Éléments de Géométrie Algébrique — English

§15. $M$-regular sequences and $\mathcal{F}$-regular sequences

15.1. $M$-regular sequences and $M$-quasi-regular sequences

(15.1.1) Let $A$ be a ring, $N$ an $A$-module; in what follows, we shall denote by $N[T_1,\cdots ,T_n]$, for short, the graded $A$-module $N{\otimes }_{A}A[T_1,\cdots ,T_n]$ (the $T_i$ indeterminates), where $N$ is graded by the trivial grading and $A[T_1,\cdots ,T_n]$ by total degree. Let $M$ be an $A$-module, $f_i$ ($1\le i\le n$) elements of $A$, $\mathfrak{J}={\sum }_{i=1}^n{f}_{i}A$ the ideal of $A$ generated by the $f_i$, and equip $M$ with the $\mathfrak{J}$-preadic filtration; one then defines a surjective graded homomorphism of graded $A$-modules, of degree 0,

(15.1.1.1)              φ : (gr_0(M))[T_1, …, T_n] → gr_•(M)

in the following way: if $\xi \in g{r}_0(M)$ is the class mod $\mathfrak{J}M$ of an element $x\in M$, then to the pair formed by $\xi$ and a polynomial $P[T_1,\cdots ,T_n]$ homogeneous of degree $k$, one associates the class mod ${\mathfrak{J}}^{k+1}M$ of $P(f_1,\cdots ,f_n)x$; it is immediate that this class depends only on $\xi$ and $P$, and if $S_k$ is the set of homogeneous polynomials of degree $k$ in $A[T_1,\cdots ,T_n]$, one has thus defined an $A$-linear map

                        φ_k : gr_0(M) ⊗_A S_k → gr_k(M),

which is surjective by the definition of ${\mathfrak{J}}^{k}M$.

We shall say in what follows that an element $f\in A$ is not a zero-divisor in an $A$-module $M$ if the homothety $f_M:x\mapsto fx$ of $M$ is injective.

Lemma (15.1.2).

Let $M$ be an $A$-module, $f$ an element of $A$, and equip $M$ with the (fA)-preadic filtration. If $f$ is not a zero-divisor in $M$, the canonical homomorphism defined in (15.1.1):

$$(\mathrm{15.1.2.1})\varphi :(M/fM)[T]\to g{r}_{\bullet }(M)$$

is bijective. The converse holds if $M$ is separated for the (fA)-preadic topology.

One has here $g{r}_k(M)=f^{k}M/f^{k+1}M$; to say that $\varphi$ is injective means that for every $x\in M$ and every $k$, the relation $f^{k}x\in f^{k+1}M$ implies $x\in fM$; it is clear that this is so if $f$ is not a zero-divisor in $M$. Conversely, observe that the homothety $f_M:x\mapsto fx$ defines by passage to the quotients homomorphisms $g{r}_k(M)\to g{r}_{k+1}(M)$, hence a graded homomorphism $g$ of degree 1 of $g{r}_{\bullet }(M)$ into itself; and to say that $\varphi$ is injective is to say that $g$ is injective. If one assumes the filtration of $M$ to be separated, it follows that $f_M$ is injective (Bourbaki, Alg. comm., chap. III, §2, n° 8, cor. 1 of th. 1).

Corollary (15.1.3).

Let $A$ be a Noetherian ring, $M$ an $A$-module of finite type. For $f$ to be not a zero-divisor in $M$, it is necessary and sufficient that $\varphi$ be bijective.

One has only to prove that the condition is sufficient; to show that $f_M$ is injective, it suffices to show that for every prime ideal $\mathfrak{p}$ of $A$, the homothety of ratio $f^{\prime }=f/1\in A^{\prime }=A_{\mathfrak{p}}$ in the $A_{\mathfrak{p}}$-module $M_{\mathfrak{p}}=M^{\prime }$ is injective (Bourbaki, Alg. comm., chap. II, §3, n° 4, th. 1). Now one has $f^{\prime k}{M}^{\prime }/f^{\prime k+1}{M}^{\prime }\cong (f^{k}M/f^{k+1}M){\otimes }_A{A}^{\prime }$, and it is immediate that the canonical homomorphism ${\varphi }^{\prime }:(M^{\prime }/f^{\prime }{M}^{\prime })[T]\to g{r}_{\bullet }(M^{\prime })$ is equal to $\varphi \otimes 1_{A^{\prime }}$; if $\varphi$ is injective, then so is ${\varphi }^{\prime }$ $(0_I,\mathrm{1.3.2})$. One is thus reduced to the case where $A$ is a Noetherian local ring, whose maximal ideal we denote by $\mathfrak{m}$. If $f\notin \mathfrak{m}$, then $f$ is invertible and there is nothing to prove. If $f\in \mathfrak{m}$, one knows that $M$ is separated for the (fA)-preadic topology $(0_I,\mathrm{7.3.5})$, and it therefore suffices to apply (15.1.2).

Definition (15.1.4).

Let $A$ be a ring, $M$ an $A$-module. One says that an element $f\in A$ is $M$-regular (resp. $M$-quasi-regular) if $f$ is not a zero-divisor in $M$ (resp. if the homomorphism (15.1.2.1) is bijective).

We have thus just proved that every $M$-regular element is $M$-quasi-regular, and that the converse holds if $A$ is Noetherian and $M$ an $A$-module of finite type. When one no longer assumes $M$ of finite type, this converse may fail: for example, if $A$ is a principal ring which is not a field, $K$ its field of fractions and $M$ the $A$-module $K/A$, one has $fM=M$ for every $f\ne 0$ in $A$, hence both members of (15.1.2.1) reduce to 0, but $f$ is then a zero-divisor in $M$.

