Éléments de Géométrie Algébrique — English

§6. Flat morphisms of locally Noetherian preschemes

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a morphism. For every $y\in Y$, the fibre $f^{-1}(y)=X{\times }_Y\mathrm{Spec}(k(y))$ is also a locally Noetherian prescheme: it is enough to check this when $Y=\mathrm{Spec}(A)$ and $X=\mathrm{Spec}(B)$ are spectra of Noetherian rings, and then $B{\otimes }_{A}k(y)$ is a ring of fractions of the quotient ring $B/\mathfrak{J}B$, hence Noetherian. We propose first in this section to relate the properties of $X$, of $Y$ and of the fibres $f^{-1}(y)$ (where $y$ runs through $Y$), under the hypothesis that the morphism $f$ is flat. The questions treated will reduce to the study of the relations between the Noetherian local rings ${\mathcal{O}}_y$, ${\mathcal{O}}_x$ and ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)$, the homomorphism ${\mathcal{O}}_y\to {\mathcal{O}}_x$ being local and making ${\mathcal{O}}_x$ into a flat ${\mathcal{O}}_y$-module. In nos. (6.11) to (6.13) (rather separate from the rest of the section, by their "absolute" rather than "relative" nature), we apply certain of the preceding results to find criteria allowing us to assert that the singular locus (or certain analogous sets) of certain preschemes are closed sets; these criteria will play an important role in §7.

6.1. Flatness and dimension

Proposition (6.1.1).

Let $A$, $B$ be two Noetherian local rings, $\mathfrak{m}$ the maximal ideal of $A$, $k=A/\mathfrak{m}$ its residue field, $\varphi :A\to B$ a local homomorphism. Suppose that for every prime ideal $\mathfrak{p}$ of $A$ distinct from $\mathfrak{m}$, and for every minimal element $\mathfrak{q}$ of the set of prime ideals of $B$ containing $\mathfrak{p}B$, ${\varphi }^{-1}(\mathfrak{q})$ is distinct from $\mathfrak{m}$ (in other words, that no irreducible component of $\mathrm{Spec}(B/\mathfrak{p}B)$ is contained in the inverse image of the closed point $\mathfrak{m}$ of $\mathrm{Spec}(A)$ in $\mathrm{Spec}(B)$). Then one has

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

We argue by induction on $n=\dim (A)$; the assertion is trivial for $n=0$, since $\mathfrak{m}B$ is then contained in the nilradical of $B$, because $\mathfrak{m}$ is the nilradical of $A$. We may therefore suppose $n>0$. Let ${\mathfrak{q}}_i$ ($1\le i\le s$) be the minimal prime ideals of $B$, and set ${\mathfrak{p}}_i={\varphi }^{-1}({\mathfrak{q}}_i)$; for every $i$, one has ${\mathfrak{p}}_i\ne \mathfrak{m}$, for otherwise (since $n>0$) there would exist a prime ideal $\mathfrak{p}\ne \mathfrak{m}$ contained in ${\mathfrak{p}}_i$, and since ${\mathfrak{q}}_i$ is minimal among prime ideals of $B$ containing $\mathfrak{p}B$, one would reach a contradiction with the hypothesis. Consequently $\mathfrak{m}$ is distinct from the union of the ${\mathfrak{p}}_i$ and of the minimal prime ideals ${\mathfrak{p}}_j^{\prime }$ ($1\le j\le r$) of $A$ (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2), and there exists $a\in \mathfrak{m}$ belonging to none of the ${\mathfrak{p}}_i$ nor the ${\mathfrak{p}}_j^{\prime }$. Set $A^{\prime }=A/aA$, $B^{\prime }=B/aB$; one has (0, 16.3.4)

  dim(A') = dim(A) − 1,           dim(B') = dim(B) − 1

by construction of $a$, and on the other hand $B^{\prime }{\otimes }_{A^{\prime }}k=B{\otimes }_{A}k=B/\mathfrak{m}B$, hence

  dim(B' ⊗_{A'} k) = dim(B ⊗_A k);

it therefore suffices to prove (6.1.1.1) for $A^{\prime }$ and $B^{\prime }$. But, by virtue of the bijective correspondence between ideals of $A$ (resp. $B$) containing $a$ and ideals of $A^{\prime }$ (resp. $B^{\prime }$), the hypotheses of the statement also hold for $A^{\prime }$ and $B^{\prime }$; one may therefore apply the inductive hypothesis, which completes the proof.

Corollary (6.1.2).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, $M\ne 0$ an $A$-module of finite type, $N\ne 0$ a $B$-module of finite type. If $N$ is a flat $A$-module (resp. if $B$ is a flat $A$-module), one has

  (6.1.2.1)        dim_B(M ⊗_A N) = dim_A(M) + dim_{B ⊗_A k}(N ⊗_A k)

(resp. (6.1.1.1)).

It suffices to prove the assertion concerning $N$; on the other hand, if $\mathfrak{b}$ is the annihilator of $N$, one may replace $B$ by $B/\mathfrak{b}$, and hence suppose that $Supp(N)=\mathrm{Spec}(B)$; the hypothesis then means that the morphism $f:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ corresponding to $\varphi$ is quasi-flat (2.3.3); one deduces (2.3.4) that for every prime ideal $\mathfrak{p}\ne \mathfrak{m}$ of $A$,

every irreducible component of $f^{-1}(V(\mathfrak{p}))$ dominates $V(\mathfrak{p})$, and consequently the condition of (6.1.1) is satisfied, whence the conclusion.

Corollary (6.1.3).

Let $A$, $B$ be two Noetherian local rings, $\mathfrak{m}$ the maximal ideal of $A$, $k$ its residue field, $\varphi :A\to B$ a local homomorphism. Suppose that $\dim (B{\otimes }_{A}k)=0$ (that is to say (0, 16.2.1), that $\mathfrak{m}B$ is an ideal of definition of $B$). Then one has $\dim (B)\le \dim (A)$. If moreover there exists a $B$-module of finite type $N$ which is a flat $A$-module and has support equal to $\mathrm{Spec}(B)$ (which holds in particular when $B$ is a flat $A$-module), one has $\dim (B)=\dim (A)$.

The first assertion follows from (5.5.1) and the second from (6.1.2).

Corollary (6.1.4).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a surjective morphism. For every closed subset $Z$ of $Y$, one has

  (6.1.4.1)            codim(f⁻¹(Z), X) ≤ codim(Z, Y).

Moreover, if $f$ is quasi-flat (2.3.3), the two sides of (6.1.4.1) are equal.

Indeed, if $y$ is a maximal point of $Z$ and $x$ a maximal point of the fibre $f^{-1}(y)$, one has $\dim ({\mathcal{O}}_x)\le \dim ({\mathcal{O}}_{Y,y})$ by virtue of (5.5.2); the inequality (6.1.4.1) follows from (5.1.2.1) and (5.1.3.1) and from the fact that $f$ is surjective. If moreover $f$ is quasi-flat, one knows (2.3.4) that every irreducible component of $f^{-1}(Z)$ dominates an irreducible component of $Z$; the maximal points of $f^{-1}(Z)$ are therefore the maximal points of the fibres $f^{-1}(y)$, where $y$ runs through the set of maximal points of $Z$ (0_I, 2.1.8), and at each of these maximal points $x$ one has $\dim ({\mathcal{O}}_x\otimes k(x))=0$; since moreover ${\mathcal{O}}_x$ is a flat ${\mathcal{O}}_y$-module, one has $\dim ({\mathcal{O}}_x)=\dim ({\mathcal{O}}_y)$ by virtue of (6.1.1), whence the equality in (6.1.4.1), in view of the fact that $f$ is surjective.

Proposition (6.1.1) admits the following partial converse:

Proposition (6.1.5).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, $M$ a $B$-module of finite type. Suppose that:

1° $A$ is a regular ring.

2° $M$ is a Cohen-Macaulay $B$-module.

3° One has dim_B(M) = dim(A) + dim_{B ⊗_A k}(M ⊗_A k).

Then $M$ is a flat $A$-module.

