Éléments de Géométrie Algébrique — English

§16. Dimension and depth in Noetherian local rings

16.1. Dimension of a ring

(16.1.1) We call dimension (or Krull dimension) of a ring $A$, and denote $\dim (A)$, the (combinatorial) dimension of its spectrum $\mathrm{Spec}(A)$ (14.2.1); since the irreducible closed subsets of $\mathrm{Spec}(A)$ are the $V(\mathfrak{p})$, where $\mathfrak{p}$ is a prime ideal of $A$ (Bourbaki, Alg. comm., chap. II, §4, n° 3, cor. 1 of prop. 14), it amounts to the same thing to say that $\dim (A)$ is the supremum of the lengths of chains (14.1.1)

$${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_n$$

of prime ideals of $A$. If $A\ne 0$, one has $\dim (A)\ge 0$. An Artinian ring $\ne 0$ (in particular a field) is of dimension 0 (Bourbaki, Alg. comm., chap. IV, §2, n° 5, prop. 7); a Dedekind ring which is not a field (in particular $\mathbb{Z}$ or a polynomial ring k[T] in one indeterminate over a field $k$) is of dimension 1 (Bourbaki, Alg. comm., chap. VII, §2). A Noetherian ring is not necessarily of finite dimension [30].

If $({\mathfrak{p}}_{\alpha })$ is the family of minimal prime ideals of $A$, one has (14.1.2.1)

(16.1.1.1)              dim(A) = sup_α dim(A/𝔭_α).

(16.1.2) For every ideal $\mathfrak{a}$ of $A$, the prime ideals of $A/\mathfrak{a}$ correspond bijectively to the prime ideals of $A$ containing $\mathfrak{a}$, hence

$$(\mathrm{16.1.2.1})\dim (A/\mathfrak{a})\le \dim (A).$$

If in addition $\mathfrak{a}$ is not contained in any minimal prime ideal ${\mathfrak{p}}_{\alpha }$ of $A$, every chain of prime ideals of $A$ containing $\mathfrak{a}$ can be completed by one of the ${\mathfrak{p}}_{\alpha }$, so one has

$$(\mathrm{16.1.2.2})1+\dim (A/\mathfrak{a})\le \dim (A).$$

(16.1.3) If $S$ is a multiplicative subset of $A$, the prime ideals of $S^{-1}A$ correspond bijectively to the prime ideals of $A$ not meeting $S$, hence one has

$$(\mathrm{16.1.3.1})\dim (S^{-1}A)\le \dim (A).$$

For every prime ideal $\mathfrak{p}$ of $A$, one has, by definition (14.2.1)

(16.1.3.2)              dim(A_𝔭) = codim(V(𝔭), Spec(A)).

This number is also called the height of the prime ideal $\mathfrak{p}$ and denoted $ht(\mathfrak{p})$. More generally, for every ideal $\mathfrak{a}$ of $A$, we call height of $\mathfrak{a}$ the number

$$(\mathrm{16.1.3.3})ht(\mathfrak{a})=codim(V(\mathfrak{a}),\mathrm{Spec}(A)).$$

If $({\mathfrak{p}}_{\lambda })$ is the family of prime ideals minimal among those that contain $\mathfrak{a}$, one has therefore (14.2.1)

$$(\mathrm{16.1.3.4})ht(\mathfrak{a})=\underset{\lambda }{\inf }ht({\mathfrak{p}}_{\lambda }).$$

One has evidently

(16.1.3.5)              dim(A) = sup_𝔪 dim(A_𝔪)

where $\mathfrak{m}$ ranges over the set of maximal ideals of $A$.

(16.1.4) For every prime ideal $\mathfrak{p}$ of $A$, one has (14.2.2.2)

(16.1.4.1)              dim(A_𝔭) + dim(A/𝔭) ≤ dim(A).

We say that $A$ is catenary (resp. equidimensional, equicodimensional, biequidimensional) when $\mathrm{Spec}(A)$ is catenary (resp. equidimensional, equicodimensional, biequidimensional). When $A$ is Noetherian and biequidimensional, the two sides of (16.1.4.1) are equal for every prime ideal of $A$ (14.3.5).

For every ideal $\mathfrak{a}$ of $A$, the prime ideals of $A/\mathfrak{a}$ correspond bijectively to the prime ideals of $A$ containing $\mathfrak{a}$, so, if $A$ is catenary, so is $A/\mathfrak{a}$. Likewise, for every multiplicative subset $S$ of $A$, the prime ideals of $S^{-1}A$ correspond bijectively to the prime ideals of $A$ not meeting $S$, so if $A$ is catenary, so is $S^{-1}A$.

In addition, since the prime ideals of $A$ contained in a prime ideal $\mathfrak{p}$ correspond bijectively to the prime ideals of $A_{\mathfrak{p}}$, for $A$ to be catenary it is necessary and sufficient that $A_{\mathfrak{p}}$ be so for every $\mathfrak{p}$. To say that $A$ is catenary therefore means ((14.3.2) and (16.1.3.2)) that for every pair of prime ideals $\mathfrak{p}$, $\mathfrak{q}$ of $A$ such that $\mathfrak{q}\subset \mathfrak{p}$, one has

(16.1.4.2)              dim(A_𝔮) + dim(A_𝔭 / 𝔮 A_𝔭) = dim(A_𝔭).

If $A$ is biequidimensional, so is $A_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$.

Proposition (16.1.5).

Let $\rho :A\to B$ be a ring homomorphism making $B$ an $A$-algebra integral over $A$. Then one has $\dim (B)\le \dim (A)$; if in addition $\rho$ is injective, one has $\dim (B)=\dim (A)$.

If $\mathfrak{n}$ is the kernel of $\rho$, $\rho (A)$ is isomorphic to $A/\mathfrak{n}$, so $\dim (\rho (A))\le \dim (A)$ by (16.1.2.1), and since $B$ is integral over $\rho (A)$, one may restrict oneself to proving the second assertion when $A\subset B$. If ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_n$ is a chain of prime ideals of $B$, the ${\mathfrak{p}}_i\cap A$ are then distinct prime ideals in $A$ (Bourbaki, Alg. comm., chap. V, §2, n° 1, cor. 1 of prop. 1), hence form a chain of prime ideals. Conversely, for every chain ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_n$ of prime ideals of $A$, one knows that there exists for each $i$ a prime ideal ${\mathfrak{p}}_i^{\prime }$ of $B$ such that ${\mathfrak{p}}_i^{\prime }\cap A={\mathfrak{p}}_i$ and that ${\mathfrak{p}}_i^{\prime }\subset {\mathfrak{p}}_{i+1}^{\prime }$ for $0\le i\le n-1$ (loc. cit., cor. 2 of th. 1). Whence the conclusion.

Proposition (16.1.6).

Let $B$ be an integral ring, $A$ a local subring of $B$, $\mathfrak{m}$ its maximal ideal. Suppose that $A$ is integrally closed and that $B$ is integral over $A$; then, for every maximal ideal $\mathfrak{n}$ of $B$, one has $\dim (B_{\mathfrak{n}})=\dim (A)$.

The first part of the reasoning of (16.1.5) shows that $\dim (B_{\mathfrak{n}})\le \dim (A)$. Conversely, consider a chain of prime ideals ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_n=\mathfrak{m}$ of $A$; one knows that $\mathfrak{n}\cap A=\mathfrak{m}$ (Bourbaki, Alg. comm., chap. V, §2, n° 1, prop. 1) and by virtue of the hypotheses made, one may apply the second Cohen-Seidenberg theorem (Bourbaki, Alg. comm., chap. V, §2, n° 4, th. 3), which proves the existence of a chain of prime ideals ${\mathfrak{p}}_0^{\prime }\subset {\mathfrak{p}}_1^{\prime }\subset \cdots \subset {\mathfrak{p}}_n^{\prime }=\mathfrak{n}$ of $B$ such that ${\mathfrak{p}}_i^{\prime }\cap A={\mathfrak{p}}_i$ for every $i$; one deduces at once that $\dim (B_{\mathfrak{n}})\ge \dim (A)$, whence the proposition.