(15.1.5) Let us now assume $M$ equipped with a filtration $(M_k)$ such that $M_0=M$, and denote by $g{r}_{\bullet }(M)={⨁}_{k\ge 0}{M}_k/M_{k+1}$ the associated graded $A$-module. With the notations of (15.1.1), let us set, for every $k\ge 0$,

(15.1.5.1)              M'_k = M_k + 𝔍 M_{k-1} + … + 𝔍^{k-1} M_1 + 𝔍^k M_0.

It is clear that $(M_k^{\prime })$ is a filtration on $M$. When $M_k={\mathfrak{L}}^{k}M$, where $\mathfrak{L}$ is an ideal of $A$, the filtration $(M_k^{\prime })$ is none other than the $(\mathfrak{J}+\mathfrak{L})$-preadic filtration. We shall denote by $g{r}_{\bullet }^{\prime }(M)$ the graded $A$-module associated with $(M_k^{\prime })$. Let us further denote by $(g{r}_{\bullet }(M){\otimes }_A(A/\mathfrak{J}))[T_1,\cdots ,T_n]$ the tensor product of graded $A$-modules (II, 2.1.2) $(g{r}_{\bullet }(M){\otimes }_A(A/\mathfrak{J})){\otimes }_{A}A[T_1,\cdots ,T_n]$. One defines a surjective graded homomorphism of graded $A$-modules, of degree 0,

(15.1.5.2)              ψ : (gr_•(M) ⊗_A (A/𝔍))[T_1, …, T_n] → gr'_•(M)

in the following way: if $\alpha \in A/\mathfrak{J}$ is the class of an element $a\in A$, $\xi \in M_k/M_{k+1}$ the class of an element $x\in M_k$ and $P[T_1,\cdots ,T_n]$ a polynomial homogeneous of degree $h$, one associates to the triple $(\alpha ,\xi ,P)$ the class mod $M_{k+h+1}^{\prime }$ of $P(f_1,\cdots ,f_n)ax$; one verifies once more at once that this element depends only on $\xi$, $\alpha$ and $P$, and this defines the graded homomorphism (15.1.5.2), which is surjective by virtue of the definition (15.1.5.1). One recovers the homomorphism (15.1.1.1) by considering the filtration on $M$ such that $M_1=0$.

Lemma (15.1.6).

One keeps the notations of (15.1.5) and one assumes $n=1$, setting $f=f_1$. If $f$ is $g{r}_{\bullet }(M)$-regular, the homomorphism (15.1.5.2) is bijective.

Let $Q_k$ (resp. $Q_k^{\prime }$) be the sub-$A$-module of terms of degree $k$ in the first (resp. second) member of (15.1.5.2). Equip $Q_k$ with the filtration

(15.1.6.1)              (Q_k)_i = ∑_{j ≤ k-i} (gr_{k-j}(M) ⊗_A (A/fA)) T^j

so that $(Q_k{)}_0=Q_k$ and $(Q_k{)}_{k+1}=0$, and for this filtration one has

                        gr_i(Q_k) = (gr_i(M) ⊗_A (A/fA)) T^{k-i}.

Equip $Q_k^{\prime }$ with the filtration formed by the $\psi ((Q_k{)}_i)=(Q_k^{\prime }{)}_i$. Since these filtrations are finite, it suffices to prove that the homomorphisms ${\psi }_{ki}:g{r}_i(Q_k)\to g{r}_i(Q_k^{\prime })$ deduced from $\psi$ are injective (Bourbaki, Alg. comm., chap. III, §3, n° 8, cor. 1 of th. 1). Now one has $g{r}_i(Q_k)=((M_i/M_{i+1}){\otimes }_A(A/fA))T^{k-i}=(M_i/(f{M}_i+M_{i+1}))T^{k-i}$; on the other hand, $(Q_k^{\prime }{)}_i$ is the image of

                        M_k + fM_{k-1} + … + f^{k-i-1} M_{i+1}

in $M_k^{\prime }/M_{k+1}^{\prime }$. One is thus reduced to writing that, for $x\in M_i$, the relation

(15.1.6.2)              f^{k-i} x ∈ f M_i + f^2 M_{i-1} + … + f^{k-i-1} M_{i+1} + M'_{k+1}

implies $x\in f{M}_i+M_{i+1}$. Now, the second member of (15.1.6.2) is contained in $M_{k+1}^{\prime }+M_{i+1}$, and by virtue of (15.1.5.1), $M_{k+1}^{\prime }$ is contained in $M_{i+1}+f^{k+1-i}M$. The hypothesis on $f$ implies that $f$ is not a zero-divisor in $M/M_h$ for every $h$ (Bourbaki, loc. cit.). The relation $f^{k-i}x\in f^{k+1-i}M+M_{i+1}$ therefore implies (reasoning

in $M/M_{i+1}$) the existence of a $y\in M$ such that $x-fy\in M_{i+1}$, and since $x\in M_i$, one has $fy\in M_i$, whence (reasoning in $M/M_i$), $y\in M_i$, and finally $x\in f{M}_i+M_{i+1}$. Q.E.D.

Definition (15.1.7).

Let $(f_i{)}_{1\le i\le n}$ be a sequence of elements of $A$. One says that it is $M$-regular if, for $1\le i\le n$, $f_i$ is $(M/({\sum }_{1\le j\le i-1}{f}_{j}M))$-regular. One says that the sequence $(f_i)$ is $M$-quasi-regular if the canonical homomorphism $\varphi$ defined in (15.1.1.1) is bijective.

When $M=A$, one simply says "regular sequence" and "quasi-regular sequence".

Proposition (15.1.8).

With the notations of (15.1.5), suppose that the sequence $(f_i{)}_{1\le i\le n}$ is $g{r}_{\bullet }(M)$-regular. Then the canonical homomorphism (15.1.5.2) is bijective.

