Éléments de Géométrie Algébrique — English

§3. Associated prime cycles and primary decompositions

In this section we mainly give the translation of the results on modules expounded in Bourbaki, Alg. comm., chap. IV, which we follow very closely. The notions that follow seem to be of interest only in the case of locally Noetherian preschemes.

3.1. Associated prime cycles of a Module

Definition (3.1.1).

Let $X$ be a prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module. We say that a point $x\in X$ is associated to $\mathcal{F}$ if the maximal ideal ${\mathfrak{m}}_x$ of ${\mathcal{O}}_x$ is associated to the ${\mathcal{O}}_x$-Module ${\mathcal{F}}_x$ (in other words, if ${\mathfrak{m}}_x$ is the annihilator of an element of ${\mathcal{F}}_x$). We denote by $Ass(\mathcal{F})$ the set of $x\in X$ associated to $\mathcal{F}$. We say that a closed irreducible subset $Z$ of $X$ is an associated prime cycle of $\mathcal{F}$ if its generic point is associated to $\mathcal{F}$. When $\mathcal{F}={\mathcal{O}}_X$, we shall also say that the points (resp. prime cycles) associated to $\mathcal{F}$ are associated to the prescheme $X$.

We say that an associated prime cycle of $\mathcal{F}$ (resp. $X$) is embedded if it is contained in another associated prime cycle of $\mathcal{F}$ (resp. $X$) (in other words, if it is not maximal in the set of these cycles).

If $X$ is locally Noetherian, the irreducible components of $X$ are nothing other than the maximal (or non-embedded) prime cycles associated to $X$.

It is clear that if $x\in Ass(\mathcal{F})$, then ${\mathcal{F}}_x\ne 0$; in other words

$$(\mathrm{3.1.1.1})Ass(\mathcal{F})\subset Supp(\mathcal{F}).$$

If $x\in X$ is associated to $\mathcal{F}$, it is evidently associated to $\mathcal{F}|U$ for every open neighbourhood $U$ of $x$, and conversely, if it is associated to $\mathcal{F}|U$ for one of these neighbourhoods, it is associated to $\mathcal{F}$.

We note finally that for a quasi-coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$, the existence of embedded associated prime cycles is a local question, since if $y$ and $z$ are two points of $Ass(\mathcal{F})$ with $y\in \overline{z}$, every neighbourhood of $y$ contains $z$.

Proposition (3.1.2).

Let $A$ be a Noetherian ring, $M$ an $A$-module, $X=\mathrm{Spec}(A)$, $\mathcal{F}=\tilde{M}$; for a point $x\in X$ to be associated to $\mathcal{F}$, it is necessary and sufficient that the prime ideal ${\mathfrak{j}}_x$ of $A$ be associated to the module $M$ (in other words, be the annihilator of an element $f\in M$).

This results from the definition (3.1.1) and from Bourbaki, loc. cit., §1, n° 2, cor. of prop. 5, applied to $S=A-{\mathfrak{j}}_x$.

Proposition (3.1.3).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module, $x$ a point of $X$, $Z$ the reduced closed sub-prescheme of $X$ having $\overline{x}$ as underlying space (I, 5.2.1). The following conditions are equivalent:

a) $x\in Ass(\mathcal{F})$.

b) There exists an open neighbourhood $U$ of $x$ and a section $f\in \Gamma (U,\mathcal{F})$ such that $U\cap Z$ is equal to $Supp({\mathcal{O}}_U\cdot f)$.

b') There exists an open neighbourhood $U$ of $x$ and a section $f\in \Gamma (U,\mathcal{F})$ such that $U\cap Z$ is an irreducible component of $Supp({\mathcal{O}}_U\cdot f)$.

c) There exists an open neighbourhood $U$ of $x$ and a sub-Module of $\mathcal{F}|U$ isomorphic to ${\mathcal{O}}_Z|U$ (${\mathcal{O}}_Z$ being identified with a quotient of ${\mathcal{O}}_X$).

c') There exists an open neighbourhood $U$ of $x$ and a coherent sub-Module $\mathcal{G}$ of $\mathcal{F}|U$ such that $U\cap Z$ is an irreducible component of $Supp(\mathcal{G})$.