(16.1.7) Let $M$ be an $A$-module; we call dimension of $M$, and denote ${\dim }_A(M)$ or $\dim (M)$, the dimension of the ring $A/Ann(M)$ (isomorphic to the ring A_M of homotheties of $M$). One has therefore $\dim (M^k)=\dim (M)$ for every $k>0$. If $M$ is of finite type, the prime ideals of $A$ containing $Ann(M)$ are exactly those belonging to $Supp(M)$ $(0_I,\mathrm{1.7.4})$ and the latter is closed in $\mathrm{Spec}(A)$, hence one has then

(16.1.7.1)              dim(M) = dim(Supp(M)) ≤ dim(A).

If $M$, $N$ are two $A$-modules such that $N\subset M$, one has

(16.1.7.2)              dim(N) ≤ dim(M),         dim(M/N) ≤ dim(M).

If $S$ is a multiplicative subset of $A$ and if $M$ is of finite type, one has

$$(\mathrm{16.1.7.3}){\dim }_{S^{-1}A}(S^{-1}M)\le {\dim }_A(M).$$

Indeed, one has $Ann(S^{-1}M)=S^{-1}(Ann(M))$ (Bourbaki, Alg. comm., chap. II, §2, n° 4) and $S^{-1}A/S^{-1}(Ann(M))\cong S^{-1}(A/Ann(M))$.

On the other hand, every maximal chain of prime ideals containing $Ann(M)$ ends at a maximal ideal $\mathfrak{m}$, so its length is equal to that of the chain of corresponding prime ideals of $A_{\mathfrak{m}}$ (which contain $Ann(M_{\mathfrak{m}})=(Ann(M){)}_{\mathfrak{m}}$); from this remark and from (16.1.7.3), one obtains

(16.1.7.4)              dim_A(M) = sup_𝔪 dim_{A_𝔪}(M_𝔪)

where $\mathfrak{m}$ ranges over the set of maximal ideals of $A$.

We say that $M$ is catenary (resp. equidimensional, equicodimensional, biequidimensional) when $A/Ann(M)$ is.

Proposition (16.1.8).

Let $A$ be a ring, $M$ an $A$-module of finite type, $\mathfrak{p}$ a prime ideal of $A$. One has then

(16.1.8.1)              dim_{A_𝔭}(M_𝔭) + dim_A(M/𝔭M) ≤ dim_A(M).

One has $Ann(M_{\mathfrak{p}})=(Ann(M){)}_{\mathfrak{p}}$, hence the prime ideals of $A_{\mathfrak{p}}$ containing $Ann(M_{\mathfrak{p}})$ correspond bijectively to the prime ideals of $A$ containing $Ann(M)$ and contained in $\mathfrak{p}$. On the other hand, if $Ann(M)=\mathfrak{n}$, $V(Ann(M/\mathfrak{p}M))=V(\mathfrak{p}+\mathfrak{n})$ $(0_I,\mathrm{1.7.5})$, hence the prime ideals of $A$ containing $Ann(M/\mathfrak{p}M)$ contain $\mathfrak{p}+\mathfrak{n}$; whence the conclusion.

This last remark shows in addition that one has

$$(\mathrm{16.1.8.2}){\dim }_{A/\mathfrak{p}}(M/\mathfrak{p}M)={\dim }_A(M/\mathfrak{p}M).$$

Proposition (16.1.9).

Let $A$ be a Noetherian ring, $\rho :A\to B$ a ring homomorphism, $M$ a $B$-module such that $M_{[\rho ]}$ is an $A$-module of finite type. Then one has

$$(\mathrm{16.1.9.1}){\dim }_A(M_{[\rho ]})={\dim }_B(M).$$

Indeed, the ring B_M of homotheties of the $B$-module $M$ identifies with a subring of $C={\mathrm{End}}_A(M_{[\rho ]})$, and $C$ is an $A$-module of finite type (Bourbaki, Alg. comm., chap. III, §3, n° 1, lemma 2), hence so is B_M; moreover the ring A_M of homotheties of the $A$-module $M_{[\rho ]}$ is a subring of B_M, canonical image of $A$ in $C$, hence B_M is finite over A_M; the conclusion thus results from (16.1.5).

(16.1.10) Let $A$ be a Noetherian ring; if $M$ is an $A$-module of finite length, one knows (Bourbaki, Alg. comm., chap. IV, §2, n° 5, prop. 7 and cor. 1 of prop. 7) that $Supp(M)$ is a finite discrete space, hence $\dim (M)=0$ if $M\ne 0$; conversely, if $M$ is an $A$-module of finite type such that $\dim (M)=0$, every point of $Supp(M)$ is closed, in other words is a maximal ideal of $A$, hence (loc. cit., prop. 7) $M$ is of finite length.

16.2. Dimension of a Noetherian semi-local ring ¹

(16.2.1) Let $A$ be a Noetherian semi-local ring, $\mathfrak{r}$ its radical; recall that an ideal of definition $\mathfrak{q}$ of $A$ is an ideal such that $\mathfrak{q}\subset \mathfrak{r}$ and that $\mathfrak{q}$ contains a power of $\mathfrak{r}$; it amounts to the same thing to say that the only prime ideals containing $\mathfrak{q}$ are maximal, or that $A/\mathfrak{q}$ is an Artinian ring. Let $M$ be an $A$-module of finite type; for every integer $n>0$, $M/{\mathfrak{q}}^{n}M$ is an $A$-module of finite length, equal to ${\sum }_{i=0}^{n-1}long(g{r}_i(M))$, where $g{r}_{\bullet }(M)$ is the graded $(A/\mathfrak{q})$-module associated with $M$ endowed with the $\mathfrak{q}$-preadic filtration. One knows that there exists a polynomial $Q(n)$ with rational coefficients such that $long(g{r}_n(M))=Q(n)$ for $n$ large enough (III, 2.5.4); one deduces at once from this that there exists a polynomial $P_{\mathfrak{q}}(M,n)$ in $n$, with rational coefficients, such that $long(M/{\mathfrak{q}}^{n}M)=P_{\mathfrak{q}}(M,n)$ for $n$ large enough; it is clear that this polynomial is unique and that its leading coefficient is > 0 if $M\ne 0$; one says that it is the Hilbert-Samuel polynomial of $M$ for the $\mathfrak{q}$-preadic filtration.

¹ This number reproduces the exposition given by Serre in a course at the Collège de France in 1958.

Lemma (16.2.2.1).

Let $(M_n)$ be a $\mathfrak{q}$-good filtration of $M$ (Bourbaki, Alg. comm., chap. III, §3, n° 1); then $long(M/M_n)$ is equal, for $n$ large, to a polynomial in $n$ whose degree and leading coefficient are the same as those of $P_{\mathfrak{q}}(M,n)$.

Indeed, there exists $n_0$ such that $M_{n+1}=\mathfrak{q}{M}_n$ for $n\ge n_0$, whence the inclusions ${\mathfrak{q}}^{n+n_0}M\subset M_{n+n_0}={\mathfrak{q}}^n{M}_{n_0}\subset {\mathfrak{q}}^{n}M\subset M_n$ for $n$ large, which entails

            P_𝔮(M, n + n_0) ≥ long(M/M_{n+n_0}) ≥ P_𝔮(M, n) ≥ long(M/M_n)

and consequently the conclusion.