The proposition is none other than (15.1.6) for $n=1$; we reason by induction on $n$. Set ${\mathfrak{J}}^{\prime \prime }={\sum }_{i=1}^{n-1}{f}_{i}A$, so that $\mathfrak{J}={\mathfrak{J}}^{\prime \prime }+f_{n}A$, and let

                        M''_k = M_k + 𝔍'' M_{k-1} + … + 𝔍''^{k-1} M_1 + 𝔍''^k M_0;

one can then write

                        M'_k = M''_k + f_n M''_{k-1} + … + f_n^{k-1} M''_1 + f_n^k M''_0.

Denote by $g{r}_{\bullet }^{\prime \prime }(M)$ the graded $A$-module associated with the filtration $(M_k^{\prime \prime })$ of $M$. One can write, up to canonical isomorphisms,

        (gr_•(M) ⊗ (A/𝔍)) ⊗_A A[T_1, …, T_n]
                = (gr_•(M) ⊗ (A/𝔍''))[T_1, …, T_{n-1}] ⊗ (A/f_n A) ⊗ A[T_n]

and consequently (15.1.5.2) factors as

(15.1.8.1)              (gr''_•(M) ⊗ (A/f_n A))[T_n] → gr'_•(M)

and

        ((gr_•(M) ⊗ (A/𝔍''))[T_1, …, T_{n-1}]) ⊗ (A/f_n A) ⊗ A[T_n]
                ──ψ' ⊗ 1──→ gr''_•(M) ⊗ (A/f_n A) ⊗ A[T_n]

where ${\psi }^{\prime }$ is the map (15.1.5.2) in which one replaces $n$ by $n-1$, $\mathfrak{J}$ by ${\mathfrak{J}}^{\prime \prime }$ and $g{r}_{\bullet }^{\prime }(M)$ by $g{r}_{\bullet }^{\prime \prime }(M)$. The induction hypothesis implies that ${\psi }^{\prime }$ is bijective, and it remains therefore to see that the same holds for (15.1.8.1). Applying (15.1.6), it suffices to show that $f_n$ is $g{r}_{\bullet }^{\prime \prime }(M)$-regular. But $g{r}_{\bullet }^{\prime \prime }(M)$ is identified with $(g{r}_{\bullet }(M)\otimes (A/{\mathfrak{J}}^{\prime \prime }))[T_1,\cdots ,T_{n-1}]$; by hypothesis, the homothety of ratio $f_n$ in $g{r}_{\bullet }(M)\otimes (A/{\mathfrak{J}}^{\prime \prime })$ is bijective; consequently the same holds for the homothety of ratio $f_n$ in $g{r}_{\bullet }^{\prime \prime }(M)$, since $A[T_1,\cdots ,T_{n-1}]$ is a free $A$-module.

Proposition (15.1.9).

Let $M$ be an $A$-module, $(f_i{)}_{1\le i\le n}$ a sequence of elements of $A$. If the sequence $(f_i)$ is $M$-regular, it is $M$-quasi-regular. The converse holds if one assumes in addition that the $A$-modules $M$, $M/f_{1}M$, …, $M/({\sum }_{1\le j\le n-1}{f}_{j}M)$ are separated for the $\mathfrak{J}$-preadic topology (where $\mathfrak{J}={\sum }_{i=1}^n{f}_{i}A$).

The first assertion follows from (15.1.8), where one takes $M_1=0$, so that the homomorphism (15.1.5.2) reduces to (15.1.1.1). Conversely, suppose the homomorphism $\varphi$ of (15.1.1.1) is bijective. Note that if one sets ${\mathfrak{J}}^{\prime \prime }={\sum }_{i=1}^{n-1}{f}_{i}A$,

and if one denotes by $g{r}_{\bullet }^{\prime \prime }(M)$ the graded $A$-module associated with $M$ filtered by the ${\mathfrak{J}}^{\prime \prime }$-preadic filtration, then $\varphi$ factors as

(15.1.9.1)              (gr''_0(M) ⊗ (A/f_n A))[T_n] → gr_•(M)

and as ${\varphi }^{\prime }\otimes 1$, where ${\varphi }^{\prime }$ is the canonical homomorphism

$$(g{r}_0^{\prime \prime }(M))[T_1,\cdots ,T_{n-1}]\to g{r}_{\bullet }^{\prime \prime }(M).$$

Since both these homomorphisms are surjective, they are bijective if $\varphi$ is assumed bijective; hence ${\varphi }^{\prime }$ is necessarily injective, and consequently bijective since it is surjective. One can then reason by induction on $n$ (the case $n=1$ resulting from (15.1.2)), since on $M$, $M/f_{1}M$, …, $M/({\sum }_{j=1}^{n-2}{f}_{j}M)$ the ${\mathfrak{J}}^{\prime \prime }$-preadic topology is finer than the $\mathfrak{J}$-preadic topology, and is consequently separated. One thus sees that the sequence $(f_i{)}_{1\le i\le n-1}$ is $M$-regular. On the other hand, let us show that the hypothesis implies that if $y\in M/{\mathfrak{J}}^{\prime \prime }M$ is such that $f_{n}y\in f_n^{k+1}(M/{\mathfrak{J}}^{\prime \prime }M)$, then one has $y\in f_n^k(M/{\mathfrak{J}}^{\prime \prime }M)$. Indeed, if $x$ is a representative of $y$, let us show that one has $f_n^{k}x\in {\mathfrak{J}}^{k+1}M$; our assertion will then result from the fact that (15.1.9.1) is injective. The hypothesis implies $f_n^{k}x\in f_n^{k+1}M+{\mathfrak{J}}^{\prime \prime }{f}_n^{k}M\subset {\mathfrak{J}}^{k+1}M$, whence, since $\varphi$ is injective, $f_n^{k}x\in {\mathfrak{J}}^{k}M$; by induction, one thus obtains $x\in \mathfrak{J}M$, whence finally $f_n^{k}x\in {\mathfrak{J}}^{k+1}M$. One finishes the reasoning as in (15.1.2): on $M/{\mathfrak{J}}^{\prime \prime }M$, the $\mathfrak{J}$-preadic filtration is identical to the $(f_{n}A)$-preadic filtration, hence is separated by hypothesis, and one concludes that the homothety of ratio $f_n$ in $M/{\mathfrak{J}}^{\prime \prime }M$ is injective, which completes the proof that the sequence $(f_i{)}_{1\le i\le n}$ is $M$-regular.