It is clear that c) implies b), since it suffices to take for $f$ the element of $\Gamma (U,\mathcal{F})$ corresponding to the unit section of ${\mathcal{O}}_Z|U$. As $U\cap Z$ is irreducible $(0_I,\mathrm{2.1.6})$, b) implies b'), and b') implies c') since ${\mathcal{O}}_X$ is coherent $(0_I,\mathrm{5.3.4})$. To see that c') implies a), we may restrict to the case where $U=X=\mathrm{Spec}(A)$ is affine, $A$ therefore Noetherian, and where $\mathcal{F}=M$, $M$ being an $A$-module, and $\mathcal{G}=N$, where $N$ is a sub-module of $M$, of finite type. We then know that the minimal elements of $Supp(\mathcal{G})$ are the maximal points of $V(Ann(N))$ $(0_I,\mathrm{1.7.4})$, and these are also the minimal elements of $Ass(N)$ (Bourbaki, Alg. comm., chap. IV, §1, n° 3, cor. 1 of prop. 7); since Ass(N) ⊂ Ass(M) = Ass(ℱ), we see that c') implies a). Finally, a) implies c) by virtue of (3.1.2), taking again $X$ affine, $\mathcal{F}=\tilde{M}$, and $Z$ defined by the ideal ${\mathfrak{j}}_x\cdot A$ (with the notation of (3.1.2)).

Corollary (3.1.4).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. Then the maximal prime cycles associated to $\mathcal{F}$ are the irreducible components of $Supp(\mathcal{F})$, and the generic points of these components are the points $x\in X$ such that ${\mathcal{F}}_x$ is an ${\mathcal{O}}_x$-Module of finite length and $\ne 0$.

Indeed, if $x$ is the generic point of one of the irreducible components $Z$ of $Supp(\mathcal{F})$, it follows from the equivalence of a) and c') in (3.1.3) that $x$ belongs to $Ass(\mathcal{F})$, and $Z$ is an associated prime cycle of $\mathcal{F}$, necessarily maximal by virtue of (3.1.1.1); the converse follows trivially from (3.1.1.1). Finally, the last assertion, being evidently local, follows from Bourbaki, Alg. comm., chap. IV, §2, n° 5, cor. 2 of prop. 7.

Corollary (3.1.5).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module. For $\mathcal{F}=0$, it is necessary and sufficient that $Ass(\mathcal{F})=\varnothing$.

The question being local, we are reduced to the case where $X$ is affine, and the conclusion follows immediately from (3.1.2) and from Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 1 of prop. 2.

Proposition (3.1.6).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module; then $Ass(\mathcal{F})$ is locally finite (that is to say, every point of $X$ admits a neighbourhood whose intersection with $Ass(\mathcal{F})$ is finite).

It suffices to consider the case where $X$ is affine, hence Noetherian, and then the proposition follows from (3.1.2) and from Bourbaki, Alg. comm., chap. IV, §1, n° 4, cor. of th. 2.

Proposition (3.1.7).

Let $X$ be a prescheme.

(i) Let $0\to {\mathcal{F}}^{\prime }\to \mathcal{F}\to {\mathcal{F}}^{\prime \prime }\to 0$ be an exact sequence of quasi-coherent ${\mathcal{O}}_X$-Modules. Then Ass(ℱ') ⊂ Ass(ℱ) ⊂ Ass(ℱ') ∪ Ass(ℱ'').

(ii) Let $\mathcal{F}$ be a quasi-coherent ${\mathcal{O}}_X$-Module, $({\mathcal{F}}_{\alpha })$ a family of quasi-coherent sub-${\mathcal{O}}_X$-Modules of $\mathcal{F}$ such that $\mathcal{F}$ is the union of the ${\mathcal{F}}_{\alpha }$. Then $Ass(\mathcal{F})={\bigcup }_{\alpha }Ass({\mathcal{F}}_{\alpha })$.

(iii) For every family $({\mathcal{F}}_{\alpha })$ of quasi-coherent ${\mathcal{O}}_X$-Modules, one has $Ass({⨁}_{\alpha }{\mathcal{F}}_{\alpha })={\bigcup }_{\alpha }Ass({\mathcal{F}}_{\alpha })$.

One is immediately reduced to the corresponding propositions for modules (Bourbaki, loc. cit., §1, n° 1, formula (1), prop. 3 and cor. 1 of prop. 3).

Proposition (3.1.8).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module, $U$ an open subset of $X$, $\mathfrak{J}$ a coherent Ideal of ${\mathcal{O}}_X$ defining a closed sub-prescheme of $X$ having $X-U$ as underlying space. The following conditions are equivalent:

a) $Ass(\mathcal{F})\subset U$.

b) For every affine open subset $V$, every section of $\mathcal{F}$ over $V$ whose restriction to $V\cap U$ is zero, is equal to 0.

c) The canonical homomorphism $\mathcal{F}\to \mathcal{H}o{m}_{{\mathcal{O}}_X}(\mathfrak{J},\mathcal{F})$ is injective.