If ${\mathfrak{q}}^{\prime }$ is a second ideal of definition, one has ${\mathfrak{q}}^{\prime }\supset {\mathfrak{q}}^m$ for some integer $m$, hence $P_{{\mathfrak{q}}^{\prime }}(M,n)\le P_{\mathfrak{q}}(M,mn)$; consequently the degree of $P_{\mathfrak{q}}(M,n)$ does not depend on the ideal of definition $\mathfrak{q}$ considered; one denotes it $d(M)$.

Lemma (16.2.2.2).

Let $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ be an exact sequence of $A$-modules of finite type. Then the polynomial $P_{\mathfrak{q}}(M,n)-P_{\mathfrak{q}}(M^{\prime },n)-P_{\mathfrak{q}}(M^{\prime \prime },n)$ has degree $\le d(M^{\prime })-1\le d(M)-1$.

Indeed, the filtration of the $M_n^{\prime }=M^{\prime }\cap {\mathfrak{q}}^{n}M$ is $\mathfrak{q}$-good (Bourbaki, Alg. comm., chap. III, §3, n° 1, prop. 1); since long(M/𝔮^n M) = long(M''/𝔮^n M'') + long(M'/M'_n), the lemma follows at once from (16.2.2.1) applied to $M^{\prime }$.

Theorem (16.2.3) (Krull-Chevalley-Samuel).

Let $A$ be a Noetherian semi-local ring, $\mathfrak{r}$ its radical, $M\ne 0$ an $A$-module of finite type, $d(M)$ the degree of the Hilbert-Samuel polynomial of $M$ (for an arbitrary ideal of definition of $A$); let on the other hand $s(M)$ be the infimum of the integers $n$ such that there exist elements $x_1,\cdots ,x_n$ of $\mathfrak{r}$ for which $M/(x_{1}M+\cdots +x_{n}M)$ is of finite length. Then one has

$$d(M)=s(M)=\dim (M).$$

Lemma (16.2.3.1).

For every $x\in \mathfrak{r}$, let ${}_{x}M$ be the submodule of $M$ formed of the elements annihilated by $x$. Then:

(i) One has $s(M)\le s(M{/}_{x}M)+1$.

(ii) Let ${\mathfrak{p}}_i$ be the prime ideals belonging to $Supp(M)$ and such that

                        dim(A/𝔭_i) = dim(M)            (1 ≤ i ≤ m).

If $x\notin {\mathfrak{p}}_i$ for every $i$, one has $\dim (M/xM)\le \dim (M)-1$.

(iii) If $\mathfrak{q}$ is an ideal of definition of $A$, the polynomial

                        P_𝔮(M, n) − P_𝔮(M/xM, n)

is of degree $\le d(M)-1$.

Assertion (i) is trivial, since if $(x_i{)}_{1\le i\le r}$ is such that for the module $N=M{/}_{x}M$, $N/(x_{1}N+\cdots +x_{r}N)$ is of finite length, it suffices to observe that this module is isomorphic to $M/(xM+x_{1}M+\cdots +x_{r}M)$.

To prove (ii), one observes that, if $\mathfrak{a}$ is the annihilator of $M$, the prime ideals containing the annihilator of $M/xM$ are those that contain $\mathfrak{a}+Ax$ $(0_I,\mathrm{1.7.5})$; none of the ${\mathfrak{p}}_i$ can therefore contain $\mathfrak{a}+Ax$, and by definition of $\dim (M)$ (16.1.7) every chain of prime ideals containing $\mathfrak{a}+Ax$ therefore has length $\le \dim (M)-1$.

Finally, to prove (iii), one notes that one has two exact sequences

            0 → _x M → M --x→ xM → 0,        0 → xM → M → M/xM → 0

and the assertion follows at once from lemma (16.2.2.2).

The proof of (16.2.3) is done in three steps:

(16.2.3.2) One has $\dim (M)\le d(M)$.

We reason by induction on $d(M)$, the case $d(M)=0$ being trivial. Suppose then $d(M)\ge 1$, and let ${\mathfrak{p}}_0\in Supp(M)$ be such that $\dim (A/{\mathfrak{p}}_0)=\dim (M)$. This entails that ${\mathfrak{p}}_0$ is a minimal element of $Supp(M)$, hence it is associated with $M$ (Bourbaki, Alg. comm., chap. IV, §1, n° 4, th. 2) and $M$ consequently contains a submodule $N$ isomorphic to $A/{\mathfrak{p}}_0$ (loc. cit., §1, n° 1); since $d(M)\ge d(N)$ by (16.2.2.2), one is reduced to proving that $\dim (N)\le d(N)$. Let then ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_n$ be a chain of distinct prime ideals in $A$, and let us show that $n\le d(N)$. This is evident if $n=0$. Otherwise, there exists $x\in {\mathfrak{p}}_1\cap \mathfrak{r}$ such that $x\notin {\mathfrak{p}}_0$, for the relation ${\mathfrak{p}}_0\supset {\mathfrak{p}}_1\cap \mathfrak{r}$ would entail ${\mathfrak{p}}_0\supset \mathfrak{r}$ since ${\mathfrak{p}}_0$ does not contain ${\mathfrak{p}}_1$, hence ${\mathfrak{p}}_0$ would be maximal, which is absurd. One has ${\mathfrak{p}}_i\in Supp(N/xN)$ for $i\ge 1$, hence $n-1\le \dim (N/xN)$ (16.1.7). Since ${}_{x}N=0$ by virtue of the choice of $x$, one has $d(N/xN)\le d(N)-1$ by (16.2.3.1, (iii)). The induction hypothesis then implies $d(N/xN)=\dim (N/xN)\ge n-1$, hence $n-1\le d(N)-1$.

(16.2.3.3) One has $d(M)\le s(M)$.

Let indeed $(x_i{)}_{1\le i\le n}$ be a family of elements of $\mathfrak{r}$ such that, on setting $\mathfrak{a}=x_{1}A+\cdots +x_{n}A$, $M/\mathfrak{a}M$ is of finite length. The ideals associated with $M/\mathfrak{a}M$ are then maximal (Bourbaki, Alg. comm., chap. IV, §2, n° 5, prop. 7) and since these are the only prime ideals containing $\mathfrak{q}=\mathfrak{a}+(\mathfrak{r}\cap Ann(M))$, $A/\mathfrak{q}$ is Artinian (loc. cit., prop. 9) and $\mathfrak{q}$ is an ideal of definition of $A$. But one has ${\mathfrak{q}}^{m}M/{\mathfrak{q}}^{m+1}M={\mathfrak{a}}^{m}M/{\mathfrak{a}}^{m+1}M$, and if $M$ is generated by $r$ elements $y_j$, ${\mathfrak{a}}^{m}M/{\mathfrak{a}}^{m+1}M$ is an $(A/\mathfrak{q})$-module generated by the canonical images of the $x_1^{k_1}{x}_2^{k_2}\cdots x_n^{k_n}{y}_j$ where $\sum k_i=m$, hence by $r\cdot ((\genfrac{}{}{0ex}{}{m+n-1}{n-1}))$ elements; its length is consequently $\le \alpha \cdot \left(\genfrac{}{}{0ex}{}{m+n-1}{n-1}\right)$, where $\alpha$ is a constant, and one deduces at once that the degree of $P_{\mathfrak{q}}(M,n)$ is $\le n$; whence $d(M)\le s(M)$.

(16.2.3.4) One has $s(M)\le \dim (M)$.