Corollary (15.1.10).

With the notations of (15.1.9), suppose that for every $A$-module $N$ of finite type, $M{\otimes }_{A}N$ is separated for the $\mathfrak{J}$-preadic topology. Then, for the sequence $(f_i)$ to be $M$-regular, it is necessary and sufficient that it be $M$-quasi-regular.

One will note that the hypothesis of (15.1.10) implies that, if the sequence $(f_i{)}_{1\le i\le n}$ is $M$-regular, the same holds for the sequence $(f_{\sigma (i)})$ for every permutation $\sigma$ of the interval [1, n].

Corollary (15.1.11).

Let $A$ be a Noetherian ring, $M$ an $A$-module of finite type, $(f_i{)}_{1\le i\le n}$ a finite sequence of elements contained in the radical of $A$. Then, for the sequence $(f_i)$ to be $M$-regular, it is necessary and sufficient that it be $M$-quasi-regular.

This is a particular case of (15.1.10), since every $A$-module of finite type is then separated for the $\mathfrak{J}$-preadic topology $(0_I,\mathrm{7.3.5})$.

Remarks (15.1.12).

(i) When $A$ is a Noetherian local ring, $M$ an $A$-module of finite type, and when the $f_i$ are contained in the maximal ideal of $A$, the notion of $M$-regular sequence was introduced by J.-P. Serre under the name $M$-sequence [17].

(ii) Even if $A$ is Noetherian and $M=A$, a sequence $(f,g)$ of two elements of $A$ which is quasi-regular is not necessarily regular. For example, the sequence (0, 1) is quasi-regular, since then $\mathfrak{J}=fA+gA=A$, hence $g{r}_{\bullet }(A)=0$; but if $A\ne 0$ it is not regular; one will note however that in this case the sequence (1, 0) is regular, since 1 is not a zero-divisor in $A$, and since $A/A=0$, 0 is not a zero-divisor in this module (cf. (15.1.1)).

(iii) In (15.1.6), one cannot in general replace the hypothesis that $f$ is $g{r}_{\bullet }(M)$-regular by the hypothesis that it is $g{r}_{\bullet }(M)$-quasi-regular. To see this, take up again the example (15.1.4) where $A$ is a principal ring with field of fractions $K\ne A$, and take $M=K$, the filtration $(M_k)$ of $M$ being defined by $M_0=M=K$, $M_1=A$, $M_k=0$ for $k\ge 2$, whence $g{r}_0(M)=K/A$, $g{r}_1(M)=A$, $g{r}_k(M)=0$ for $k\ge 2$. If $f\ne 0$ in $A$, one has seen that $f$ is not $g{r}_{\bullet }(M)$-regular; but since one has $f^{k}g{r}_{\bullet }(M)/f^{k+1}g{r}_{\bullet }(M)=f^{k}A/f^{k+1}A$, it is immediate that the homomorphism (15.1.2.1) (where $M$ is replaced by $g{r}_{\bullet }(M)$) is injective, hence $f$ is $g{r}_{\bullet }(M)$-quasi-regular. But with the notations of (15.1.6), one then has $M_k^{\prime }=K$ for every $k\ge 0$, hence $g{r}_{\bullet }^{\prime }(M)=0$, while the first member of (15.1.5.2) is not reduced to 0.

Proposition (15.1.13).

Let $A$ be a ring, $M$, $N$ two $A$-modules, $(f_i{)}_{1\le i\le n}$ a sequence of elements of $A$. If the sequence $(f_i)$ is $M$-regular and if $N$ is a flat $A$-module, then the sequence $(f_i)$ is $(M{\otimes }_{A}N)$-regular. The converse holds if $N$ is a faithfully flat $A$-module.

Indeed (without hypothesis on $N$), $(M{\otimes }_{A}N)/({\sum }_{j=1}^{i-1}{f}_j(M{\otimes }_{A}N))$ is isomorphic to $(M/({\sum }_{j=1}^{i-1}{f}_{j}M)){\otimes }_{A}N$; it then suffices to apply $(0_I,\mathrm{6.1.1})$ and $(0_I,\mathrm{6.4.1})$ to the homothety of ratio $f_i$ in $M/({\sum }_{j=1}^{i-1}{f}_{j}M)$.

Corollary (15.1.14).

Let $A$ be a ring, $M$ an $A$-module, $(f_i{)}_{1\le i\le n}$ a finite sequence of elements of $A$. Let $\rho :A\to B$ be a ring homomorphism, $M^{\prime }=M_{(B)}=M{\otimes }_{A}B$, $f_i^{\prime }=\rho (f_i)$ for $1\le i\le n$. If $B$ is a flat $A$-module and if the sequence $(f_i)$ is $M$-regular, then the sequence $(f_i^{\prime })$ is $M^{\prime }$-regular; the converse holds if $B$ is a faithfully flat $A$-module.