The question being local, we may suppose $X=\mathrm{Spec}(A)$ affine, $A$ being a Noetherian ring, $\mathcal{F}=M$, $M$ an $A$-module, and $\mathfrak{J}=\mathfrak{J}$, where $\mathfrak{J}$ is an ideal of $A$. The homomorphism $M\to {\mathrm{Hom}}_A(\mathfrak{J},M)$ associates to $m\in M$ the homomorphism $x\mapsto xm$ from $\mathfrak{J}$ to $M$; to say that it is not injective means that there exists $m\ne 0$ in $M$ such that $\mathfrak{J}m=0$.

Let us first show that c) implies a): indeed, if $Ass(\mathcal{F})$ then met $X-U$, there would be a prime ideal $\mathfrak{p}\in Ass(M)$ containing $\mathfrak{J}$, hence an element $m\ne 0$ of $M$ such that $\mathfrak{p}m=0$. Secondly, b) implies c): indeed, if one then had $\mathfrak{J}m=0$ for some $m\ne 0$ in $M$, then for every prime ideal $\mathfrak{q}⊉\mathfrak{J}$, there exists $a\in \mathfrak{J}$ not in $\mathfrak{q}$, hence the relation $am=0$ entails that the canonical image of $m$ in $M_{\mathfrak{q}}$ is 0; in other words $m$ would be a section $\ne 0$ of $\mathcal{F}$ over $X$ whose restriction to $U$ would be zero. Finally, a) entails b). To see this, let us note that the canonical homomorphism $M\to {\prod }_{\mathfrak{p}\in Ass(M)}{M}_{\mathfrak{p}}$ is injective: indeed, if $N$ is the kernel of this homomorphism, one has $Ass(N)\subset Ass(M)$; if there existed $\mathfrak{p}\in Ass(N)$, there would be an element $n\in N$ whose annihilator would be $\mathfrak{p}$; but by definition of $N$, there is an element $s\notin \mathfrak{p}$ such that $sn=0$, which is absurd; one concludes that $Ass(N)=\varnothing$, whence $N=0$. Now condition a) entails that if $m\in M$ is a section of $\mathcal{F}$ whose restriction to $U$ is zero, the canonical image of $m$ in $M_{\mathfrak{p}}$ is zero for every $\mathfrak{p}\in Ass(M)$, hence $m=0$. Q.E.D.

Corollary (3.1.9).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module, $f$ a section of ${\mathcal{O}}_X$ over $X$. For $f$ to be $\mathcal{F}$-regular (0, 15.2.2), it is necessary and sufficient that $Ass(\mathcal{F})\subset X_f$.

Indeed, it is immediate that to say that $f$ is $\mathcal{F}$-regular means that the canonical homomorphism $\mathcal{F}\to \mathcal{H}o{m}_{{\mathcal{O}}_X}(f{\mathcal{O}}_X,\mathcal{F})$ is injective, and it suffices to apply (3.1.8) to the Ideal $\mathfrak{J}=f{\mathcal{O}}_X$.

Proposition (3.1.10).

Let $X$, $Y$ be locally Noetherian preschemes, $f:X\to Y$ an integral morphism. Then, for every quasi-coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$, one has $f(Ass(\mathcal{F}))=Ass(f_{\ast }(\mathcal{F}))$.

The question being local on $Y$ and the morphism $f$ being affine, one is immediately reduced to the case where $Y$ is affine; in other words, to the

Lemma (3.1.10.1).

Let $A$, $B$ be two Noetherian rings, $\rho :A\to B$ a ring homomorphism making $B$ into an integral $A$-algebra, $M$ a $B$-module. Then the prime ideals $\mathfrak{p}\in Ass(M_{[\rho ]})$ are the inverse images by $\rho$ of the prime ideals $\mathfrak{q}\in Ass(M)$.