We proceed by induction on $n=\dim (A)$. If $n=0$, $A$ is a field since it is regular (0, 17.1.4) and the assertion is trivial. Suppose $n>0$; let $\mathfrak{m}$ be the maximal ideal of $A$, and let $x\in \mathfrak{m}$ be an element not belonging to ${\mathfrak{m}}^2$; one then knows (0, 17.1.8) that $A^{\prime }=A/xA$ is regular and $\dim (A^{\prime })=\dim (A)-1$. Set

  B' = B/xB,            M' = M/xM = M ⊗_A A';

one has therefore

  B' ⊗_{A'} k = B ⊗_A k,           M' ⊗_{A'} k = M ⊗_A k

and by virtue of (5.5.1.2) one has

  dim_{B'}(M') ≤ dim_B(M) + dim_{B' ⊗_{A'} k}(M' ⊗_{A'} k);

one therefore concludes from (0, 16.3.4) that one has

  dim_B(M) ≤ dim_{B'}(M') + 1 ≤ (dim(A') + dim_{B ⊗_A k}(M ⊗_A k)) + 1 = dim(A) + dim_{B ⊗_A k}(M ⊗_A k).

Since by hypothesis the extreme members of these inequalities are equal, one necessarily has: (i) ${\dim }_{B^{\prime }}(M^{\prime })={\dim }_B(M)-1$, and since $M$ is by hypothesis a Cohen-Macaulay $B$-module, this means that $x$ is $M$-regular (0, 16.1.9 and 16.5.5); (ii) dim_{B'}(M') = dim(A') + dim_{B' ⊗_{A'} k}(M' ⊗_{A'} k); since $M^{\prime }$ is a Cohen-Macaulay $B^{\prime }$-module by virtue of (i) and of (0, 16.1.9 and 16.5.5) and since $A^{\prime }$ is regular, the inductive hypothesis proves that $M^{\prime }$ is a flat $A^{\prime }$-module; one then deduces from (0_III, 10.2.7) that $M$ is a flat $A$-module, since $x$ is $M$-regular by (i).

6.2. Flatness and projective dimension

Proposition (6.2.1).

(i) Let $A$, $B$ be two rings, $\varphi :A\to B$ a homomorphism such that $B$ is a flat $A$-module. Then, for every $A$-module $M$, one has

  (6.2.1.1)               dim. proj_B(M ⊗_A B) ≤ dim. proj_A(M).

(ii) Suppose moreover that $A$ is a Noetherian ring, $B$ is a faithfully flat $A$-module and $M$ is an $A$-module of finite type; then

  (6.2.1.2)               dim. proj_B(M ⊗_A B) = dim. proj_A(M).

(i) One may restrict to the case where $n=\dim .pro{j}_A(M)$ is finite; there exists therefore a left resolution

  0 → P_n → P_{n−1} → ⋯ → P_0 → M → 0

of $M$ by projective $A$-modules; since $B$ is a flat $A$-module, the sequence

  0 → P_n ⊗_A B → P_{n−1} ⊗_A B → ⋯ → P_0 ⊗_A B → M ⊗_A B → 0

is exact; moreover the $P_i{\otimes }_{A}B$ are projective $B$-modules; whence the conclusion.

(ii) Suppose $\dim .pro{j}_B(M{\otimes }_{A}B)=m$, and consider an exact sequence

  0 → R → P_{m−1} → P_{m−2} → ⋯ → P_0 → M → 0

where the $P_i$ ($0\le i\le m-1$) are projective $A$-modules of finite type; since $A$ is Noetherian, $R$ is also an $A$-module of finite type. Since $B$ is a flat $A$-module, one also has an exact sequence

  0 → R ⊗_A B → P_{m−1} ⊗_A B → ⋯ → P_0 ⊗_A B → M ⊗_A B → 0

and the hypothesis on $M{\otimes }_{A}B$ implies that $R{\otimes }_{A}B$ is a projective $B$-module (0, 17.2.1). Since $B$ is a faithfully flat $A$-module, one concludes that $R$ is a projective $A$-module (Bourbaki, Alg. comm., chap. I, §3, n° 6, prop. 12); hence $\dim .pro{j}_A(M)\le m$, which completes the proof.

Corollary (6.2.2).

Let $f:X\to Y$ be a flat morphism of preschemes, $\mathcal{E}$ a quasi-coherent ${\mathcal{O}}_Y$-Module. If $\dim .proj(\mathcal{E})\le n$ (0, 17.2.14), one has $\dim .proj(f\ast (\mathcal{E}))\le n$.

Proposition (6.2.3).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, $N$ a $B$-module of finite type. Suppose that $B$ and $N$ are flat $A$-modules. Then one has

  (6.2.3.1)               dim. proj_B(N) = dim. proj_{B ⊗_A k}(N ⊗_A k).

Consider in effect a left resolution of $N$ by free $B$-modules of finite type

  ⋯ → L_i → L_{i−1} → ⋯ → L_0 → N → 0.

Since the $L_i$ and $N$ are flat $A$-modules ($B$ being a flat $A$-module), it follows from (2.1.10) that the $Z_i(L_{\bullet })$ are flat $A$-modules, that $L_{\bullet }{\otimes }_{A}k$ is a left resolution of the $(B{\otimes }_{A}k)$-module $N{\otimes }_{A}k$, and that one has $Z_i(L_{\bullet }{\otimes }_{A}k)=Z_i(L_{\bullet }){\otimes }_{A}k$ for every $i$. Note that since $B$ is Noetherian, the $Z_i(L_{\bullet })$ are $B$-modules of finite type, hence the $Z_i(L_{\bullet }{\otimes }_{A}k)$ are $(B{\otimes }_{A}k)$-modules of finite type. Now, to say that a $B$-module (resp. a $(B{\otimes }_{A}k)$-module) of finite type is flat amounts to saying that it is free or also projective (0_III, 10.1.3). Since the $Z_i(L_{\bullet })$ are flat $A$-modules, it follows on the other hand from (0_III, 10.2.5) (where one takes $C=B$) that, for $Z_i(L_{\bullet })$ to be a flat $B$-module, it is necessary and sufficient that $Z_i(L_{\bullet }{\otimes }_{A}k)=Z_i(L_{\bullet }){\otimes }_{A}k$ be a flat $(B{\otimes }_{A}k)$-module. The smallest integer $n$ such that $Z_{n-1}(L_{\bullet })$ is a free $B$-module is therefore also the smallest integer such that $Z_{n-1}(L_{\bullet }{\otimes }_{A}k)$ is a free $(B{\otimes }_{A}k)$-module, which proves the proposition (0, 17.2.1).

6.3. Flatness and depth

Proposition (6.3.1).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, $M$ an $A$-module of finite type, $N$ a $B$-module of finite type. If $N$ is a flat $A$-module $\ne 0$, one has

  (6.3.1.1)            prof_B(M ⊗_A N) = prof_A(M) + prof_{B ⊗_A k}(N ⊗_A k).

One may restrict to the case where $M\ne 0$, otherwise both sides of (6.3.1.1) are equal to $+\infty$. We proceed by induction on the integer $n$ equal to the second member of (6.3.1.1) (which is finite by virtue of the hypotheses, of (0, 16.4.6.2) and of Nakayama's lemma applied to $N$). Suppose first $n=0$; then, if $\mathfrak{m}$ and $\mathfrak{n}$ denote the maximal ideals of $A$ and $B$ respectively, $\mathfrak{m}$ is a prime ideal associated with $M$ and $\mathfrak{n}/\mathfrak{m}B$ a prime ideal of $B/\mathfrak{m}B=B{\otimes }_{A}k$ associated with $N{\otimes }_{A}k$ (0, 16.4.6, (i)). One then concludes from (3.3.1) that $\mathfrak{n}$ is a prime ideal of $B$ associated with $M{\otimes }_{A}N$, hence (0, 16.4.6, (i)) that $pro{f}_B(M{\otimes }_{A}N)=0$. Suppose therefore $n>0$, and distinguish two cases:

a) Suppose $pro{f}_A(M)>0$. Let $x\in \mathfrak{m}$ be an $M$-regular element, and set

  A' = A/xA,           B' = B/xB,           M' = M/xM,           N' = N/xN;

one has

  B' ⊗_{A'} k = B ⊗_A k,          N' ⊗_{A'} k = N ⊗_A k

since $x\in \mathfrak{m}$, and

  M' ⊗_{A'} N' = M ⊗_A N/x(M ⊗_A N);

moreover, since $N$ is a flat $A$-module, the hypothesis that $x$ is $M$-regular entails that $x$ is also $(M{\otimes }_{A}N)$-regular (0_I, 6.1.1). One has consequently (0, 16.4.6, (ii) and 16.4.8)

  prof_{A'}(M') = prof_A(M) − 1,            prof_{B'}(M' ⊗_{A'} N') = prof_B(M ⊗_A N) − 1

and

  prof_{B' ⊗_{A'} k}(N' ⊗_{A'} k) = prof_{B ⊗_A k}(N ⊗_A k).