Let us first note that $n=\dim (M)$ is finite by (16.2.3.2). We reason by induction on $n$; the proposition is evident for $n=0$, since $M$ is then of finite length (16.1.10). Suppose $n\ge 1$, and let ${\mathfrak{p}}_i$ ($1\le i\le m$) be the prime ideals of $Supp(M)$ such that $\dim (A/{\mathfrak{p}}_i)=n$; the definition of the ${\mathfrak{p}}_i$ shows that they are minimal elements of $Supp(M)$, hence finite in number (Bourbaki, Alg. comm., chap. IV, §1, n° 4, th. 2). Since $n\ge 1$, the ${\mathfrak{p}}_i$ are not maximal, and there exists therefore $x\in \mathfrak{r}$ such that $x\notin {\mathfrak{p}}_i$ for every $i$ (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2). By virtue of (16.2.3.1), one has $s(M)\le s(M/xM)+1$ and $\dim (M)\ge \dim (M/xM)+1$; the induction hypothesis entails that $s(M/xM)\le \dim (M/xM)$, whence the conclusion.

Corollary (16.2.4).

Under the hypotheses of (16.2.3), one has

$$(\mathrm{16.2.4.1}){\dim }_A(M)={\dim }_{\hat{A}}(\hat{M}).$$

Indeed, $M/{\mathfrak{r}}^{n}M$ and $\hat{M}/{\hat{\mathfrak{r}}}^n\hat{M}$ are isomorphic, hence $d(M)=d(\hat{M})$.

Corollary (16.2.5).

The dimension of a Noetherian semi-local ring $A$ is finite and equal to the minimum number of elements of the radical of $A$ generating an ideal of definition.

This is the equality $s(M)=\dim (M)$ for $M=A$.

In particular:

Corollary (16.2.6).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field; one has

$$(\mathrm{16.2.6.1})\dim (A)\le r{g}_k(\mathfrak{m}/{\mathfrak{m}}^2).$$

Indeed, one knows that if the $x_i$ ($1\le i\le n$) are elements of $\mathfrak{m}$ whose classes mod ${\mathfrak{m}}^2$ form a basis of the $k$-vector space $\mathfrak{m}/{\mathfrak{m}}^2$, the $x_i$ generate $\mathfrak{m}$ (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 4).

Proposition (16.2.7).

Let $k$ be a field, $B$ a graded $k$-algebra of finite type with positive degrees generated by its homogeneous elements of degree 1 and such that $B_0=k$; let $E$ be a graded $B$-module of finite type, so that (III, 2.5.4), for $n$ large enough, $r{g}_k(E_n)=r(n)$ is of the form $\Phi (n)-\Phi (n-1)$, where $\Phi$ is a polynomial in $n$ of degree $d$. Then there exists in $B$ a family $(t_i{)}_{1\le i\le d}$ of $d$ homogeneous elements of degrees $\ge 1$ such that $E/({\sum }_i{t}_{i}E)$ is of finite rank over $k$.

Let $\mathfrak{q}$ be the maximal ideal ${⨁}_{n\ge 1}{B}_n$ of $B$, and, for every $n$, let $E^n$ be the sub-$B$-module ${⨁}_{h\ge n+1}{E}_h$ of $E$. Every element of $B-\mathfrak{q}$ determines an injective homothety in the Artinian $B$-module $E/E^n$, and this homothety is therefore bijective (Bourbaki, Alg., chap. VIII, §2, n° 2, lemma 3), which entails that $(E/E^n{)}_{\mathfrak{q}}$ identifies with $E/E^n$; since the residue field of $B_{\mathfrak{q}}$ is $k$, one deduces from the hypothesis that $lon{g}_{B_{\mathfrak{q}}}((E/E^n{)}_{\mathfrak{q}})$ is a polynomial in $n$ of degree $d$. On the other hand, since $E$ is a $B$-module of finite type and $B$ is generated by its homogeneous elements of degree 1, the filtration $(E^n)$ on $E$ is $\mathfrak{q}$-good (Bourbaki, Alg. comm., chap. III, §1, n° 3, lemma 1), hence the filtration $((E^n{)}_{\mathfrak{q}})$ on $E_{\mathfrak{q}}$ is $\mathfrak{q}{B}_{\mathfrak{q}}$-good. Since $B_{\mathfrak{q}}$ is a Noetherian local ring, one concludes from (16.2.3) and (16.2.2.1) that one has $\dim (E_{\mathfrak{q}})=d$. We then reason by induction on $d$, the lemma being evident if $d=0$. Suppose $d\ge 1$, and observe that the minimal prime ideals ${\mathfrak{p}}_i$ ($1\le i\le r$) in $Ass(E)$ are graded (Bourbaki, Alg. comm., chap. IV, §3, n° 1, prop. 1), and that $\mathfrak{q}$ is distinct from the union of the ${\mathfrak{p}}_i$ since $d\ge 1$; there is therefore a homogeneous element $t_1\in \mathfrak{q}$ not belonging to any of the ${\mathfrak{p}}_i$ (Bourbaki, Alg. comm., chap. III, §1, n° 4, prop. 8). Since the prime ideals $({\mathfrak{p}}_i{)}_{\mathfrak{q}}$ are the minimal elements of $Ass(E_{\mathfrak{q}})$ (Bourbaki, Alg. comm., chap. IV, §1, n° 2, prop. 5), one has $\dim (E_{\mathfrak{q}}/t_1{E}_{\mathfrak{q}})=d-1$ (16.3.4). It then suffices to apply the induction hypothesis to the graded $B$-module $E/t_{1}E$ to obtain the conclusion.

16.3. Systems of parameters in a Noetherian local ring

Proposition (16.3.1).

Let $A$ be a Noetherian ring, $\mathfrak{p}$ a prime ideal of $A$, $n$ an integer. The following conditions are equivalent:

a) $ht(\mathfrak{p})=\dim (A_{\mathfrak{p}})\le n$;

b) there exist $n$ elements $x_i$ of $A$ ($1\le i\le n$) such that $\mathfrak{p}$ is minimal in the set of prime ideals containing the ideal $\mathfrak{a}$ generated by the $x_i$ (in other words $V(\mathfrak{p})$ is an irreducible component of $V(\mathfrak{a})$).

Condition b) entails that $\mathfrak{a}{A}_{\mathfrak{p}}$ is an ideal of definition of $A_{\mathfrak{p}}$, whence a) by virtue of (16.2.5). Conversely, if a) is satisfied, there exists an ideal of definition $\mathfrak{b}$ of $A_{\mathfrak{p}}$ generated by $n$ elements $x_i/s$, with $s\in A-\mathfrak{p}$. By virtue of the bijective correspondence between prime ideals of $A_{\mathfrak{p}}$ and prime ideals of $A$ contained in $\mathfrak{p}$, the ideal $\mathfrak{a}$ of $A$ generated by the $x_i$ satisfies b).

For $n=1$, one obtains the particular case:

Corollary (16.3.2) (Hauptidealsatz).

For a prime ideal in a Noetherian ring to be of height $\le 1$, it is necessary and sufficient that it be minimal in the set of prime ideals containing a suitable principal ideal.

Corollary (16.3.3) (Artin-Tate).

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

a) $A$ is a semi-local ring of dimension $\le 1$.

b) There exists $f\ne 0$ in $A$ such that $A_f$ is a field.