Indeed, if $\mathfrak{q}\in Ass(M)$, $\mathfrak{q}$ is the annihilator in $B$ of an element $x\in M$, hence ${\rho }^{-1}(\mathfrak{q})$ is the annihilator in $A$ of $x$. Conversely, let $\mathfrak{p}\in Ass(M_{[\rho ]})$, so that $\mathfrak{p}$ is the inverse image by $\rho$ of the annihilator $\mathfrak{b}$ in $B$ of an element $x\in M$; it follows from the first theorem of Cohen-Seidenberg that there exists a prime ideal $\mathfrak{q}$ of $B$ containing $\mathfrak{b}$ and whose inverse image is $\mathfrak{p}$ (Bourbaki, Alg. comm., chap. V, §2, n° 1, cor. 2 of th. 1); on considering one of the prime ideals minimal among those contained in $\mathfrak{q}$ and containing $\mathfrak{b}$, we may evidently suppose that $\mathfrak{q}$ itself is one of these minimal ideals. But as $B\cdot x\subset M$ is isomorphic to $B/\mathfrak{b}$, we know that one then has $\mathfrak{q}\in Ass(B/\mathfrak{b})\subset Ass(M)$ (Bourbaki, Alg. comm., chap. IV, §1, n° 4, th. 2).

Corollary (3.1.11).

Under the hypotheses of (3.1.10), for $\mathcal{F}$ to be without embedded associated prime cycle, it suffices that the same be true of $f_{\ast }(\mathcal{F})$.

Suppose indeed that $f_{\ast }(\mathcal{F})$ has no embedded associated prime cycle. Note that if $A$ is an integral algebra over a field $k$, all the prime ideals of $A$ are maximal (Bourbaki, Alg. comm., chap. V, §2, n° 1, prop. 1); it follows from (I, 6.2.2) that the fibres of $f$ are discrete spaces. If $x$, $x^{\prime }$ are two distinct points of $Ass(\mathcal{F})$, neither of them can be adherent to the other if $f(x)=f(x^{\prime })$; and if $f(x)\ne f(x^{\prime })$, (3.1.10) and the hypothesis entail that neither of the two points $f(x)$, $f(x^{\prime })$ can be adherent to the other, hence the same is true of $x$ and $x^{\prime }$.

Remark (3.1.12).

Under the hypotheses of (3.1.10), it can on the other hand happen that $\mathcal{F}$ is without embedded associated prime cycle, but not $f_{\ast }(\mathcal{F})$. Take for example $Y=\mathrm{Spec}(k[T])$ where $k$ is a field ("affine line"), and $X$ the sum of $X_1=Y$ and $X_2=\mathrm{Spec}(k)$, the morphism $X_2\to Y$ corresponding to the canonical homomorphism $k[T]\to k[T]/\mathfrak{m}$, where $\mathfrak{m}$ is the maximal ideal (T). It is clear that the morphism $f:X\to Y$ is finite; if one takes $\mathcal{F}={\mathcal{O}}_X$, then $\mathcal{F}$ is without embedded associated prime cycle, but $f_{\ast }(\mathcal{F})=\tilde{M}$, where $M$ is the k[T]-module direct sum of k[T] and $k$, hence $Ass(M)$ is formed of the generic point (0) of $Y$ and the point $\mathfrak{m}$.

Proposition (3.1.13).

Let $X$ be a locally Noetherian prescheme, $U$ an open subset of $X$, $i:U\to X$ the canonical injection. For every quasi-coherent ${\mathcal{O}}_U$-Module $\mathcal{F}$, one has $Ass(i_{\ast }(\mathcal{F}))=Ass(\mathcal{F})$.

Recall that $i_{\ast }(\mathcal{F})$ is a quasi-coherent ${\mathcal{O}}_X$-Module (1.2.2 and 1.7.4); as $i_{\ast }(\mathcal{F})|U=\mathcal{F}$, one has $(Ass(i_{\ast }(\mathcal{F})))\cap U=Ass(\mathcal{F})$, and it therefore remains to prove that $Ass(i_{\ast }(\mathcal{F}))\subset U$. But by (3.1.8), this relation means that for every affine open $V$ of $X$, every section of $i_{\ast }(\mathcal{F})$ over $V$ which is zero in $U\cap V$, is zero, a condition trivially verified since $\Gamma (V,i_{\ast }(\mathcal{F}))=\Gamma (U\cap V,\mathcal{F})=\Gamma (U\cap V,i_{\ast }(\mathcal{F}))$.

3.2. Irredundant decompositions

Proposition (3.2.1).

Let $X$ be a locally Noetherian prescheme, $U$ a dense open subset of $X$. The following conditions are equivalent:

a) $X$ is reduced.

b) The induced sub-prescheme on $U$ is reduced and $X$ is without embedded prime cycle.

c) $X$ is without embedded prime cycle and for every generic point $x$ of an irreducible component of $X$, one has $long({\mathcal{O}}_x)=1$.