The equality (6.3.1.1) is therefore a consequence of the same relation for $A^{\prime }$, $B^{\prime }$, $M^{\prime }$ and $N^{\prime }$; but since $N$ is a flat $A$-module, $N^{\prime }=N{\otimes }_A{A}^{\prime }$ is a flat $A^{\prime }$-module (0_I, 6.2.1); one may consequently apply the inductive hypothesis, which proves (6.3.1.1) in this case.

b) Suppose $pro{f}_{B{\otimes }_{A}k}(N{\otimes }_{A}k)>0$. Consider an element $y\in \mathfrak{m}$ which is $(N{\otimes }_{A}k)$-regular; it follows from (0_III, 10.2.4) that $y$ is then $N$-regular and that

  N' = N/yN

is a flat $A$-module, since $N$ is supposed to be a flat $A$-module. Applying (0_I, 6.1.2) to the exact sequence of $A$-modules

  0 → N →^y N → N' → 0

one concludes isomorphisms

  (M ⊗_A N)/y(M ⊗_A N) ⥲ M ⊗_A N'         and          N' ⊗_A k ⥲ (N ⊗_A k)/y(N ⊗_A k)

and moreover that $y$ is $(M{\otimes }_{A}N)$-regular. One has consequently

  prof_{B ⊗_A k}(N' ⊗_A k) = prof_{B ⊗_A k}(N ⊗_A k) − 1

and

  prof_B(M ⊗_A N') = prof_B(M ⊗_A N) − 1;

the inductive hypothesis shows that the relation (6.3.1.1) is valid for $A$, $B$, $M$ and $N^{\prime }$, and from what precedes, one deduces (6.3.1.1) for $A$, $B$, $M$ and $N$.

Corollary (6.3.2).

Under the hypotheses of (6.3.1), suppose moreover $M\ne 0$ and $N\ne 0$; then one has

  (6.3.2.1)            coprof_B(M ⊗_A N) = coprof_A(M) + coprof_{B ⊗_A k}(N ⊗_A k).

This follows at once from (6.3.1.1) and (6.1.2) and from the definition of codepth (0, 16.4.9).

Corollary (6.3.3).

Under the hypotheses of (6.3.1), suppose moreover $M\ne 0$ and $N\ne 0$; for $M{\otimes }_{A}N$ to be a Cohen-Macaulay $B$-module, it is necessary and sufficient that $M$ be a Cohen-Macaulay $A$-module and that $N{\otimes }_{A}k$ be a Cohen- Macaulay $(B{\otimes }_{A}k)$-module.

This follows from corollary (6.3.2) and from the definition of Cohen-Macaulay modules by the fact that their codepth is zero (0, 16.5.1), taking into account that the condition $N\ne 0$ is equivalent to $N{\otimes }_{A}k\ne 0$ by virtue of Nakayama's lemma.

Corollary (6.3.4).

Under the hypotheses of (6.3.1), suppose moreover that $N{\otimes }_{A}k$ is a $B$-module of finite length; then one has

  (6.3.4.1)               prof_B(M ⊗_A N) = prof_A(M).

If moreover $M\ne 0$ and $N\ne 0$, one has

  (6.3.4.2)               coprof_B(M ⊗_A N) = coprof_A(M).

Indeed, it amounts to the same to say that $N{\otimes }_{A}k$ is a $B$-module of finite length or a $(B{\otimes }_{A}k)$-module of finite length, and one knows (0, 16.2.3) that modules of finite length $\ne 0$ are of dimension 0 and of depth 0.

Corollary (6.3.4) will be applicable in particular when $B{\otimes }_{A}k$ is a ring of finite length, that is to say when $\mathfrak{m}B$ is an ideal of definition of the ring $B$.

Corollary (6.3.5).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism.

(i) If, at a point $x\in X$, one has $coprof({\mathcal{O}}_x)\le n$, then, setting $y=f(x)$, one has $coprof({\mathcal{O}}_y)\le n$ and $coprof({\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y))\le n$; in particular, if ${\mathcal{O}}_x$ is a Cohen-Macaulay ring, so are ${\mathcal{O}}_y$ and ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)$.

(ii) Suppose conversely that ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)$ is a Cohen-Macaulay ring. Then, if $coprof({\mathcal{O}}_y)\le n$ (resp. if ${\mathcal{O}}_y$ is a Cohen-Macaulay ring), one has $coprof({\mathcal{O}}_x)\le n$ (resp. ${\mathcal{O}}_x$ is a Cohen-Macaulay ring).

It suffices to apply (6.3.2) for $M=A={\mathcal{O}}_y$ and $N=B={\mathcal{O}}_x$.

Corollary (6.3.6).

Let $A$ be a Cohen-Macaulay ring. Then $A^{\prime }=A[T_1,\cdots ,T_n]$ ($T_i$ indeterminates) is a Cohen-Macaulay ring.

Indeed, if one sets $Y=\mathrm{Spec}(A)$, $X=\mathrm{Spec}(A^{\prime })$ and if $f:X\to Y$ is the canonical morphism, $f$ is flat (since $A^{\prime }$ is a free $A$-module (2.1.2)) and for every $y\in Y$, $k(y)[T_1,\cdots ,T_n]$ is a regular ring (0, 17.3.7), hence a fortiori Cohen-Macaulay; it therefore suffices to apply (6.3.5).

Corollary (6.3.7).

Every quotient of a Cohen-Macaulay ring is universally catenary.

This follows from (6.3.6), from (5.6.1) and from (0, 16.5.12).

Proposition (6.3.8).

Let $A$ be a Noetherian local ring; suppose that there exists an $A$-module $M$ of finite type which is a Cohen- Macaulay $A$-module and whose support is equal to $\mathrm{Spec}(A)$. Let $B$ be a quotient ring of $A$, and let $f:\mathrm{Spec}(\hat{B})\to \mathrm{Spec}(B)$ be the canonical morphism; then, for every $x\in \mathrm{Spec}(B)$, $f^{-1}(x)$ is a Cohen-Macaulay scheme.

Since $B$ is an $A$-module of finite type, one has $\hat{B}=B{\otimes }_A\hat{A}$ (0_I, 7.3.3), hence $f$ is the restriction to $\mathrm{Spec}(\hat{B})$ of the canonical morphism $\mathrm{Spec}(\hat{A})\to \mathrm{Spec}(A)$, and the fibres of these two morphisms at a point of $\mathrm{Spec}(B)$ are therefore the same. One is therefore reduced to proving the proposition when $B=A$. So let $\mathfrak{p}={\mathfrak{j}}_x$ be a prime ideal of $A$; since by hypothesis $\mathfrak{p}\in Supp(M)$, one knows (0, 16.5.6 and 16.5.9) that there exists an $M$-regular sequence $(t_i{)}_{1\le i\le r}$ of elements of $\mathfrak{p}$ such that $N=M/({\sum }_{i=1}^r{t}_{i}M)$ is a Cohen-Macaulay $A$-module, $\mathfrak{p}$ a minimal associated prime ideal of $N$ and dim(N) = dim(M/𝔭M) = dim(A/𝔭). The same reasoning as at the start shows that one may replace $M$ by $N$ and $A$ by $A/({\sum }_{i=1}^r{t}_{i}A)$, and consequently suppose that $\mathfrak{p}$ is a minimal prime ideal of $A$.