Condition a) is equivalent to saying that $A$ has only a finite number of prime ideals ${\mathfrak{m}}_i\ne 0$, and that these ideals are all maximal (0 being a prime ideal). Since the product of the ${\mathfrak{m}}_i$ is not zero, as $A$ is integral, an element $f\ne 0$ of this product belongs to all the ${\mathfrak{m}}_i$, hence (0) is the only prime ideal of the integral ring $A_f$, in other words $A_f$ is a field. (One notes that this part of the proof does not use the fact that $A$ is Noetherian.) Let us now prove that b) entails a); the hypothesis b) means that every prime ideal $\mathfrak{p}\ne 0$ of $A$ contains $f$. Let ${\mathfrak{p}}_i$ ($1\le i\le n$) be the prime ideals minimal among those that contain $f$ (which are finite in number since $A/fA$ is Noetherian); it suffices to prove that the ${\mathfrak{p}}_i$ are maximal ideals, for then every prime ideal $\ne 0$ necessarily contains one of the ${\mathfrak{p}}_i$, hence is equal to it. Suppose then that there exists a maximal ideal $\mathfrak{m}$ containing one of the ${\mathfrak{p}}_i$ and distinct from the ${\mathfrak{p}}_i$; $\mathfrak{m}$ is then not contained in the union of the ${\mathfrak{p}}_i$ (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2), in other words there exists $g\in \mathfrak{m}$ not belonging to any of the ${\mathfrak{p}}_i$. Let $\mathfrak{q}$ be one of the prime ideals minimal among those that contain $g$; it follows from the Hauptidealsatz (16.3.2) that $\mathfrak{q}$ is of height 1, hence $\ne 0$; consequently it contains $f$, hence also one of the ${\mathfrak{p}}_i$, and since $g\in \mathfrak{q}$ and $g$ belongs to no ${\mathfrak{p}}_i$, $\mathfrak{q}$ is distinct from the ${\mathfrak{p}}_i$; it would therefore be of height $\ge 2$, which is contradictory.

Proposition (16.3.4).

Let $A$ be a Noetherian semi-local ring, $\mathfrak{r}$ its radical, $M$ an $A$-module of finite type, ${\mathfrak{p}}_i$ the elements of $Supp(M)$ such that $\dim (A/{\mathfrak{p}}_i)=\dim (M)$. Then the ${\mathfrak{p}}_i$ are minimal elements of $Ass(M)$, and one has $\dim (M/{\mathfrak{p}}_{i}M)=\dim (M)$. For every $x\in \mathfrak{r}$, one has

$$(\mathrm{16.3.4.1})\dim (M/xM)\ge \dim (M)-1.$$

In addition, for the two sides of (16.3.4.1) to be equal, it is necessary and sufficient that $x$ not belong to any of the ${\mathfrak{p}}_i$.

The first assertion is immediate, for the ${\mathfrak{p}}_i$ are by definition minimal elements of $Supp(M)$ (16.1.6), and the latter are also minimal elements of $Ass(M)$ (Bourbaki, Alg. comm., chap. IV, §1, n° 4, th. 2). In addition, $Supp(M/{\mathfrak{p}}_{i}M)$ is the set of prime ideals containing ${\mathfrak{p}}_i$ $(0_I,\mathrm{1.7.5})$, hence dim(M/𝔭_i M) = dim(A/𝔭_i) = dim(M).

Set $N=M/xM$. If $y_1,\cdots ,y_r$ are elements of $\mathfrak{r}$ such that $N/(y_{1}N+\cdots +y_{r}N)$ is of finite length, this means that $M/(xM+y_{1}M+\cdots +y_{r}M)$ is of finite length, whence the inequality (16.3.4.1) by virtue of (16.2.3).

The fact that the two sides of (16.3.4.1) are equal when $x$ does not belong to any of the ${\mathfrak{p}}_i$ results from (16.3.4.1) and from (16.2.3.1). Conversely, if

$\dim (M/xM)=\dim (M)-1$, none of the ${\mathfrak{p}}_i$ can belong to $Supp(M/xM)$, and the prime ideals of this support are those that contain $Ann(M)+Ax$; since the ${\mathfrak{p}}_i$ contain $Ann(M)$, they cannot contain $x$.

Corollary (16.3.5).

With the notations of (16.3.4), for every ideal $\mathfrak{a}\subset {\mathfrak{p}}_i$, one has $\dim (M/\mathfrak{a}M)=\dim (M)$ and if one sets $N=M/\mathfrak{a}M$, the relation $\dim (M/xM)=\dim (M)-1$ entails $\dim (N/xN)=\dim (N)-1$.

Indeed, $Supp(N)$ is the set of prime ideals containing $\mathfrak{a}+Ann(M)$, hence ${\mathfrak{p}}_i$ belongs to $Supp(N)$ and since dim(N) ≤ dim(M) = dim(A/𝔭_i), one has $\dim (A/{\mathfrak{p}}_i)=\dim (N)$; in addition the prime ideals $\mathfrak{q}\in Supp(N)$ such that $\dim (A/\mathfrak{q})=\dim (N)$ are some of the ${\mathfrak{p}}_j$; if $x$ does not belong to any of the ${\mathfrak{p}}_j$, one has therefore $\dim (N/xN)=\dim (N)-1$ by virtue of (16.3.4).

Definition (16.3.6).

Let $A$ be a Noetherian semi-local ring, $\mathfrak{r}$ its radical, $M$ an $A$-module of finite type, and set $n=\dim (M)$. We call system of parameters for $M$ any system $(x_i{)}_{1\le i\le n}$ of $n$ elements of $\mathfrak{r}$ such that $M/(x_{1}M+\cdots +x_{n}M)$ is of finite length.

It has been seen (16.2.3) that such systems always exist.

It has been seen in the course of the proof of (16.2.3.3) that if $\mathfrak{j}={\sum }_i{x}_{i}A$ is such that $M/\mathfrak{j}M$ is of finite length, $\mathfrak{q}=\mathfrak{j}+Ann(M)$ is an ideal of definition of $A$, and conversely, if this is so, $M/{\mathfrak{q}}^{n}M$ is a module of finite type over the Artinian ring $A/\mathfrak{q}$, hence is of finite length. It therefore amounts to the same to say that $(x_i{)}_{1\le i\le n}$ is a system of parameters for $M$ or for $A/Ann(M)$ (or again that their images in $A/Ann(M)$ form a system of parameters for this ring). If $(x_i{)}_{1\le i\le n}$ is a system of parameters for $M$, so is $(x_i^k)$ for every integer $k\ge 1$, since one may restrict oneself, by virtue of what precedes, to the case where $M=A$, and if $\mathfrak{j}$ is an ideal of definition of $A$, the ideal ${\sum }_{i=1}^n{x}_i^{k}A$ contains ${\mathfrak{j}}^{kn}$, hence is also an ideal of definition.

Proposition (16.3.7).

Let $A$ be a Noetherian semi-local ring, $\mathfrak{r}$ its radical, $x_1,\cdots ,x_k$ elements of $\mathfrak{r}$, $M$ an $A$-module of finite type. One has then

(16.3.7.1)              dim(M/(x_1 M + ⋯ + x_k M)) ≥ dim(M) − k.

In addition, for the two sides of (16.3.7.1) to be equal, it is necessary and sufficient that there exist elements $x_i$ ($k+1\le i\le n=\dim (M)$) of $\mathfrak{r}$ such that $(x_i{)}_{1\le i\le n}$ is a system of parameters for $M$.

The inequality (16.3.7.1) results from (16.3.4.1) by induction on $k$. If the two sides of (16.3.7.1) are equal, and if $x_{k+1},\cdots ,x_n$ is a system of parameters for $M/(x_{1}M+\cdots +x_{k}M)$, it is clear that $M/(x_{1}M+\cdots +x_{k}M+x_{k+1}M+\cdots +x_{n}M)$ is of finite length, hence $(x_i{)}_{1\le i\le n}$ is a system of parameters for $M$; conversely, if there exist $x_i$ ($k+1\le i\le n$) having this property and if one sets $N=M/(x_{1}M+\cdots +x_{k}M)$, it is clear that $N/(x_{k+1}N+\cdots +x_{n}N)$ is of finite length, hence $\dim (N)\le n-k$, and the two sides are equal by virtue of (16.3.7.1).