Indeed, $M^{\prime }/({\sum }_{j=1}^{i-1}{f}_j^{\prime }{M}^{\prime })$ is identified with $(M/({\sum }_{j=1}^{i-1}{f}_{j}M)){\otimes }_{A}B$, and the homothety of ratio $f_i^{\prime }$ in the $B$-module $M^{\prime }/({\sum }_{j=1}^{i-1}{f}_j^{\prime }{M}^{\prime })$ with the homothety of ratio $f_i$ in the $A$-module $(M/({\sum }_{j=1}^{i-1}{f}_{j}M)){\otimes }_{A}B$; the assertion therefore follows from (15.1.13).

Proposition (15.1.15).

Let $A$ be a ring, $B$ a commutative $A$-algebra, $(f_i{)}_{1\le i\le n}$ a finite sequence of elements of $B$, $M$ a $B$-module. Suppose that the sequence $(f_i)$ is $M$-regular and that each of the $A$-modules $M/({\sum }_{j=1}^{i-1}{f}_{j}M)$ ($1\le i\le n$) is flat. Then, for every ring homomorphism $\rho :A\to A^{\prime }$, if one sets $B^{\prime }=B{\otimes }_A{A}^{\prime }$, $M^{\prime }=M{\otimes }_A{A}^{\prime }$ and $f_i^{\prime }=f_i\otimes 1$ ($1\le i\le n$), the sequence $(f_i^{\prime })$ is $M^{\prime }$-regular and the $A^{\prime }$-modules $M^{\prime }/({\sum }_{j=1}^{i-1}{f}_j^{\prime }{M}^{\prime })$ ($1\le i\le n$) are flat.

Consider the sequence (exact by hypothesis)

                        0 → M ──f_1──→ M → M/f_1 M → 0

(the arrow $M\to M$ being the homothety of ratio $f_1$ in $M$). The hypothesis that $M/f_{1}M$ is $A$-flat implies that the sequence

                        0 → M ⊗_A A' ──f_1 ⊗ 1──→ M ⊗_A A' → (M/f_1 M) ⊗_A A' → 0

is exact $(0_I,\mathrm{6.1.2})$. This shows that the homothety of ratio $f_1^{\prime }$ in $M^{\prime }$ is injective. Set next, for $1\le i\le n-1$, $M_i=M/({\sum }_{j=1}^i{f}_{j}M)$, $M_i^{\prime }=M^{\prime }/({\sum }_{j=1}^i{f}_j^{\prime }{M}^{\prime })$; one has $M_i^{\prime }=M_i{\otimes }_A{A}^{\prime }$, $M_i/f_{i+1}{M}_i=M_{i+1}$, $M_i^{\prime }/f_{i+1}^{\prime }{M}_i^{\prime }=M_{i+1}^{\prime }$; it suffices to replace in the preceding reasoning $M$ and $f_1$ by $M_i$ and $f_{i+1}$ to conclude that the homothety of ratio $f_{i+1}^{\prime }$ in $M_i^{\prime }$ is injective, by virtue of the flatness of $M_i/f_{i+1}{M}_i$. The last assertion follows from $(0_I,\mathrm{6.2.1})$.

Proposition (15.1.16).

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. Let $(f_i{)}_{1\le i\le n}$ be a sequence of elements of the maximal ideal of $B$, and for every $i$, let $g_i$ be the image of $f_i$ in $B{\otimes }_{A}k$. The following conditions are equivalent:

a) The sequence $(f_i)$ is $M$-regular and the quotients $M_i=M/({\sum }_{j=1}^i{f}_{j}M)$ are flat $A$-modules for $1\le i\le n$.

b) The sequence $(f_i)$ is $M$-regular and the $A$-module $M_n$ is flat.

c) The $A$-module $M$ is flat and the sequence $(g_i{)}_{1\le i\le n}$ is $(M{\otimes }_{A}k)$-regular.

d) The $A$-module $M$ is flat, and for every ring homomorphism $\rho :A\to A^{\prime }$, if one sets $B^{\prime }=B{\otimes }_A{A}^{\prime }$, $M^{\prime }=M{\otimes }_A{A}^{\prime }$, $f_i^{\prime }=f_i\otimes 1$ ($1\le i\le n$), the sequence $(f_i^{\prime })$ is $M^{\prime }$-regular.

It is clear that a) implies b); to show that b) implies d), it suffices, by virtue of (15.1.15), to show that b) implies that the $M_i$ are flat $A$-modules for $0\le i\le n-1$ (setting $M_0=M$). Since $M_{i+1}=M_i/f_{i+1}{M}_i$, one is reduced, by descending induction, to proving that if $h$ is an element of the maximal ideal of $B$ such that the homothety of ratio $h$ in $M$ is injective and $M/hM$ is a flat $A$-module, then $M$ is a flat $A$-module; but this is none other than $(0_{III},\mathrm{10.2.7})$. It is clear that c) is the particular case of d) obtained by taking $A^{\prime }=k$. It therefore remains to prove that c) implies a). Since $M$ is a flat $A$-module and a $B$-module of finite type, it follows from c) and $(0_{III},\mathrm{10.2.4})$ that the homothety of ratio $f_1$ in $M$ is injective and that its cokernel $M/f_{1}M=M_1$ is a flat $A$-module. We reason by induction on $i$ and suppose it proved that $M_i$ is a flat $A$-module; since the homothety of ratio $g_{i+1}$ in $M_i{\otimes }_{A}k$ is injective by hypothesis, it again follows from $(0_{III},\mathrm{10.2.4})$ that the homothety of ratio $f_{i+1}$ in $M_i$ is injective and that $M_i/f_{i+1}{M}_i=M_{i+1}$ is a flat $A$-module, which completes the proof. One will note that the proof that c) implies a) does not use the fact that the $f_i$ belong to the maximal ideal of $B$.

Corollary (15.1.17).

The hypotheses on $A$ and $B$ being those of (15.1.16), let $M$ be a $B$-module of finite type, flat over $A$, $R$ a sub-$B$-module of $M$ such that $N=M/R$ is a flat $A$-module. Let $(f_i{)}_{1\le i\le n}$ be a sequence of elements of the maximal ideal of $B$ having the following properties:

(i) $f_{i}M\subset R$ for $1\le i\le n$.