Set for simplicity $A^{\prime }=\hat{A}$, $M^{\prime }=\hat{M}=M{\otimes }_A{A}^{\prime }$ (0_I, 7.3.3); one knows (0, 16.5.2) that $M^{\prime }$ is a Cohen-Macaulay $A^{\prime }$-module, which entails that for every prime ideal ${\mathfrak{p}}^{\prime }$ of $A^{\prime }$, $M_{{\mathfrak{p}}^{\prime }}^{\prime }$ is a Cohen-Macaulay $A_{{\mathfrak{p}}^{\prime }}^{\prime }$-module (0, 16.5.10). Taking (I, 3.6.5) into account, one sees that if one sets $A^{\prime \prime }=A^{\prime }{\otimes }_A{A}_{\mathfrak{p}}$ and $M^{\prime \prime }=M^{\prime }{\otimes }_A{A}_{\mathfrak{p}}$, M'' is a Cohen-Macaulay A''-module. For every prime ideal ${\mathfrak{p}}^{\prime \prime }$ of A'' over $\mathfrak{p}$, $M_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }=M_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}{A}_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }$ is therefore a Cohen-Macaulay $A_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }$-module (0, 16.5.10); on the other hand, $M_{\mathfrak{p}}$ is by hypothesis a Cohen-Macaulay $A_{\mathfrak{p}}$-module and $A_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }$ is a flat $A_{\mathfrak{p}}$-module since $A^{\prime }$ is a flat $A$-module (0_I, 7.3.3 and 6.3.2). One concludes from (6.3.3) that, if $k$ is the residue field of $A_{\mathfrak{p}}$, $k{\otimes }_{A_{\mathfrak{p}}}{A}_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }$ is a Cohen-Macaulay ring. But since $A_{\mathfrak{p}}$ is of dimension 0, the prime ideals of A'' correspond bijectively to those of $k{\otimes }_{A_{\mathfrak{p}}}{A}^{\prime \prime }$ (I, 3.5.7), and if ${\mathfrak{q}}^{\prime \prime }$ is the prime ideal of $k{\otimes }_{A_{\mathfrak{p}}}{A}^{\prime \prime }$ corresponding to ${\mathfrak{p}}^{\prime \prime }$, the local rings $(k{\otimes }_{A_{\mathfrak{p}}}{A}^{\prime \prime }{)}_{{\mathfrak{q}}^{\prime \prime }}$ and $k{\otimes }_{A_{\mathfrak{p}}}{A}_{{\mathfrak{p}}^{\prime \prime }}^{\prime \prime }$ are isomorphic. Consequently (0, 16.5.13), the ring $k{\otimes }_{A_{\mathfrak{p}}}{A}^{\prime \prime }$ is a Cohen-Macaulay ring. Q.E.D.

6.4. Flatness and property $(S_n)$

Proposition (6.4.1).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a morphism, $\mathcal{E}$ a coherent ${\mathcal{O}}_Y$-Module, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module that is $f$-flat.

(i) Let $x\in Supp(\mathcal{F})$ be such that $\mathcal{E}{\otimes }_Y\mathcal{F}$ has property $(S_k)$ at the point $x$; then $\mathcal{E}$ has property $(S_k)$ at the point $y=f(x)$.

(ii) Suppose that for every $y\in f(Supp(\mathcal{F}))$, ${\mathcal{F}}_y=\mathcal{F}{\otimes }_{{\mathcal{O}}_y}k(y)$ has property $(S_k)$; then, if for a point $y\in f(Supp(\mathcal{F}))$, $\mathcal{E}$ has property $(S_k)$ at the point $y$, $\mathcal{E}{\otimes }_Y\mathcal{F}$ has property $(S_k)$ at every point of $f^{-1}(y)$.

(i) One knows (Err_III, 30) that there exists a closed sub-prescheme $X^{\prime }$ of $X$ with underlying space $Supp(\mathcal{F})$ and a coherent ${\mathcal{O}}_{X^{\prime }}$-Module ${\mathcal{F}}^{\prime }$ such that $\mathcal{F}=j_{\ast }({\mathcal{F}}^{\prime })$, where $j$ is the canonical injection $X^{\prime }\to X$; one may replace $X$ by $X^{\prime }$ and $\mathcal{F}$ by ${\mathcal{F}}^{\prime }$, in other words suppose that $Supp(\mathcal{F})=X$. Let then $y^{\prime }$ be a generization of $y$ in $Y$; one knows (2.3.4) that there is a generization $x^{\prime }$ of $x$ in $X$ such that $y^{\prime }=f(x^{\prime })$; one may moreover suppose that $x^{\prime }$ is a generic point of an irreducible component of $f^{-1}(y^{\prime })$; one has consequently $\dim ({\mathcal{F}}_{x^{\prime }}{\otimes }_{{\mathcal{O}}_{y^{\prime }}}k(y^{\prime }))=0$. By virtue of (6.1.2), one therefore has, setting $\mathcal{G}=\mathcal{E}{\otimes }_Y\mathcal{F}$,

$$\dim ({\mathcal{G}}_{x^{\prime }})=\dim ({\mathcal{E}}_{y^{\prime }})$$

and by virtue of (6.3.1)

$$prof({\mathcal{G}}_{x^{\prime }})=prof({\mathcal{E}}_{y^{\prime }})$$

taking into account that depth is always at most equal to dimension. By hypothesis, one has

  prof(𝒢_{x'}) ≥ inf(k, dim(𝒢_{x'}))

hence

  prof(ℰ_{y'}) ≥ inf(k, dim(ℰ_{y'}))

which proves the first assertion.

(ii) Since for every generization $x^{\prime }$ of $x\in f^{-1}(y)$, $y^{\prime }=f(x^{\prime })$ is a generization of $y$, one may restrict to verifying that if $x\in Supp(\mathcal{F})$ and $f(x)=y$, one has prof(𝒢_x) ≥ inf(k, dim(𝒢_x)); it follows from (6.1.2) and (6.3.1) that one has

  dim(𝒢_x) = dim(ℰ_y) + dim(ℱ_x ⊗_{𝒪_y} k(y))

  prof(𝒢_x) = prof(ℰ_y) + prof(ℱ_x ⊗_{𝒪_y} k(y)).

By hypothesis, one has

  prof(ℰ_y) ≥ inf(k, dim(ℰ_y))

  prof(ℱ_x ⊗_{𝒪_y} k(y)) ≥ inf(k, dim(ℱ_x ⊗_{𝒪_y} k(y)))

whence, adding term by term,

  prof(𝒢_x) ≥ inf(k, dim(ℰ_y)) + inf(k, dim(ℱ_x ⊗_{𝒪_y} k(y))) ≥
            ≥ inf(k, dim(ℰ_y) + dim(ℱ_x ⊗_{𝒪_y} k(y))) = inf(k, dim(𝒢_x))

which proves (ii).

Corollary (6.4.2).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism, $\mathcal{E}$ a coherent ${\mathcal{O}}_Y$-Module. Suppose that for every $y\in Y$, the prescheme $f^{-1}(y)$ has property $(S_k)$. Then, for $f\ast (\mathcal{E})$ to satisfy $(S_k)$ at a point $x$, it is necessary and sufficient that $\mathcal{E}$ have property $(S_k)$ at the point $f(x)$.

Remark (6.4.3).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, $M$ a $B$-module of finite type which is a flat $A$-module, and suppose moreover that the $A$-module $A$ and the $(B{\otimes }_{A}k)$-module $M{\otimes }_{A}k$ have property $(S_n)$; can one then conclude that the $B$-module $M$ has property $(S_n)$? We do not know this, even when $n=1$, $M=B$ and $B=\hat{A}$; in other words, we do not know whether in general the completion of a Noetherian local ring satisfying $(S_n)$ also satisfies $(S_n)$, even for $n=1$. Setting $Y=\mathrm{Spec}(A)$, $X=\mathrm{Spec}(B)$ and $\mathcal{F}=\tilde{M}$, it would suffice, by virtue of (6.4.1, (ii)), to show that for every $y\in Y$, ${\mathcal{F}}_y$ has property $(S_n)$ (and not only when $y$ is the closed point of $Y$); it would also suffice to show that the set $U$ of $x\in X$ such that ${\mathcal{F}}_x\otimes k(f(x))$ satisfies $(S_n)$ is open in $X$ (since it contains the closed point of $X$ by hypothesis). We shall show below (12.1.4) that $U$ is open when $B$ is a local ring of an $A$-algebra of finite type and $M=B$. On the other hand, for $B=\hat{A}$ and $M=B$, we saw in (6.3.8) that the preschemes $f^{-1}(y)$ are Cohen-Macaulay preschemes (in other words satisfy $(S_n)$ for every $n$) when one supposes that $A$ is a quotient of a Cohen-Macaulay local ring. One concludes that when $A$ is a quotient of a Cohen-Macaulay local ring (or more generally of a local ring satisfying the hypotheses of (6.3.8)), for $A$ to satisfy $(S_n)$, it is necessary and sufficient that its completion Â satisfy $(S_n)$. It would remain to be seen whether this property persists without restriction on $A$.