Proposition (16.3.8).

Let $A$ be a Noetherian semi-local ring, $X=\mathrm{Spec}(A)$ its spectrum, $\mathfrak{a}$ an ideal of $A$ distinct from $A$ such that $codim(V(\mathfrak{a}),X)=r>0$. There exist then $r$ elements $x_1,\cdots ,x_r$ of $\mathfrak{a}$, forming part of a system of parameters of $A$, such that

                        codim(V(𝔞), V(A x_1 + ⋯ + A x_r)) = 0.

Let ${\mathfrak{p}}_i$ ($1\le i\le m$) be the minimal prime ideals of the ring $A$, ${\mathfrak{q}}_j$ ($1\le j\le n$) the prime ideals minimal among those that contain $\mathfrak{a}$; one has (14.2.1) codim(V(𝔞), X) = inf_j codim(V(𝔮_j), X) = inf_j dim(A_{𝔮_j}) (16.1.3.2). The hypothesis $r>0$ entails that $\mathfrak{a}$ is contained in none of the ${\mathfrak{p}}_i$; consequently there exists $x\in \mathfrak{a}$ not belonging to any of the ${\mathfrak{p}}_i$ (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2). One has as above codim(V(𝔞), V(Ax)) = inf_j dim((A/Ax)_{𝔮_j}) = inf_j dim(A_{𝔮_j}/x A_{𝔮_j}); but since $x/1$ is not contained in any of the minimal prime ideals of $A_{{\mathfrak{q}}_j}$, one has $\dim (A_{{\mathfrak{q}}_j}/x{A}_{{\mathfrak{q}}_j})=\dim (A_{{\mathfrak{q}}_j})-1$ (16.3.4), whence $codim(V(\mathfrak{a}),V(Ax))=r-1$. For the same reason, one has $\dim (A/Ax)=\dim (A)-1$. We now reason by induction on $r$; if $r>1$, there exist $r-1$ elements $x_i^{\prime }$ ($2\le i\le r$) of $A^{\prime }=A/Ax$, forming part of a system of parameters of this ring, belonging to $\mathfrak{a}/Ax$ and such that $codim(V(\mathfrak{a}/Ax),V(A^{\prime }{x}_2^{\prime }+\cdots +A^{\prime }{x}_r^{\prime }))=0$. Let $(x_i^{\prime }{)}_{r+1\le i\le s}$ be a system of elements of $A^{\prime }$ such that $(x_i^{\prime }{)}_{2\le i\le s}$ is a system of parameters of $A^{\prime }$; for every $i$ such that $2\le i\le s$, let $x_i$ be an element of the class $x_i^{\prime }$ in $A$, with $x_i\in \mathfrak{a}$ for $2\le i\le r$; it is immediate that $A/(Ax+A{x}_2+\cdots +A{x}_s)$ is of finite length, and since $\dim (A)=s$, one sees that $x_1=x$ and the $x_i$ of indices $i\ge 2$ answer the conditions of the statement.

Proposition (16.3.9).

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.

(i) One has

(16.3.9.1)              dim(B) ≤ dim(A) + dim(B ⊗_A k).

(ii) Let $M\ne 0$ be an $A$-module of finite type, $N\ne 0$ a $B$-module of finite type. One has

(16.3.9.2)              dim_B(M ⊗_A N) ≤ dim_A(M) + dim_{B ⊗_A k}(N ⊗_A k).

Let $m$ be the dimension of $A$, $(s_i{)}_{1\le i\le m}$ a system of parameters of $A$ (16.3.6), so that if $\mathfrak{a}={\sum }_i{s}_{i}A$, $A/\mathfrak{a}$ is an $A$-module of finite length. Then the ring $A/\mathfrak{a}$ is Artinian, so the maximal ideal $\mathfrak{m}/\mathfrak{a}$ of this ring is nilpotent; the ideal $\mathfrak{m}B/\mathfrak{a}B$ of $B/\mathfrak{a}B$, image of $(\mathfrak{m}/\mathfrak{a}){\otimes }_{A}B$, is itself also nilpotent, and consequently one has $\dim (B/\mathfrak{a}B)=\dim (B/\mathfrak{m}B)$ (the spectra of these two rings having the same underlying space). On the other hand $B/\mathfrak{a}B=B/(s_{1}B+\cdots +s_{m}B)$, and the images of the $s_i$ in $B$ belong to the maximal ideal of $B$. Consequently (16.3.7), one has

                        dim(B) ≤ m + dim(B ⊗_A k)

which is nothing other than (16.3.9.1).

To prove (16.3.9.2), note that if $\mathfrak{r}$ (resp. $\mathfrak{s}$) is the annihilator of $M$ (resp. $N$)

in $A$ (resp. $B$), one has ${\dim }_A(M)=\dim (A/\mathfrak{r})$, ${\dim }_B(N)=\dim (B/\mathfrak{s})$ by definition (16.1.7); on the other hand, $Supp(M{\otimes }_{A}N)$ is a closed subset of $\mathrm{Spec}(B)$ which identifies (I, 9.1.13) with Supp(M) ×_{Spec(A)} Supp(N) = Spec((B/𝔰) ⊗_A (A/𝔯)) = Spec(B/(𝔰 + 𝔯B)). Since $N{\otimes }_{A}k$ is a $B$-module of finite type, one has ${\dim }_{B{\otimes }_{A}k}(N{\otimes }_{A}k)={\dim }_B(N{\otimes }_{A}k)$ (16.1.9), and $Supp(N{\otimes }_{A}k)$ is equal to $\mathrm{Spec}(B/(\mathfrak{s}+\mathfrak{m}B))$ by the same reasoning as above. Applying (16.3.9.1) to $A/\mathfrak{r}$ and to $B/(\mathfrak{s}+\mathfrak{r}B)$ one gets

            dim(B/(𝔰 + 𝔯B)) ≤ dim(A/𝔯) + dim(B/(𝔰 + 𝔪 B))

which is nothing other than (16.3.9.2).

Corollary (16.3.10).

Under the hypotheses of (16.3.9), if $\mathfrak{m}B$ is an ideal of definition of $B$, one has $\dim (B)\le \dim (A)$; if in addition $A$ is integral and $\dim (A)=\dim (B)$, $\varphi$ is injective.

The first assertion results from (16.3.9) since then $\dim (B/\mathfrak{m}B)=0$. One may moreover replace $A$ by $\varphi (A)$, hence $\dim (B)\le \dim (\varphi (A))$; if $\mathfrak{a}=Ker(\varphi )\ne 0$, and if $A$ is integral, one has dim(φ(A)) = dim(A/𝔞) < dim(A) (16.1.2.2), hence one cannot then have $\dim (A)=\dim (B)$ unless $\mathfrak{a}=0$.

16.4. Depth and codepth ¹

Proposition (16.4.1).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $M$ an $A$-module of finite type. Then every $M$-regular sequence $(x_i{)}_{1\le i\le r}$ (15.1.7) formed of elements of $\mathfrak{m}$ forms part of a system of parameters for $M$.

We reason by induction on $r$; consider the $A$-module

                        N = M/(x_1 M + ⋯ + x_{r-1} M);

by hypothesis $x_1,\cdots ,x_{r-1}$ form part of a system of parameters for $M$, hence $\dim (N)=\dim (M)-(r-1)$ (16.3.7). Since by hypothesis the homothety of ratio $x_r$ in $N$ is injective, $x_r$ does not belong to any of the prime ideals associated with $N$, hence (16.3.4) one has $\dim (N/x_{r}N)=\dim (N)-1$, which is also written

                        dim(M/(x_1 M + ⋯ + x_r M)) = dim(M) − r;

the conclusion follows therefore from (16.3.7).