The prime cycles associated to $X$ are then identical to the irreducible components of $X$.

It is clear that if $X$ is reduced, the same is true of the sub-prescheme induced on $U$. Moreover, the existence of embedded prime cycles being local, we may restrict to the case where $X=\mathrm{Spec}(A)$ is affine, $A$ Noetherian. If $A$ is reduced, we know that the minimal prime ideals of $A$ form a reduced primary decomposition of (0) (Bourbaki, Alg. comm., chap. IV, §2, n° 5, prop. 10) and are the elements of $Ass(A)$, hence there exist no embedded prime ideals associated to $A$, which shows that a) implies b). It is immediate that b) entails c), since a generic point $x$ of an irreducible component of $X$ belongs to $U$, hence ${\mathcal{O}}_x$ is a field. Finally, c) entails a): it suffices indeed to note that if $\mathcal{N}$ is the Nilradical of ${\mathcal{O}}_X$, which is a coherent Ideal, $Supp(\mathcal{N})$ cannot contain any of the generic points of the irreducible components of $X$ by hypothesis; if $Supp(\mathcal{N})$ were not empty and if $x$ were one of the maximal points of this closed set, the criterion (3.1.3, c')) would show that $x\in Ass({\mathcal{O}}_X)$, and $\overline{x}$ would therefore be an embedded prime cycle of $X$, contrary to the hypothesis; hence $\mathcal{N}=0$.

Definition (3.2.2).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. We say that $\mathcal{F}$ is reduced if it satisfies the two following conditions: 1° $\mathcal{F}$ is without embedded associated prime cycle; 2° for every maximal point $x$ of $Supp(\mathcal{F})$, one has $long({\mathcal{F}}_x)=1$.

Condition 1° means that the associated prime cycles of $\mathcal{F}$ are the irreducible components of $Supp(\mathcal{F})$ (3.1.4), and condition 2° means that for every generic point $x$ of such a component one has $long({\mathcal{F}}_x)=1$.

For an affine scheme $X$, this definition gives in particular the notion of reduced module on a Noetherian ring $A$; an $A$-module of finite type $M$ is said to be reduced if it has no embedded associated prime ideals and if, for every $\mathfrak{p}\in Ass(M)$, $lon{g}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})=1$. Returning to the case of a locally Noetherian prescheme $X$ and of a coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$, we say that $\mathcal{F}$ is reduced at a point $x\in X$ if ${\mathcal{F}}_x$ is a reduced ${\mathcal{O}}_x$-module; that signifies again that, on the local scheme $\mathrm{Spec}({\mathcal{O}}_x)$, ${\tilde{\mathcal{F}}}_x$ is reduced; it therefore amounts to the same to say that $x$ belongs to no embedded associated prime cycle of $\mathcal{F}$ and that $long({\mathcal{F}}_z)=1$ for every maximal point $z$ of $Supp(\mathcal{F})$ such that $x\in \overline{z}$. It is clear that if $\mathcal{F}$ is a reduced coherent ${\mathcal{O}}_X$-Module, it is reduced at every point of $X$; conversely, if $\mathcal{F}$ is reduced at a point $x$, there exists an open neighbourhood $U$ of $x$ such that $\mathcal{F}|U$ is a reduced ${\mathcal{O}}_U$-Module: it suffices indeed to take $U$ meeting no embedded associated prime cycle of $\mathcal{F}$ (such a neighbourhood exists since these cycles form a locally finite set of closed parts

of $X$). To say that ${\mathcal{O}}_X$ is reduced at a point $x$ amounts to saying that $X$ is reduced at the point $x$.

Proposition (3.2.3).

Let $X$ be a locally Noetherian prescheme, $U$ an open subset of $X$, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module such that $U\cap Supp(\mathcal{F})$ is dense in $Supp(\mathcal{F})$. The following conditions are equivalent:

a) $\mathcal{F}$ is reduced.

b) $\mathcal{F}|U$ is reduced and $\mathcal{F}$ is without embedded associated prime cycle.

c) There exist a reduced closed sub-prescheme $X^{\prime }$ of $X$ and a coherent ${\mathcal{O}}_{X^{\prime }}$-Module ${\mathcal{F}}^{\prime }$ torsion-free and of rank 1 on every irreducible component of $X^{\prime }$ such that, if $j:X^{\prime }\to X$ is the canonical injection, one has $j_{\ast }({\mathcal{F}}^{\prime })=\mathcal{F}$.