6.5. Flatness and property $(R_n)$

Proposition (6.5.1).

Let $A$, $B$ be two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism for which $B$ is a flat $A$-module. Then:

(i) If $B$ is regular, $A$ is regular.

(ii) If $A$ and $B{\otimes }_{A}k$ are regular, $B$ is regular.

This proposition is given for the record, having already been proved in (0, 17.3.3).

Corollary (6.5.2).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism.

(i) If $X$ is regular at a point $x$, $Y$ is regular at the point $f(x)$.

(ii) If for $y\in f(X)$, $Y$ is regular at the point $y$ and if $f^{-1}(y)$ is a prescheme regular at a point $x$, then $X$ is regular at the point $x$.

Proposition (6.5.3).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism.

(i) If $X$ has property $(R_k)$ at a point $x$ (resp. if $X$ has property $(R_k)$), $Y$ has property $(R_k)$ at the point $f(x)$ (resp. $Y$ has property $(R_k)$ at every point of $f(X)$).

(ii) Suppose that for every $y\in f(X)$, the prescheme $f^{-1}(y)$ has property $(R_k)$; then, if for a point $y\in f(X)$, $Y$ has property $(R_k)$ at the point $y$, $X$ has property $(R_k)$ at every point of $f^{-1}(y)$.

(i) Set $y=f(x)$ and let $y^{\prime }$ be a generization of $y$; as in the proof of (6.4.1), there is a generic point $x^{\prime }$ of an irreducible component of $f^{-1}(y^{\prime })$ which is a generization of $x$; hence one has $\dim ({\mathcal{O}}_{x^{\prime }}{\otimes }_{{\mathcal{O}}_{y^{\prime }}}k(y^{\prime }))=0$ and by virtue of the hypothesis and of (6.1.2), $\dim ({\mathcal{O}}_{y^{\prime }})=\dim ({\mathcal{O}}_{x^{\prime }})$. The hypothesis entails either that $\dim ({\mathcal{O}}_x)>k$, in which case $\dim ({\mathcal{O}}_{y^{\prime }})>k$, or that ${\mathcal{O}}_x$ is a regular ring, and then ${\mathcal{O}}_{y^{\prime }}$ is a regular ring by virtue of (6.5.1).

(ii) Since, for every generization $x^{\prime }$ of $x\in f^{-1}(y)$, $y^{\prime }=f(x^{\prime })$ is a generization of $y$, one may restrict to verifying that if $\dim ({\mathcal{O}}_x)\le k$, then ${\mathcal{O}}_x$ is a regular ring. Now, one has, by virtue of (6.1.2)

  dim(𝒪_x) = dim(𝒪_y) + dim(𝒪_x ⊗_{𝒪_y} k(y))

hence if $\dim ({\mathcal{O}}_x)\le k$, one has a fortiori $\dim ({\mathcal{O}}_y)\le k$ and $\dim ({\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y))\le k$, and the hypothesis entails that ${\mathcal{O}}_y$ and ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)$ are regular rings. One then concludes from (6.5.1) that ${\mathcal{O}}_x$ is a regular ring.

Corollary (6.5.4).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism.

(i) If $X$ is normal at a point $x$, $Y$ is normal at the point $f(x)$.

(ii) If, for every $y\in f(X)$, $f^{-1}(y)$ is a normal prescheme and if $Y$ is normal at a point $y\in f(X)$, then $X$ is a normal prescheme at every point of $f^{-1}(y)$.

This follows at once from Serre's normality criterion (5.8.6) and from (6.4.1) applied for $k=2$ and (6.5.3) applied for $k=1$.

Remarks (6.5.5).

(i) Let $A$, $B$ be two Noetherian local rings, $\varphi :A\to B$ a local homomorphism such that $B$ is a flat $A$-module. Let $k$ be the residue field of $A$, and suppose that the two rings $A$ and $B{\otimes }_{A}k$ satisfy property $(R_i)$ (5.8.2); then it does not necessarily follow that $B$ satisfies $(R_i)$, even in the particular case where $i=0$ or $i=1$ and where $B$ is the completion Â of $A$. Nagata has indeed given an example where $A$ is normal (hence satisfies (R_1)) but where Â is not even reduced (hence does not satisfy (R_0)) [30]. One cannot apply proposition (6.5.3) to this case because the fibres $f^{-1}(y)$ do not necessarily satisfy property $(R_i)$ at points distinct from the closed point of $Y=\mathrm{Spec}(A)$. We shall however show below (7.8.3, (v)) that such phenomena do not occur for the Noetherian local rings which one most often encounters in applications.

(ii) The property of being integral for a prescheme does not behave at all like the properties we have just examined in this no. and the preceding ones: it can happen that $f:X\to Y$ is a flat morphism of finite type, that all the fibres $f^{-1}(y)$ are regular (and even geometrically regular (6.7.6)) and that $Y$ is integral, without $X$ being even locally integral. For example, let $k$ be an algebraically closed field of characteristic 0, and let $A$ be the integral ring $k[U,V]/(UV-(U+V{)}^3)$ (whose spectrum is therefore a "cubic with a double point"); $A$ is not integrally closed, and if $u$, $v$ are the canonical images of $U$, $V$ in $A$, the integral closure of $A$ is the ring $C=A[t]$, with $t=(u-v)/(u+v)$, which satisfies the equation $t^2=1+u+v$, whence one gets u = ½(t³ + t² − t), v = ½(−t³ + t² + t) and consequently $C=k[t]$, isomorphic to the ring of polynomials in one indeterminate over $k$. If $\mathfrak{m}=Au+Av$, the maximal ideal of $A$ (corresponding to the "double point" of the cubic), there are two maximal ideals ${\mathfrak{n}}_1$, ${\mathfrak{n}}_2$ of $C$, generated respectively by $t-1$ and $t+1$. Let then $B$ be the sub-ring of the product $C\times C$ formed of the pairs of polynomials $(f,g)$ such that $f(1)=g(-1)$ and $f(-1)=g(1)$ ($\mathrm{Spec}(B)$ is the scheme obtained by "gluing" two copies of $\mathrm{Spec}(C)$, the point ${\mathfrak{n}}_1$ (resp. ${\mathfrak{n}}_2$) of one of the two copies being "glued" to the point ${\mathfrak{n}}_2$ (resp. ${\mathfrak{n}}_1$) of the other; cf. chap. V, where this operation will be discussed in general). There are therefore two maximal ideals ${\mathfrak{r}}_1$, ${\mathfrak{r}}_2$ of $B$ above the maximal ideal $\mathfrak{m}$ of $A$. Moreover, since the process of "gluing" commutes with localization and completion, one verifies easily that the canonical homomorphisms ${\hat{A}}_{\mathfrak{m}}\to {\hat{B}}_{{\mathfrak{r}}_1}$ and ${\hat{A}}_{\mathfrak{m}}\to {\hat{B}}_{{\mathfrak{r}}_2}$ are bijective, and consequently (Bourbaki, Alg. comm., chap. III, §3, n° 5, prop. 10) $B_{{\mathfrak{r}}_1}$ and $B_{{\mathfrak{r}}_2}$ are flat $A_{\mathfrak{m}}$-modules, having moreover the same residue field as $A_{\mathfrak{m}}$. For every other maximal ideal $\mathfrak{p}$ of $A$, it is immediate that there are two maximal ideals ${\mathfrak{q}}_1$, ${\mathfrak{q}}_2$ of $B$ above $\mathfrak{p}$, and that the homomorphisms $A_{\mathfrak{p}}\to B_{{\mathfrak{q}}_1}$ and $A_{\mathfrak{p}}\to B_{{\mathfrak{q}}_2}$ are bijective. One sees thus that the morphism $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is flat and finite, and that all its fibres are geometrically regular (it is even étale, as we shall see later (17.6.3)); however it is immediate that $B$ is not integral.