For an $A$-module $M\ne 0$, an $M$-regular sequence of elements of $\mathfrak{m}$ has therefore at most $\dim (M)$ elements.

Lemma (16.4.2).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field, $M$ an $A$-module. For an $M$-regular sequence $(x_i{)}_{1\le i\le n}$ of elements of $\mathfrak{m}$ to be maximal, it is necessary and sufficient that one have ${\mathrm{Hom}}_A(k,M/(x_{1}M+\cdots +x_{n}M))\ne 0$.

To say that $(x_i)$ is maximal means that, for no $x\in \mathfrak{m}$, the homothety of ratio $x$ in $N=M/(x_{1}M+\cdots +x_{n}M)$ is injective, or again that $\mathfrak{m}$ is contained in the union of the prime ideals associated with $N$ (Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 2 of prop. 2), hence equal to one of these, since it is a maximal ideal

¹ In the 3rd part of chap. III, we develop the notion of depth from the cohomological point of view and in a more general framework.

(Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2); in other words, there is an element $z\in N$ whose annihilator is $\mathfrak{m}$, and consequently the submodule Az of $N$ is isomorphic to $A/\mathfrak{m}=k$, whence the lemma.

Lemma (16.4.3).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field, $M$ an $A$-module, $(x_i{)}_{1\le i\le n}$ an $M$-regular sequence of elements of $\mathfrak{m}$. Then the $A$-modules ${\mathrm{Hom}}_A(k,M/(x_{1}M+\cdots +x_{n}M))$ and $Ex{t}_A^n(k,M)$ are isomorphic, and for $n\ge 1$, they are also isomorphic to $Ex{t}_A^{n-1}(k,M/x_{1}M)$.

We reason by induction on $n$, the proposition being evident for $n=0$. Set $N=M/x_{1}M$; the sequence $(x_i{)}_{2\le i\le n}$ is $N$-regular, hence

            Hom_A(k, M/(x_1 M + ⋯ + x_n M)) = Hom_A(k, N/(x_2 N + ⋯ + x_n N))

is isomorphic to $Ex{t}_A^{n-1}(k,N)$ by virtue of the induction hypothesis. Consider then the exact sequence $0\to M--x_1\to M\to M/x_{1}M=N\to 0$; from it one deduces the exact sequence of Exts

(16.4.3.1)      Ext_A^{n-1}(k, M) → Ext_A^{n-1}(k, N) → Ext_A^n(k, M) --x_1→ Ext_A^n(k, M).

By virtue of the induction hypothesis, $Ex{t}_A^{n-1}(k,M)$ is isomorphic to

                        Hom_A(k, M/(x_1 M + ⋯ + x_{n-1} M)),

which is zero, since $(x_i{)}_{1\le i\le n-1}$ is not a maximal $M$-regular sequence (16.4.2). On the other hand, since $x_1\in \mathfrak{m}$, the homothety of ratio $x_1$ in the $A$-module ${\mathrm{Hom}}_A(k,T)$ is zero for every $A$-module $T$, hence so is the homothety of ratio $x_1$ in every $A$-module $Ex{t}_A^i(k,T)$; the assertions of the lemma then follow at once from the exact sequence (16.4.3.1).

Corollary (16.4.4).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k=A/\mathfrak{m}$ its residue field, $M$ an $A$-module of finite type $\ne 0$. All maximal $M$-regular sequences of elements of $\mathfrak{m}$ have the same number of elements, which is the smallest integer $n$ such that $Ex{t}_A^n(k,M)\ne 0$.

Definition (16.4.5).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $M$ an $A$-module of finite type. We call depth of $M$, and denote $pro{f}_{\mathfrak{m}}(M)$ or $prof(M)$, the supremum of the number of elements of an $M$-regular sequence formed of elements of $\mathfrak{m}$.

One has therefore $prof(M)=+\infty$ if and only if $M=0$, and, by (16.4.1)

(16.4.5.1)              prof(M) ≤ dim(M)            if M ≠ 0.

One has evidently $prof(M^k)=prof(M)$ for every $k>0$.

Proposition (16.4.6).

Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $M$ an $A$-module of finite type.

(i) For $prof(M)=0$, it is necessary and sufficient that $\mathfrak{m}\in Ass(M)$ (or again that there exist a submodule of $M$ isomorphic to the residue field $k=A/\mathfrak{m}$ of $A$).

(ii) For every $M$-regular element $x$ belonging to $\mathfrak{m}$, one has

$$(\mathrm{16.4.6.1})prof(M/xM)=prof(M)-1.$$

(iii) If $M\ne 0$, one has

(16.4.6.2)              prof(M) ≤ inf_{𝔭 ∈ Ass(M)} dim(A/𝔭) ≤ dim(M).

(iv) One has $pro{f}_A(M)=pro{f}_{\hat{A}}(\hat{M})$.

(i) To say that $prof(M)=0$ means that there exists no $x\in \mathfrak{m}$ such that the homothety of ratio $x$ in $M$ is injective, in other words that every $x\in \mathfrak{m}$ belongs to a prime ideal associated with $M$ (Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 2 of prop. 2); but $\mathfrak{m}$ cannot be the union of the prime ideals associated with $M$ unless it is equal to one of them (Bourbaki, Alg. comm., chap. II, §1, n° 1, prop. 2).

(ii) If $(x_i{)}_{1\le i\le r}$ is an $(M/xM)$-regular sequence formed of elements of $\mathfrak{m}$, the sequence formed of $x$ and the $x_i$ is $M$-regular, and it follows from the definition that it is maximal when the sequence $(x_i{)}_{1\le i\le r}$ is a maximal $(M/xM)$-regular sequence; whence the conclusion.

(iii) We reason by induction on $n=prof(M)$, the assertion being evident for $n=0$. We shall use the following lemma:

Lemma (16.4.6.3).

Let $A$ be a Noetherian ring, $M$ an $A$-module of finite type, $t\in A$ an $M$-regular element. Then, for every $\mathfrak{p}\in Ass(M)$, every prime ideal minimal among those containing $\mathfrak{p}+At$ is associated with $M/tM$.

One knows that there exists an exact sequence

$$0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$$

such that $Ass(M^{\prime })=\mathfrak{p}$, $Ass(M^{\prime \prime })=Ass(M)-\mathfrak{p}$ (Bourbaki, Alg. comm., chap. IV, §1, n° 1, prop. 4); since $t$ is $M$-regular, it is also $M^{\prime }$-regular and M''-regular (loc. cit., §1, n° 1, cor. 2 of prop. 2). One deduces that the sequence

$$0\to M^{\prime }/t{M}^{\prime }\to M/tM\to M^{\prime \prime }/t{M}^{\prime \prime }\to 0$$

is exact (15.1.18). Consequently, one has $Ass(M^{\prime }/t{M}^{\prime })\subset Ass(M/tM)$. But the prime ideals of $A$ containing $\mathfrak{p}$ are the points of $Supp(M^{\prime })$ (loc. cit., §1, n° 3, prop. 7), hence the prime ideals containing $\mathfrak{p}+At$ are the points of the support of $M^{\prime }/t{M}^{\prime }=M^{\prime }{\otimes }_A(A/tA)$ $(0_I,\mathrm{1.7.5})$; a minimal element of the set of these prime ideals therefore belongs to $Ass(M^{\prime }/t{M}^{\prime })$ (Bourbaki, loc. cit., §1, n° 3, cor. 1 of prop. 7), and a fortiori to $Ass(M/tM)$.

