rg_{k′}(V ⊗_k k′) - rg_{k′} ψ(V ⊗_k k′)

and to say that the two extreme terms of this inequality are equal means that $\psi$ is injective. Condition a) therefore necessarily implies that in each of the relations (17.3.3.3), (17.3.3.4) and (17.3.3.5), the two sides are equal; now, equality in (17.3.3.4) (resp. (17.3.3.5)) means that $A$ (resp. $B{\otimes }_{A}k$) is regular (17.1.1); one has therefore indeed proved that a) implies e).

Proposition (17.3.4).

Let $A$ be a regular local ring of dimension $n$, $\mathfrak{m}$ its maximal ideal. For every non-zero $A$-module of finite type $M$, one has

(17.3.4.1)              prof(M) + dim. proj(M) = n.

We argue by induction on $m=prof(M)$. If $m=0$, one knows (16.4.6, (i)) that there exists a submodule $N$ of $M$ isomorphic to $k=A/\mathfrak{m}$; applying the exact sequence of Tor to the exact sequence $0\to N\to M\to M/N\to 0$, one obtains an exact sequence

                        Tor^A_{n+1}(M/N, k) → Tor^A_n(N, k) → Tor^A_n(M, k),

and the hypothesis that $A$ is regular implies $To{r}_{n+1}^A(M/N,k)=0$ ((17.3.1.1) and (17.2.6)), hence $To{r}_n^A(k,k)$ is isomorphic to a sub-$A$-module of $To{r}_n^A(M,k)$; applying again (17.3.1.1) and (17.2.6), one sees that $To{r}_n^A(M,k)\ne 0$, hence $\dim .proj(M)\ge n$ (17.2.6); but since $\dim .proj(M)\le n$ by (17.3.1.1), one has indeed $\dim .proj(M)=n$. Suppose now $m>0$, and let $x$ be an $M$-regular element belonging to $\mathfrak{m}$; one knows then (16.4.6, (i)) that one has $prof(M/xM)=m-1$, and on the other hand (17.2.7) dim. proj(M/xM) = dim. proj(M) + 1; the induction hypothesis at once proves the relation (17.3.4.1).

Corollary (17.3.5).

(i) Let $A$ be a regular local ring of dimension $n$; for an $A$-module $M\ne 0$ of finite type to be free, it is necessary and sufficient that $M$ be a Cohen-Macaulay $A$-module of dimension $n$.

(ii) Let $A$ be a regular local ring, $B$ a local ring, $\rho :A\to B$ a local homomorphism making $B$ an $A$-module of finite type. For $B$ to be a free $A$-module, it is necessary and sufficient that $B$ be a Cohen-Macaulay ring and that $\rho$ be injective (or, what comes to the same thing (16.1.5), that $\dim (B)=\dim (A)$).

(i) This follows from (17.3.4.1) and from the fact that for an $A$-module of finite type, it comes to the same to say that this module is projective or free (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 5); the free $A$-modules of finite type $M$ are therefore characterized by the relation $\dim .proj(M)=0$ (17.2.2, (i)).

(ii) To say that $B$ is a Cohen-Macaulay ring is equivalent to saying that $B$ is a Cohen-Macaulay $A$-module (16.5.3), hence it suffices to apply (i), since $\dim B=\dim A$ (16.1.5).

(17.3.6) One says that a Noetherian ring $A$ is regular if for every prime ideal $\mathfrak{p}$ of $A$, the local ring $A_{\mathfrak{p}}$ is regular; when $A$ itself is a local ring, it follows from (17.3.2) that this definition is equivalent to that of (17.1.2). For $A$ to be regular it suffices, by virtue of (17.3.2), that $A_{\mathfrak{m}}$ be regular for every maximal ideal $\mathfrak{m}$ of $A$. Moreover, it follows at once from this definition that for every multiplicative subset $S$ of $A$, $S^{-1}A$ is regular.

Proposition (17.3.7).

If $A$ is a regular Noetherian ring, every polynomial ring $A[T_1,\cdots ,T_n]$ is regular.