6.6. Transitivity properties

Proposition (6.6.1).

For a locally Noetherian prescheme $Z$, denote by $P(Z)$ any one of the following properties:

a) $Z$ is a Cohen-Macaulay prescheme.

b) $Z$ satisfies $(S_n)$.

c) $Z$ is regular.

d) $Z$ satisfies $(R_n)$.

e) $Z$ is normal.

f) $Z$ is reduced.

Let then $X$, $Y$, $Z$ be three locally Noetherian preschemes, $f:X\to Y$, $g:Y\to Z$ two morphisms.

(i) Suppose that $f$ and $g$ are flat and that for every $y\in f(X)$ (resp. every $z\in g(Y)$), the prescheme $f^{-1}(y)$ (resp. $g^{-1}(z)$) satisfies $P$. Then $h=g\circ f$ is flat and for every $z\in h(X)$, $h^{-1}(z)$ satisfies $P$.

(ii) Suppose the following conditions hold: $f$ is faithfully flat, $h=g\circ f$ is flat, for every $y\in Y$ the prescheme $f^{-1}(y)$ satisfies $P$, and for every $z\in h(X)$ the prescheme $h^{-1}(z)$ satisfies $P$. Then $g$ is flat, and for every $z\in g(Y)$, $g^{-1}(z)$ satisfies $P$.

(i) One already knows (2.1.6) that $h$ is flat; on the other hand, for every $z\in Z$, $f_z=f\otimes 1_{k(z)}:h^{-1}(z)\to g^{-1}(z)$ is flat (2.1.4), and for every $y\in g^{-1}(z)$, $f_z^{-1}(y)$ is isomorphic to $f^{-1}(y)$ by transitivity of fibres (I, 3.6.4). The conclusion then follows respectively from (6.3.5, (ii)), (6.4.2), (6.5.2, (ii)), (6.5.3, (ii)), (6.5.4, (ii)) and (3.3.5, (ii)).

(ii) One already knows that $g$ is flat (2.2.13); moreover, for every $z\in Z$, $f_z$ is faithfully flat (2.2.13). The conclusion then follows respectively from (6.3.5, (i)), (6.4.2), (6.5.2, (i)), (6.5.3, (i)), (6.5.4, (ii)) and (2.1.13).

Remark (6.6.2).

Suppose $h$ flat, $f$ faithfully flat, and suppose that for every $z\in Z$, $h^{-1}(z)$ is of codepth $\le n$ (5.7.1); then it follows from the reasoning of (6.6.1, (ii)) and from (6.3.2) that $g^{-1}(z)$ is of codepth $\le n$ for every $z\in g(Y)$ and that for every $y\in Y$, $f^{-1}(y)$ is of codepth $\le n$.

6.7. Application to base changes in algebraic preschemes

Proposition (6.7.1).

Let $k$ be a field, $X$ a locally Noetherian $k$-prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. Let $k^{\prime }$ be an extension of $k$; set $X^{\prime }=X{\otimes }_k{k}^{\prime }$, ${\mathcal{F}}^{\prime }=\mathcal{F}{\otimes }_k{k}^{\prime }$ and let $p:X^{\prime }\to X$ be the canonical projection. Suppose either that $X$ is locally of finite type over $k$, or that $k^{\prime }$ is an extension of finite type of $k$, so that in both cases $X^{\prime }$ is locally Noetherian. Let $x^{\prime }$ be a point of $X^{\prime }$, $x=p(x^{\prime })$. Then:

(i) One has $coprof({\mathcal{F}}_x)=coprof({\mathcal{F}}_{x^{\prime }}^{\prime })$; in particular, for ${\mathcal{F}}_x$ to be a Cohen-Macaulay ${\mathcal{O}}_x$-module, it is necessary and sufficient that ${\mathcal{F}}_{x^{\prime }}^{\prime }$ be a Cohen-Macaulay ${\mathcal{O}}_{x^{\prime }}$-module.

(ii) For $\mathcal{F}$ to satisfy property $(S_n)$ at the point $x$, it is necessary and sufficient that ${\mathcal{F}}^{\prime }$ satisfy $(S_n)$ at the point $x^{\prime }$.

One knows that $p$ is faithfully flat (2.2.13); the two assertions are therefore respective consequences of (6.3.2) and (6.4.2), provided one proves that $p^{-1}(x)=\mathrm{Spec}(k(x){\otimes }_k{k}^{\prime })$ is a Cohen-Macaulay prescheme; since one of the two fields $k(x)$, $k^{\prime }$ is an extension of finite type of $k$ by hypothesis, one is reduced to proving the

Lemma (6.7.1.1).

Let $K$, $L$ be two extensions of a field $k$, one of which is of finite type (so that the ring $K{\otimes }_{k}L$ is Noetherian). Then $K{\otimes }_{k}L$ is a Cohen-Macaulay ring.

Suppose for example that $L$ is an extension of finite type of $k$, so that $L$ is a finite extension of a pure extension $k(\mathbf{t})$ of $k$ ($\mathbf{t}$ being a system $(t_i{)}_{1\le i\le m}$ of indeterminates). If one sets $A=K{\otimes }_{k}k(\mathbf{t})$, $B=K{\otimes }_{k}L$, $B$ is a flat $A$-module (0_I, 6.2.1) and of finite type; the morphism $h:\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is hence finite, and since every Artinian prescheme is Cohen-Macaulay, to prove that $B$ is a Cohen-Macaulay ring, it suffices to prove that $A$ is a Cohen-Macaulay ring (6.3.5). Now, if $S$ is the set of elements $\ne 0$ of $k[\mathbf{t}]$, one has $A=S^{-1}{A}^{\prime }$, where $A^{\prime }=K{\otimes }_{k}k[\mathbf{t}]=K[\mathbf{t}]$; by virtue of (0, 16.5.13), it suffices to prove that $A^{\prime }$ is a Cohen- Macaulay ring; but this follows from the fact that $A^{\prime }$ is regular (0, 17.3.7).

Corollary (6.7.2).

For ${\mathcal{O}}_x$ to be a Cohen-Macaulay ring (resp. for $X$ to satisfy $(S_n)$ at the point $x$), it is necessary and sufficient that ${\mathcal{O}}_{x^{\prime }}$ be a Cohen-Macaulay ring (resp. that $X^{\prime }$ satisfy $(S_n)$ at the point $x^{\prime }$).

Corollary (6.7.3).

Let $X$, $Y$ be two locally Noetherian $k$-preschemes, at least one of which is locally of finite type over $k$. Let $\mathcal{F}$ (resp. $\mathcal{G}$) be a coherent ${\mathcal{O}}_X$-Module (resp. a coherent ${\mathcal{O}}_Y$-Module). If $\mathcal{F}$ and $\mathcal{G}$ satisfy property $(S_n)$, so does $\mathcal{F}{\otimes }_k\mathcal{G}$.

The hypothesis entails that $X{\times }_{k}Y$ is locally Noetherian; let $p:X{\times }_{k}Y\to X$ and $q:X{\times }_{k}Y\to Y$ be the canonical projections, which are flat morphisms. Suppose for example that $X$ is locally of finite type over $k$; one may write $\mathcal{F}{\otimes }_k\mathcal{G}=p\ast (\mathcal{F}){\otimes }_{\mathcal{O}}q\ast (\mathcal{G})$, and since $\mathcal{F}$ is flat relative to the structure morphism $X\to \mathrm{Spec}(k)$, $p\ast (\mathcal{F})$ is $q$-flat (2.1.4). Apply criterion (6.4.1, (ii)) to the morphism $q$; it suffices to see that for every $y\in Y$, $p\ast (\mathcal{F}){\otimes }_{{\mathcal{O}}_y}k(y)$ has property $(S_n)$; but $p\ast (\mathcal{F}){\otimes }_{{\mathcal{O}}_y}k(y)=\mathcal{F}{\otimes }_{k}k(y)$, and since $X$ is locally of finite type over $k$, the conclusion follows from (6.7.1, (ii)).