(ii) If $g_i$ is the image of $f_i$ in $B{\otimes }_{A}k$, the sequence $(g_i{)}_{1\le i\le n}$ is $(M{\otimes }_{A}k)$-regular.

(iii) The sum of the sub-$(B{\otimes }_{A}k)$-modules $g_i(M{\otimes }_{A}k)$ of $M{\otimes }_{A}k$ is the canonical image of $R{\otimes }_{A}k$ in $M{\otimes }_{A}k$ (equal to the kernel of the canonical homomorphism $M{\otimes }_{A}k\to N{\otimes }_{A}k$).

Then the sequence $(f_i)$ is $M$-regular and one has $R={\sum }_{i=1}^n{f}_{i}M$.

Since $N$ is a flat $A$-module, it follows from the exactness of the sequence $0\to R\to M\to N\to 0$ that the sequence $0\to R{\otimes }_{A}k\to M{\otimes }_{A}k\to N{\otimes }_{A}k\to 0$ is exact $(0_I,\mathrm{6.1.2})$; one therefore has $R{\otimes }_{A}k={\sum }_i{g}_i(M{\otimes }_{A}k)$. Since $M$ is a $B$-module of finite type and $B$ is Noetherian, $R$ is also a $B$-module of finite type, whence $R={\sum }_i{f}_{i}M$ by virtue of Nakayama's lemma, the $f_i$ belonging to the maximal ideal of $B$.

Lemma (15.1.18).

Let $A$ be a ring, $M$ an $A$-module equipped with a separated and exhaustive filtration $(M_k{)}_{k\ge 0}$, and let $f\in A$ be a $g{r}_{\bullet }(M)$-regular element. Then:

(i) $f$ is $M$-regular.

(ii) For every $k$, one has $M_k\cap fM=f{M}_k$, and the canonical image $M_k^{\prime \prime }$ of $M_k$ in $M^{\prime \prime }=M/fM$ is therefore canonically isomorphic to $M_k/f{M}_k$.

(iii) The sequence

        0 → M_{k+1}/fM_{k+1} → M_k/fM_k → gr_k(M)/f · gr_k(M) → 0

is exact, and if one equips M'' with the quotient filtration $(M_k^{\prime \prime })$, $g{r}_{\bullet }(M^{\prime \prime })$ is therefore canonically isomorphic to $g{r}_{\bullet }(M)/f\cdot g{r}_{\bullet }(M)$.

Assertion (i) follows from the hypothesis on $f$ and from Bourbaki, Alg. comm., chap. III, §2, n° 8, cor. 1 of th. 1. Since $g{r}_{\bullet }(M/M_k)$ is a direct summand of $g{r}_{\bullet }(M)$, $f$ is $g{r}_{\bullet }(M/M_k)$-regular for every $k$, hence $(M/M_k)$-regular by (i), which establishes (ii). To prove (iii), it suffices to see that the map $M_{k+1}/f{M}_{k+1}\to M_k/f{M}_k$ is injective: now, if $x\in M_{k+1}$ is such that $x\in f{M}_k$, it follows from (ii) that $x\in f{M}_{k+1}$, which completes the proof of the lemma.

Proposition (15.1.19).

Let $A$ be a ring, $M$ an $A$-module equipped with a filtration $(M_k{)}_{k\ge 0}$ such that $M_0=M$, $(f_i{)}_{1\le i\le n}$ a $g{r}_{\bullet }(M)$-regular sequence of elements of $A$. Suppose in addition that for $0\le i\le n-1$, the quotient filtration of $(M_k)$ on $M/({\sum }_{j\le i}{f}_{j}M)$ is separated. Then the sequence $(f_i)$ is $M$-regular, and if one sets $M^{\prime }=M/({\sum }_{i\le n}{f}_{i}M)$ and equips $M^{\prime }$ with the quotient filtration of $(M_k)$, then $g{r}_{\bullet }(M^{\prime })$ is canonically isomorphic to $g{r}_{\bullet }(M)/({\sum }_{i\le n}{f}_{i}g{r}_{\bullet }(M))$.

For $n=1$, the proposition follows from Lemma (15.1.18). We reason by induction on $n$, setting $N=M/({\sum }_{i\le n-1}{f}_{i}M)$ and denoting by $(N_k)$ the quotient filtration of $(M_k)$ on $N$, which is separated by hypothesis; in addition $g{r}_{\bullet }(N)$ is isomorphic to

                        gr_•(M)/(∑_{i ≤ n-1} f_i gr_•(M)).

By hypothesis $f_n$ is therefore $g{r}_{\bullet }(N)$-regular, and it therefore follows from Lemma (15.1.18) applied to $N$ that $f_n$ is $N$-regular and $g{r}_{\bullet }(M^{\prime })$ isomorphic to $g{r}_{\bullet }(N)/f_{n}g{r}_{\bullet }(N)$, which proves the proposition.

The proposition (15.1.19) will apply in particular for filtrations satisfying one or the other of the following hypotheses:

1° The filtration $(M_k)$ is finite and separated (since this implies $M_k=0$ for $k$ large enough);

2° $A$ is a Noetherian ring, $\mathfrak{J}$ an ideal contained in the radical of $A$, $M$ an $A$-module of finite type and $(M_k)$ the $\mathfrak{J}$-preadic filtration $(0_I,\mathrm{7.3.5})$.

Corollary (15.1.20).