It is evidently enough to prove that the polynomial ring $B=A[T]$ is regular; since $B$ is a free $A$-module, for every prime ideal $\mathfrak{q}$ of $B$, $B_{\mathfrak{q}}$ is a flat $A_{\mathfrak{p}}$-module, where $\mathfrak{p}=\mathfrak{q}\cap A$ $(0_I,\mathrm{6.3.1})$; it therefore suffices, by virtue of (17.1.10) ¹, to prove that $B_{\mathfrak{q}}/\mathfrak{p}{B}_{\mathfrak{q}}$ is regular, and since this ring is a local ring at a prime ideal of $A_{\mathfrak{p}}[T]/\mathfrak{p}{A}_{\mathfrak{p}}[T]=k[T]$ (where $k$ is the residue field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$), it suffices to prove that $C=k[T]$ is regular; now, $C$ being principal, the local rings at the prime ideals of $C$ are discrete valuation rings or a field (for the ideal (0)), hence regular (17.1.4), which completes the proof.

Corollary (17.3.8).

If $A$ is a regular ring, every formal power series ring $A[[T_1,\cdots ,T_n]]$ is regular.

Let $\mathfrak{J}$ be the ideal generated by the $T_i$ in the polynomial ring $A[T_1,\cdots ,T_n]$; since the latter is regular by (17.3.7), and $A[[T_1,\cdots ,T_n]]$ is the completion of $A[T_1,\cdots ,T_n]$ for the $\mathfrak{J}$-preadic topology (Bourbaki, Alg. comm., chap. III, §2, n° 12, Example 1), the conclusion follows from the:

Lemma (17.3.8.1).

Let $A$ be a regular ring, $\mathfrak{J}$ an ideal of $A$, Â the separated completion of $A$ for the $\mathfrak{J}$-preadic topology. Then Â is regular.

It suffices indeed (17.3.6) to see that for every maximal ideal $\mathfrak{n}$ of Â, the local ring ${\hat{A}}_{\mathfrak{n}}$ is regular; one knows (Bourbaki, Alg. comm., chap. III, §3, n° 4, prop. 8) that $\mathfrak{n}$

is of the form $\mathfrak{m}\hat{A}$, where $\mathfrak{m}$ is a maximal ideal of $A$ containing $\mathfrak{J}$, and that the canonical homomorphism $A\to \hat{A}$ gives an injective homomorphism $A_{\mathfrak{m}}\to {\hat{A}}_{\mathfrak{n}}$, such that the $\mathfrak{n}{\hat{A}}_{\mathfrak{n}}$-preadic topology of ${\hat{A}}_{\mathfrak{n}}$ induces on $A_{\mathfrak{m}}$ the $\mathfrak{m}{A}_{\mathfrak{m}}$-preadic topology, and such that $A_{\mathfrak{m}}$ is dense in ${\hat{A}}_{\mathfrak{n}}$ for the $\mathfrak{n}{\hat{A}}_{\mathfrak{n}}$-preadic topology; consequently the completions of the Noetherian local rings $A_{\mathfrak{m}}$ and ${\hat{A}}_{\mathfrak{n}}$ are canonically isomorphic. Now, by hypothesis $A_{\mathfrak{m}}$ is regular, hence so is its completion (17.1.5), and since the completion of ${\hat{A}}_{\mathfrak{n}}$ is regular, so is ${\hat{A}}_{\mathfrak{n}}$ by (17.1.5).

Corollary (17.3.9).

Let $A$ be a Noetherian ring, quotient of a regular Noetherian ring $B$. If $C$ is an $A$-algebra of finite type, every ring of fractions $S^{-1}C$ of $C$ is a quotient of a regular ring.

Indeed, $C$ is a quotient of a polynomial ring $A[T_1,\cdots ,T_n]$; since $A=B/\mathfrak{J}$, where $\mathfrak{J}$ is an ideal of $B$, one has $A[T_1,\cdots ,T_n]=B[T_1,\cdots ,T_n]/\mathfrak{J}B[T_1,\cdots ,T_n]$; by virtue of (17.3.7), one can therefore restrict to the case where $C$ is a quotient of a regular ring $B^{\prime }$; but if $S^{\prime }$ is the inverse image of $S$ in $B^{\prime }$, $S^{\prime }$ is a multiplicative subset of $B^{\prime }$ and $C$ is a quotient ring of $S^{\prime -1}{B}^{\prime }$, so that ultimately everything reduces to showing that when $C$ is regular, so is $S^{-1}C$, which was seen in (17.3.6).

The importance of quotients of regular rings lies among others in the preceding property and in the fact that they are catenary rings (16.5.12).

All the rings of importance in applications to algebraic geometry are quotients of regular rings.