Proposition (6.7.4).

For a locally Noetherian prescheme $Z$, and a point $z\in Z$, let $P(Z,z)$ be one of the following properties:

a) $Z$ is regular at the point $z$.

b) $Z$ satisfies $(R_n)$ at the point $z$.

c) $Z$ is normal at the point $z$.

d) $Z$ is reduced at the point $z$.

This being so, under the hypotheses and with the notation of (6.7.1), if $P(X^{\prime },x^{\prime })$ is true, so is $P(X,x)$. The converse is true if $k^{\prime }$ is a separable extension of $k$.

The case where $P$ is property d) has been listed only for the record, having already been treated (2.1.13 and 4.6.1). The same reasoning as at the start of (6.7.1) shows that the first assertion follows respectively from (6.5.2, (i)), (6.5.3, (i)) and (6.5.4, (i)); likewise, the second assertion will follow from (6.5.2, (ii)), (6.5.3, (ii)) and (6.5.4, (ii))

provided one proves that $p^{-1}(x)$ is a regular prescheme; in other words, one is reduced to proving the

Lemma (6.7.4.1).

Let $K$, $L$ be two extensions of a field $k$, one of which is of finite type. If $L$ is a separable extension of $k$, the ring $K{\otimes }_{k}L$ is regular.

Let us distinguish two cases:

A) $L$ is an extension of finite type of $k$. Then, with the notation of (6.7.1.1), one may suppose that $L$ is a finite separable extension of $k(\mathbf{t})$. For every $s\in \mathrm{Spec}(A)$, $k(s){\otimes }_{k(\mathbf{t})}L$ is then a direct composite of a finite number of fields, finite separable extensions of $k(s)$, hence a regular ring; it follows then from (6.5.2, (ii)) that it suffices to prove that the ring $A$ is regular; since $A=S^{-1}{A}^{\prime }$, it suffices to prove that $A^{\prime }$ is regular (0, 17.3.6); but it was seen that this was indeed so in the proof of (6.7.1.1).

B) Let $(L_{\lambda })$ be the filtered family of sub-extensions of $L$ which are of finite type over $k$; by virtue of A), each of the rings $C_{\lambda }=K{\otimes }_k{L}_{\lambda }$ is regular; on the other hand, for $L_{\lambda }\subset L_{\mu }$, $L_{\mu }$ is a flat $L_{\lambda }$-module, hence $C_{\mu }$ is a flat $C_{\lambda }$-module (0_I, 6.2.1). Since $C=K{\otimes }_{k}L=\underrightarrow{\lim }(K{\otimes }_k{L}_{\lambda })$ is Noetherian, one may apply (5.13.7), and $C$ is therefore regular.

Remarks (6.7.5).

(i) In the proof of the fact that $P(X,x)$ entails $P(X^{\prime },x^{\prime })$, one cannot dispense with the hypothesis that $k^{\prime }$ is a separable extension of $k$. This has already been seen (4.6.1) when $P$ is property d); but even when $X$ is geometrically integral (4.6.2), it can happen that $X$ is regular without $X^{\prime }$ being normal.

Take for example $X$ to be a normal algebraic curve over $k$ (II, 7.4.2); the local rings of $X$ being integrally closed and of dimension 1 are discrete valuation rings, hence regular (II, 7.1.6), and $X$ is therefore a regular $k$-scheme (and a fortiori satisfies $(R_n)$ for every $n\ge 0$). To say that $X$ is geometrically integral means then that the field $K$ of rational functions on $X$ is a separable and primary extension of $k$ (4.6.3). Now, take $k$ to be a non-perfect field of characteristic $p>2$, and let $a\in k$ be an element not belonging to $k^p$. Let $B$ be the polynomial ring k[S, T] in two indeterminates $S$, $T$; the polynomial $P(S,T)=T^2-S^p+a$ is irreducible in $B$, for one verifies at once that $S^p-a$ is not a square in the ring k[S]; hence $X=\mathrm{Spec}(A)$, where $A=B/PB$, is an irreducible affine curve over $k$. To see that the scheme $X$ is regular, it suffices to show that it is normal (II, 7.4.5); now $A=k[S][t]$, where $t$ is a root of the polynomial $P$ regarded as a polynomial in $T$ over k[S], so that the field $K=R(X)$ of rational functions on $X$ is the field of fractions $k(S)[T]/(P)$ of $A$, a quadratic extension of $k(S)$, hence separable over $k(S)$, and a fortiori over $k$. Since 2 is invertible in $k$, one verifies at once that $A$ is the integral closure of k[S] in $K$, hence is integrally closed, which shows that $X$ is regular. Moreover, if an element $f+tg$ of $K$ (with $f$, $g$ in $k(S)$) is algebraic over $k$, so are its norm and its trace over $k(S)$, and since $k(S)$ is a pure extension of $k$, one concludes easily that one must have $f\in k$ and $g=0$, in other words $k$ is algebraically closed in $K$, and a fortiori $K$ is a primary extension

of $k$ (4.3.1). However, if $k^{\prime }=k(a^{1/p})$, $X^{\prime }=X{\otimes }_k{k}^{\prime }$ is not normal, for in k'[S] one may write $S^p-a=(S-a^{1/p}{)}^p$, and $X^{\prime }$ is therefore isomorphic to $\mathrm{Spec}(A^{\prime })$, where $A^{\prime }=k^{\prime }[S][t^{\prime }]$, $t^{\prime }$ being a root of the polynomial $T^2-S^p$. Now, $A^{\prime }$ is not integrally closed, for the element $t^{\prime \prime }=t^{\prime }/S$ of the field of fractions $K^{\prime }=k^{\prime }(S)[t^{\prime }]$ of $A^{\prime }$ satisfies the integral dependence equation $t^{\prime \prime 2}-S^{p-2}=0$ over $A^{\prime }$ and does not belong to $A^{\prime }$. In the classical theory, this is expressed by saying that the "genus" over $k^{\prime }$ of the field $K^{\prime }$ of rational functions of $X^{\prime }$ is strictly less than that of $K$ over $k$.

(ii) As recalled in (i), it follows from (4.6.1) that when $X$ is an algebraic prescheme over $k$ that is not geometrically reduced, there are finite radicial extensions $k^{\prime }$ of $k$ such that $X^{\prime }=X{\otimes }_k{k}^{\prime }$ is not reduced. It is interesting to give an example of this fact where $X$ is a regular scheme over $k$, such that $k$ is algebraically closed in the field $K$ of rational functions of $X$. Let $k$ be a field of characteristic $p>0$, in which there exist two elements $a$, $b$ forming a $p$-free family over $k^p$. Again denoting by $B$ the ring k[S, T], let us consider this time the polynomial $P(S,T)=T^p-a{S}^p-b$; since $a{S}^p+b$ is not a $p$-th power in $k(S)$, $P$ is irreducible as a polynomial of k(S)[T], and the scheme $X_0=\mathrm{Spec}(B/PB)$ is therefore an integral affine curve over $k$, whose field of rational functions is $K=k(S)[t]$, where $t$ is a root of $P$; let us show that $k$ is algebraically closed in $K$. Suppose indeed that $K$ contains an element $z$ algebraic over $k$ and not in $k$; one would then also have $z\notin k(S)$, hence $[k(S)[z]:k(S)]>1$; since $[K:k(S)]=p$, one would have $K=k(S)[z]$, and since $K$ is radicial over $k(S)$, one would have $z^p\in k(S)$, hence $c=z^p\in k$ since $z$ is algebraic over $k$; but one would have $t^p=a{S}^p+b\in k^p(S^p)(c)$ hence $a$ and $b$ would belong to $k^p(c)$, which is absurd. Let then $X$ be the normalization of the curve X_0 in $K$, which is therefore a normal (and consequently regular) curve over $k$. If $k^{\prime }=k(a^{1/p},b^{1/p})$, it is clear that $a{S}^p+b$ is a $p$-th power in k'(S), hence $K{\otimes }_k{k}^{\prime }$ is not reduced, nor a fortiori the scheme $X{\otimes }_k{k}^{\prime }$.