This lemma being established, we return to the proof of (iii). Let $x\in \mathfrak{m}$ be an $M$-regular element, so that $prof(M/xM)=n-1$ by (ii); for every $\mathfrak{p}\in Ass(M)$, it follows from the lemma that there is an ideal ${\mathfrak{p}}^{\prime }$ associated with $M/xM$ and containing $\mathfrak{p}+Ax$; but as $x$ is $M$-regular, one has $x\notin \mathfrak{p}$, whence ${\mathfrak{p}}^{\prime }\ne \mathfrak{p}$ and consequently $\dim (A/{\mathfrak{p}}^{\prime })\le \dim (A/\mathfrak{p})-1$. Now the induction hypothesis shows that $n-1\le \dim (A/{\mathfrak{p}}^{\prime })$; one concludes that $n\le \dim (A/\mathfrak{p})$ for every $\mathfrak{p}\in Ass(M)$.

(iv) One may restrict oneself to the case where $M\ne 0$. Recall that $\hat{M}=M{\otimes }_A\hat{A}$ up to a canonical isomorphism $(0_I,\mathrm{7.3.3})$; since Â is a flat $A$-module, one has canonical isomorphisms $Ex{t}_A^i(k,M){\otimes }_A\hat{A}\cong Ex{t}_{\hat{A}}^i(k,M{\otimes }_A\hat{A})$ since $k=k{\otimes }_A\hat{A}$ up to a canonical isomorphism $(0_{III},\mathrm{12.3.5})$. Assertion (iv) thus results from (16.4.4) and from the fact that Â is a faithfully flat $A$-module $(0_I,\mathrm{7.3.5})$.

(16.4.6.4) One notes that if $\mathfrak{p}$ is a prime ideal of $A$, one may have either $pro{f}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})<pro{f}_A(M)$ or $pro{f}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})>pro{f}_A(M)$; for an example of the first case,

it suffices to take for $A$ a Noetherian local integral ring of dimension $\ge 1$; one has $prof(A)\ge 1$ but $prof(K)=0$ for the field of fractions $K$ of $A$. For an example of the second case, consider a Noetherian local integral ring A_0 of dimension $\ge 2$, with maximal ideal ${\mathfrak{m}}_0$ and let $k=A_0/{\mathfrak{m}}_0$ be its residue field; let $A=A_0\oplus \mathfrak{j}$ be the trivial extension of A_0 by the A_0-module $k$ (18.2.3), the ideal $\mathfrak{j}$ being A_0-isomorphic to $k$ (so that multiplication in $A$ is given by $(x,\xi )(x^{\prime },{\xi }^{\prime })=(x{x}^{\prime },x{\xi }^{\prime }+x^{\prime }\xi )$). It is clear that $\mathfrak{j}$ is the nilradical of $A$ and $A/\mathfrak{j}$ is isomorphic to A_0, so that every prime ideal $\mathfrak{p}$ of $A$ is of the form ${\mathfrak{p}}_0\oplus \mathfrak{j}$, where ${\mathfrak{p}}_0$ is a prime ideal of A_0; in particular $\mathfrak{m}={\mathfrak{m}}_0\oplus \mathfrak{j}$ is the unique maximal ideal of $A$. As every element of ${\mathfrak{m}}_0$ annihilates $\mathfrak{j}$, one has $prof(A)=prof(A_{\mathfrak{m}})=0$; on the other hand, if ${\mathfrak{p}}_0\ne {\mathfrak{m}}_0$, the elements of $\mathfrak{j}$ are annihilated by those of ${\mathfrak{m}}_0-{\mathfrak{p}}_0$, hence $A_{\mathfrak{p}}$ is canonically isomorphic to $(A_0{)}_{{\mathfrak{p}}_0}$, and as A_0 is integral, $prof(A_{\mathfrak{p}})=prof((A_0{)}_{{\mathfrak{p}}_0})\ge 1$ if ${\mathfrak{p}}_0\ne 0$.

Corollary (16.4.7).

For a Noetherian reduced local ring $A$ to be of depth 0, it is necessary and sufficient that $A$ be a field.

Indeed, since the intersection of the minimal prime ideals of $A$ is 0, $A$ has no embedded associated prime ideals; by virtue of (16.4.6, (i)), if $prof(A)=0$, the maximal ideal $\mathfrak{m}$ of $A$ is also a minimal prime ideal, hence is the only prime ideal of $A$, and since $A$ is reduced, $\mathfrak{m}=0$.

Proposition (16.4.8).

Let $A$, $B$ be two Noetherian local rings, $\rho :A\to B$ a local homomorphism, $M$ a $B$-module such that $M_{[\rho ]}$ is an $A$-module of finite type. Then one has

$$(\mathrm{16.4.8.1})pro{f}_A(M_{[\rho ]})=pro{f}_B(M).$$

One may restrict oneself to the case $M\ne 0$; suppose that $pro{f}_A(M_{[\rho ]})=n$. If $(x_i{)}_{1\le i\le n}$ is a maximal $(M_{[\rho ]})$-regular sequence formed of elements of the maximal ideal of $A$, the $\rho (x_i)$ constitute an $M$-regular sequence formed of elements of the maximal ideal of $B$, and one has therefore, on setting $N=M/(\rho (x_1)M+\cdots +\rho (x_n)M)$, $pro{f}_B(N)=pro{f}_B(M)-n$ (16.4.6, (ii)); similarly $N_{[\rho ]}=M_{[\rho ]}/(x_1{M}_{[\rho ]}+\cdots +x_n{M}_{[\rho ]})$ and $pro{f}_A(N_{[\rho ]})=0$; one is thus reduced to proving the proposition when $n=0$. Let $k$ be the residue field of $A$; since $k$ is a quotient of $A$, the $A$-module $P={\mathrm{Hom}}_A(k,M_{[\rho ]})$ is a submodule of ${\mathrm{Hom}}_A(A,M_{[\rho ]})=M_{[\rho ]}$ and the hypothesis entails that $P\ne 0$ (16.4.6, (i)); in addition, $P$ is an $A$-module of finite type (since $M_{[\rho ]}$ is an $A$-module of finite type), hence a $k$-vector space of finite type, being annihilated by the maximal ideal of $A$, and finally an $A$-module of finite length. On the other hand, $P$ is also a submodule of the $B$-module $M$, and since it is of finite length as an $A$-module, it is a fortiori so as a $B$-module. Since it is $\ne 0$, it contains a simple sub-$B$-module, that is to say isomorphic to the residue field of $B$; one concludes then from (16.4.6, (i)) that one has indeed $pro{f}_B(M)=0$.

Definition (16.4.9).

Let $A$ be a Noetherian local ring, $M$ an $A$-module of finite type. If $M\ne 0$, we call codepth of $M$ and denote $copro{f}_{\mathfrak{m}}(M)$ or $coprof(M)$, the finite integer ${\dim }_A(M)-pro{f}_A(M)\ge 0$ (16.4.5.1). If $M=0$, we set $coprof(M)=0$.

One has $coprof(M^k)=coprof(M)$ for every $k>0$.

Proposition (16.4.10).

Let $A$ be a Noetherian local ring, $M$ an $A$-module of finite type.

(i) For every $M$-regular element $x$ belonging to the maximal ideal of $A$, one has $coprof(M/xM)=coprof(M)$.

(ii) One has $copro{f}_A(M)=copro{f}_{\hat{A}}(\hat{M})$.

Indeed, (i) results from (16.3.4) and (16.4.6, (ii)), and (ii) from (16.2.4) and (16.4.6, (iv)).

We shall show later (IV, 6.11.5) that for every prime ideal $\mathfrak{p}$ of $A$, one has $copro{f}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})\le copro{f}_A(M)$.

Proposition (16.4.11).

Under the hypotheses of (16.4.8), one has

$$(\mathrm{16.4.11.1})copro{f}_A(M_{[\rho ]})\ge copro{f}_B(M).$$