Let $A$ be a ring, $M$ an $A$-module, $(f_i{)}_{1\le i\le n}$ an $M$-regular sequence of elements of $A$. Then, whatever the integers ${\nu }_i\ge 1$, the sequence $(f_i^{{\nu }_i})$ is $M$-regular and $M/({\sum }_{i\le n}{f}_i^{{\nu }_i}M)$ has a composition series whose quotients are isomorphic to $M/({\sum }_{i\le n}{f}_{i}M)$.

We reason by induction on $n$; for $n=1$ the assertions are immediate since $f^{\nu }M/f^{\nu +1}M$ is isomorphic to $M/fM$, the homothety of ratio $f$ in $M$ being injective by hypothesis. Set then $N=M/({\sum }_{i\le n-1}{f}_i^{{\nu }_i}M)$; by the induction hypothesis, $N$ admits a finite filtration $(N_k)$ for which the $g{r}_k(N)$ are isomorphic to $M/({\sum }_{i\le n-1}{f}_{i}M)$; by hypothesis $f_n$ is therefore $g{r}_{\bullet }(N)$-regular; it consequently follows from (15.1.19) that $f_n$ is $N$-regular. Applying the case $n=1$ proved at the beginning, one concludes that $f_n^{{\nu }_n}$ is $N$-regular and that $N/f_n^{{\nu }_n}N$ admits a composition series whose quotients are isomorphic to $N/f_{n}N$; but for the filtration of $N/f_{n}N$ quotient of the filtration $(N_k)$, $g{r}_k(N/f_{n}N)$ is isomorphic to $g{r}_k(N)/f_{n}g{r}_k(N)$ by (15.1.19) applied to $N$, hence to $M/({\sum }_{i\le n}{f}_{i}M)$, which proves the corollary.

Proposition (15.1.21).

Let $A$, $B$ be two Noetherian local rings, $\varphi :A\to B$ a local homomorphism, $M$ a $B$-module of finite type, $(f_i{)}_{1\le i\le n}$ an $A$-regular sequence formed of elements of the maximal ideal of $A$. The following properties are equivalent:

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

b) If $\mathfrak{J}$ is the ideal ${\sum }_{i\le n}{f}_{i}A$, $M/\mathfrak{J}M$ is a flat $A/\mathfrak{J}$-module and the sequence $(f_i)$ is $M$-regular.

The fact that a) implies b) follows from (15.1.13) and from $(0_{III},\mathrm{10.2.2})$. Conversely, if the sequence $(f_i)$ is $M$-regular, it is $M$-quasi-regular (15.1.9), hence the homomorphism (15.1.1.1) is bijective. But since the sequence $(f_i{)}_{1\le i\le n}$ is also assumed to be $A$-regular, it is $A$-quasi-regular (15.1.9), and the corresponding canonical homomorphism (15.1.1.1) $(A/\mathfrak{J})[T_1,\cdots ,T_n]\to g{r}_{\bullet }(A)$ is bijective. One concludes that the canonical homomorphism $(0_{III},\mathrm{10.1.1.2})$

                        gr_0(M) ⊗_{A/𝔍} gr_•(A) → gr_•(M)

is bijective. The conclusion then follows from $(0_{III},\mathrm{10.2.2})$, the $f_i$ being in the maximal ideal of $A$ and the homomorphism $\varphi$ being local.

15.2. $\mathcal{F}$-regular sequences

(15.2.1) Let $X$ be a ringed space, $\mathcal{F}$ an ${\mathcal{O}}_X$-Module, $(f_i{)}_{1\le i\le n}$ a sequence of sections of ${\mathcal{O}}_X$ over $X$, $\mathcal{J}$ the Ideal of ${\mathcal{O}}_X$ generated by the $f_i$ (image of ${\mathcal{O}}_X^n$ by the homomorphism ${\mathcal{O}}_X^n\to {\mathcal{O}}_X$ defined canonically by the $f_i$ $(0_I,\mathrm{5.1.1})$). The sub-${\mathcal{O}}_X$-Modules ${\mathcal{J}}^k\mathcal{F}$ define on $\mathcal{F}$ a filtration, and one further sets $g{r}_k(\mathcal{F})={\mathcal{J}}^k\mathcal{F}/{\mathcal{J}}^{k+1}\mathcal{F}$; if $S_k$ is the set of homogeneous polynomials of total degree $k$ in the ring $\mathbb{Z}[T_1,\cdots ,T_n]$, $S_k$ can be considered as a simple sheaf on $X$, and one defines as in (15.1.1) a canonical homomorphism

(15.2.1.1)              φ_k : gr_0(ℱ) ⊗_ℤ S_k → gr_k(ℱ)

in the following way: for every open set $U$ of $X$ (or of a basis of the topology of $X$), one considers the homomorphism (15.1.1.1)

                        φ_{k, U} : gr_0(Γ(U, ℱ)) ⊗_ℤ S_k → gr_k(Γ(U, ℱ))

defined by the $n$ sections $f_i|U\in \Gamma (U,{\mathcal{O}}_X)$, the $\Gamma (U,{\mathcal{O}}_X)$-module $\Gamma (U,\mathcal{F})$ being filtered by the submodules $(\Gamma (U,\mathcal{J}){)}^k\Gamma (U,\mathcal{F})$. The ${\mathcal{O}}_X$-Module $g{r}_k(\mathcal{F})$ being associated with the presheaf $U\mapsto g{r}_k(\Gamma (U,\mathcal{F}))$ for every $k$, the ${\varphi }_{k,U}$ define a homomorphism of ${\mathcal{O}}_X$-Modules (15.2.1.1). By virtue of the exactness properties of inductive limits of modules and of their commutation with tensor products, it follows from what precedes that for every $x\in X$, the homomorphism $({\varphi }_k{)}_x$ on the fibres of the two members of (15.2.1.1) at the point $x$ is identified with the homomorphism of (15.1.1)

(15.2.1.2)              gr_0(ℱ_x) ⊗_ℤ S_k → gr_k(ℱ_x)

defined by the sequence $((f_i{)}_x{)}_{1\le i\le n}$ of elements of ${\mathcal{O}}_x$.