Definition (6.7.6).

Let $k$ be a field, $X$ a locally Noetherian $k$-prescheme. We say that $X$ is geometrically regular at a point $x$ (resp. has the geometric property $(R_n)$ at a point $x$, resp. is geometrically normal at a point $x$) if, for every finite extension $k^{\prime }$ of $k$, $X^{\prime }=X_{(k^{\prime })}=X{\otimes }_k{k}^{\prime }$ is regular (resp. has property $(R_n)$, resp. is normal) at every point $x^{\prime }$ whose projection in $X$ is $x$. We say that $X$ is geometrically regular (resp. has the geometric property $(R_n)$ (or also is geometrically regular in codimension $\le n$), resp. is geometrically normal) if $X$ is geometrically regular (resp. has the geometric property $(R_n)$, resp. is geometrically normal) at every point.

We say that an algebra $A$ over $k$ is a geometrically regular ring (resp. geometrically normal, resp. geometrically reduced, resp. having the geometric property $(R_n)$) if $\mathrm{Spec}(A)$ has the same property.

If $X=\mathrm{Spec}(K)$, where $K$ is an extension of $k$, it amounts to the same to say that $X$ is geometrically regular, or geometrically normal, or geometrically reduced, or that $K$ is a separable extension of $k$: this follows from (4.6.1) and from the fact that if $K$ is a separable extension of $k$ and $k^{\prime }$ a finite extension of $k$, $K{\otimes }_k{k}^{\prime }$ is a direct composite of a finite number of fields (Bourbaki, Alg., chap. VIII, §7, n° 3, cor. 1 of th. 1).

Proposition (6.7.7).

Let $k$ be a field, $X$ a locally Noetherian $k$-prescheme, $x$ a point of $X$; denote by $Q(k^{\prime })$ the relation "$k^{\prime }$ is an extension of $k$, and $P(X_{(k^{\prime })},x^{\prime })$ is true for every point $x^{\prime }$ of $X_{(k^{\prime })}$ whose projection in $X$ is $x$", where $P(Z,z)$ is one of the properties a), b), c) of the statement (6.7.4). Then the following properties are equivalent:

a) $Q(k^{\prime })$ is true for every finite extension $k^{\prime }$ of $k$.

b) $Q(k^{\prime })$ is true for every finite radicial extension $k^{\prime }$ of $k$.

c) $Q(k^{\prime })$ is true for every extension of finite type $k^{\prime }$ of $k$.

Suppose moreover that $X$ is locally of finite type over $k$; then the three preceding properties are also equivalent to the following:

d) $Q(k^{\prime })$ is true for every extension $k^{\prime }$ of $k$.

e) $Q(k^{\prime })$ is true for a perfect extension $k^{\prime }$ of $k$.

f) $Q(k^{\prime })$ is true for every extension $k^{\prime }=k^{p^{-s}}$, where $p$ is the characteristic exponent of $k$, and $s>0$.

To prove the equivalence of a), b) and c), it suffices evidently to establish that b) entails c). So let $K$ be an extension of finite type of $k$, which is therefore the field of fractions of a $k$-algebra of finite type $A$; set $Y=\mathrm{Spec}(A)$. One knows (4.6.6) that there exists a finite radicial extension $k^{\prime }$ of $k$ such that $Y^{\prime }=(Y{\otimes }_k{k}^{\prime }{)}_{red}$ is separable over $k^{\prime }$; if ${\eta }^{\prime }$ is the generic point of $Y^{\prime }$, $K^{\prime }=k({\eta }^{\prime })$ is a separable extension of $k^{\prime }$ by (4.6.1). This being so, $P(X_{(k^{\prime })},x^{\prime })$ is true by hypothesis for every point $x^{\prime }$ of $X_{(k^{\prime })}$ above $x$, hence it follows from (6.7.4) that $P(X_{(K^{\prime })},x^{\prime \prime })$ is true for every point x'' of $X_{(K^{\prime })}$ above $x$, since $X_{(K^{\prime })}=(X_{(k^{\prime })}{)}_{(K^{\prime })}$, since x'' is above a point $x^{\prime }$ of $X_{(k^{\prime })}$, itself above $x$, and since $K^{\prime }$ is separable over $k^{\prime }$. But one also has $X_{(K^{\prime })}=(X_{(K)}{)}_{(K^{\prime })}$ and for every $x_0\in X_{(K)}$ above $x$, there exists $x^{\prime \prime }\in X_{(K^{\prime })}$ above $x_0$; it follows then from (6.7.4) that $P(X_{(K)},x_0)$ is true.

Suppose now $X$ locally of finite type over $k$, so that for every extension $K$ of $k$, $X_{(K)}$ is locally Noetherian. Since a radicial extension of $k$ is isomorphic to a sub-extension of an arbitrary perfect extension of $k$, it follows at once from (6.7.4) that e) implies f); likewise, every finite radicial extension of $k$ is contained in an extension $k^{p^{-s}}\subset k^{p^{-\infty }}$, hence f) entails b); d) trivially entails e), and finally e) entails d). Indeed, if $K$ is any extension of $k$, there exists an extension $K^{\prime }$ of $k$ containing (up to $k$-isomorphism) $k^{\prime }$ and $K$; since $K^{\prime }$ is a separable extension of $k^{\prime }$, one deduces from (6.7.4) that $Q(k^{\prime })$ is true, then that $Q(K)$ is true, reasoning as in the first part of the proof. It therefore suffices to prove that b) entails e). We shall take $k^{\prime }=k^{p^{-\infty }}$ in e).

The question being local on $X$, one may moreover restrict to the case where $X$ is affine and of finite type over $k$, replacing $X$ by a neighbourhood of $x$; so let $X=\mathrm{Spec}(B)$, where $B$ is a $k$-algebra of finite type; moreover, by definition of property $P$ in the cases considered, one may also replace $X$ by the local scheme of $X$ at the point $x$, hence suppose that $B$ is a Noetherian local ring. Set $B^{\prime }=B{\otimes }_k{k}^{\prime }$; $k^{\prime }$ is the inductive limit of its finite sub-extensions $k_{\lambda }^{\prime }$, and if one sets $B_{\lambda }^{\prime }=B{\otimes }_k{k}_{\lambda }^{\prime }$, the morphism $f_{\lambda }:\mathrm{Spec}(B^{\prime })\to \mathrm{Spec}(B_{\lambda }^{\prime })$ is a homeomorphism of $\mathrm{Spec}(B^{\prime })$ onto $\mathrm{Spec}(B_{\lambda }^{\prime })$ for every $\lambda$;

indeed $f_{\lambda }$ is bijective by virtue of (I, 3.5.2, 3.5.7 and 3.5.8); on the other hand, it is closed by (II, 6.1.10). One may now apply (5.13.6), which proves (by virtue of hypothesis b)) that b) entails e) when $P(Z,z)$ is property $(R_n)$ at the point $z$. The case where $P(Z,z)$ is the property of being regular (i.e. of having property $(R_n)$ for every $n$) follows trivially. Finally, taking into account Serre's normality criterion (5.8.6), b) entails again e) when $P(Z,z)$ is the property of being normal at the point $z$, for this amounts to saying that $Z$ has at the point $z$ properties (S_2) and (R_1): one applies then what precedes for $n=1$, and (6.7.2, (ii)) for $n=2$.

Corollary (6.7.8).

Let $k$ be a field; for a locally Noetherian $k$-prescheme $X$ and a point $x\in X$, denote by $P(X,x)$ one of the following properties:

(i) $coprof({\mathcal{O}}_x)\le n$.

(ii) ${\mathcal{O}}_x$ is a Cohen-Macaulay ring.

(iii) $X$ satisfies property $(S_n)$ at the point $x$.

(iv) $X$ is geometrically regular at the point $x$.

(v) $X$ has the geometric property $(R_n)$ at the point $x$.

(vi) $X$ is geometrically normal at the point $x$.

(vii) $X$ is geometrically reduced (i.e. separable) at the point $x$.