(15.2.2) With the notations of (15.2.1), we shall say that the sequence $(f_i{)}_{1\le i\le n}$ is $\mathcal{F}$-regular if, denoting by $f_i\mathcal{F}$ the sub-${\mathcal{O}}_X$-Module of $\mathcal{F}$ image of the endomorphism of $\mathcal{F}$ defined by $f_i$, the endomorphism of $\mathcal{F}/({\sum }_{j=1}^{i-1}{f}_j\mathcal{F})$ defined by $f_i$ is injective for $1\le i\le n$; we shall say that the sequence $(f_i)$ is $\mathcal{F}$-quasi-regular if the homomorphisms (15.2.1.1) are injective. It follows from these definitions and from the remarks of (15.2.1) that for the sequence $(f_i)$ to be $\mathcal{F}$-regular (resp. $\mathcal{F}$-quasi-regular), it is necessary and sufficient that, for every $x\in X$, the sequence $((f_i{)}_x{)}_{1\le i\le n}$ be ${\mathcal{F}}_x$-regular (resp. ${\mathcal{F}}_x$-quasi-regular). Every $\mathcal{F}$-regular sequence is therefore $\mathcal{F}$-quasi-regular (15.1.9); the converse holds if $\mathcal{F}$ is an ${\mathcal{O}}_X$-Module of finite type, if the ${\mathcal{O}}_x$ are Noetherian rings for every $x\in X$ and if in addition the $(f_i{)}_x$ belong to the radical of ${\mathcal{O}}_x$ for every $x\in X$ (15.1.11).

If $X=\mathrm{Spec}(A)$ is an affine scheme, and $\mathcal{F}=\tilde{M}$, it follows from (I, 1.3) that for the sequence $(f_i)$ to be $\mathcal{F}$-regular (resp. $\mathcal{F}$-quasi-regular) it is necessary and sufficient that it be $M$-regular (resp. $M$-quasi-regular).

When $\mathcal{F}={\mathcal{O}}_X$, one will suppress the mention of $\mathcal{F}$ in the preceding definitions.

Remark (15.2.3).

When $X=\mathrm{Spec}(A)$ and $\mathfrak{J}={\sum }_{i=1}^n{f}_{i}A$, the sequence $(f_i)$ is quasi-regular for every $A$-module $M$ in every open set $U$ not meeting $V(\mathfrak{J})$, since at every point $x\in U$, one has ${\mathfrak{J}}_x={\mathcal{O}}_x$ and both members of (15.2.1.2) are zero. The notion of $\tilde{M}$-regular sequence is therefore of interest only in the neighbourhood of points of $V(\mathfrak{J})$.

Proposition (15.2.4).

Let $X$ be a ringed space, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module, $x$ a point of $X$, $(f_i{)}_{1\le i\le n}$ a sequence of elements of $\Gamma (X,{\mathcal{O}}_X)$. If the sequence $((f_i{)}_x{)}_{1\le i\le n}$ is ${\mathcal{F}}_x$-regular, there exists an open neighbourhood $U$ of $x$ such that the sequence $(f_i|U{)}_{1\le i\le n}$ is $(\mathcal{F}|U)$-regular.

Indeed, for $1\le i\le n$, the ${\mathcal{O}}_X$-Module $\mathcal{F}/({\sum }_{j=1}^{i-1}{f}_j\mathcal{F})$ is coherent $(0_I,\mathrm{5.3.4})$, hence the same holds for the kernel of the endomorphism of this ${\mathcal{O}}_X$-Module defined by $f_i$ $(0_I,\mathrm{5.3.4})$; since the fibre at the point $x$ of this kernel is zero, the same holds for its restriction to a neighbourhood of $x$ $(0_I,\mathrm{5.2.2})$.

Proposition (15.2.5).

Let $u=(\psi ,\theta ):X\to Y$ be a flat morphism of ringed spaces $(0_I,\mathrm{6.7.1})$; let $\mathcal{F}$ be an ${\mathcal{O}}_Y$-Module, $\mathcal{G}=u\ast (\mathcal{F})$, $(f_i{)}_{1\le i\le n}$ a sequence of elements of $\Gamma (Y,{\mathcal{O}}_Y)$, $(g_i{)}_{1\le i\le n}$ the sequence of their images $g_i=\theta (f_i)\in \Gamma (Y,u_{\ast }({\mathcal{O}}_X))=\Gamma (X,{\mathcal{O}}_X)$. If the sequence $(f_i)$ is $\mathcal{F}$-regular, the sequence $(g_i)$ is $\mathcal{G}$-regular. The converse holds if $u$ is a faithfully flat morphism $(0_I,\mathrm{6.7.8})$.

Let $x$ be a point of $X$, $y=u(x)=\psi (x)$; one has

                        𝒢_x = ψ_x^*(ℱ_y) ⊗_{ψ_x^*(𝒪_y)} 𝒪_x      and      (g_i)_x = θ_x^♯(ψ_x^*((f_i)_y));

since by hypothesis ${\theta }_x^{♯}:{\psi }_x^{\ast }({\mathcal{O}}_y)\to {\mathcal{O}}_x$ makes ${\mathcal{O}}_x$ a flat ${\psi }_x^{\ast }({\mathcal{O}}_y)$-module, and since the functor ${\psi }^{\ast }$ is exact, the first assertion follows from the corresponding assertion of (15.1.14); the same holds for the second, taking into account that the hypothesis that $u$ is faithfully flat means that $\psi$ is surjective and that ${\theta }_x^{♯}$ makes ${\mathcal{O}}_x$ a faithfully flat ${\psi }_x^{\ast }({\mathcal{O}}_y)$-module.