Moreover, when this is so, the sub-prescheme $X^{\prime }$ is defined by the Ideal $\mathfrak{J}$ of ${\mathcal{O}}_X$ annihilator of $\mathcal{F}$.

To see that c) implies a), one may, by virtue of (3.1.3), restrict to the case where $X^{\prime }=X$ is integral, with generic point $x$, and one may moreover suppose $X=\mathrm{Spec}(A)$ affine and $\mathcal{F}=\tilde{M}$, where $A$ is therefore integral and $M$ is a torsion-free $A$-module of rank 1; the annihilator of every element of $M$ then being reduced to 0, one has $Ass(\mathcal{F})=x$ and ${\mathcal{F}}_x$ is isomorphic to ${\mathcal{O}}_x$, the field of fractions of $A$, so the conditions of definition (3.2.2) are satisfied. As the existence of embedded associated prime cycles is local, it is clear that if $\mathcal{F}$ has no such cycles and if $U\cap Supp(\mathcal{F})$ is dense in $Supp(\mathcal{F})$, then $Ass(\mathcal{F})=Ass(\mathcal{F}|U)$, hence a) and b) are equivalent. If a) is satisfied, take for $X^{\prime }$ the reduced closed sub-prescheme of $X$ whose underlying space is $Supp(\mathcal{F})$ (I, 5.2.1), and let ${\mathcal{F}}^{\prime }=j\ast (\mathcal{F})$; a point $x$ of $Ass(\mathcal{F})$ is necessarily a maximal point of $X^{\prime }$, and as $long({\mathcal{F}}_x)=1$, ${\mathcal{F}}_x^{\prime }$ is isomorphic to ${\mathcal{F}}_x$, hence to the field $k(x)={\mathcal{O}}_{X^{\prime },x}$, which proves that ${\mathcal{F}}^{\prime }$ is torsion-free and of rank 1 (I, 7.4.6 and 7.4.2). Finally, the last assertion is trivial, since for every $y\in X^{\prime }$, the annihilator of the ${\mathcal{O}}_{X^{\prime },y}$-Module ${\mathcal{F}}_y^{\prime }$ is zero.

Definition (3.2.4).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module. We say that $\mathcal{F}$ is irredundant if $Ass(\mathcal{F})$ is reduced to a single element $x$; if $\mathcal{F}$ is of finite type, we say that $\mathcal{F}$ is integral if moreover $\mathcal{F}$ is reduced (in other words if $long({\mathcal{F}}_x)=1$). We say that a quasi-coherent sub-${\mathcal{O}}_X$-Module $\mathcal{G}$ of a quasi-coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$ is primary in $\mathcal{F}$ if $\mathcal{F}/\mathcal{G}$ is irredundant.

For an affine scheme $X$, this definition gives in particular the notion of integral module on a Noetherian ring $A$; an $A$-module $M$ is said to be integral if it is of finite type, if $M$ is primary (that is, $Ass(M)$ is reduced to a single prime ideal $\mathfrak{p}$) and if moreover $lon{g}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})=1$. Returning to the case of an arbitrary locally Noetherian prescheme $X$ and of a coherent ${\mathcal{O}}_X$-Module $\mathcal{F}$, we say that $\mathcal{F}$ is integral at a point $x\in X$ if ${\mathcal{F}}_x$ is an integral ${\mathcal{O}}_x$-module: that means again that $x$ belongs to only a single associated prime cycle (necessarily non-embedded) of $\mathcal{F}$ and that $long({\mathcal{F}}_z)=1$ at its generic point $z$. It is clear that if $\mathcal{F}$ is an integral coherent ${\mathcal{O}}_X$-Module, it is integral at every point of $X$; conversely, if $\mathcal{F}$ is integral at a point $x$, there exists an open neighbourhood $U$ of $x$ such that $\mathcal{F}|U$ is

an integral ${\mathcal{O}}_U$-Module: it suffices indeed to take $U$ such that $\mathcal{F}|U$ is a reduced ${\mathcal{O}}_U$-Module (3.2.2).

We say that the locally Noetherian prescheme $X$ is irredundant if ${\mathcal{O}}_X$ is irredundant (which implies that $X$ is irreducible); for $X$ to be integral, it is necessary and sufficient that the ${\mathcal{O}}_X$-Module ${\mathcal{O}}_X$ be integral ((I, 2.1.8) and (3.2.1)). If ${\mathcal{O}}_X$ is integral at a point $x$, that is, if the ring ${\mathcal{O}}_x$ is integral, we say that $X$ is integral at the point $x$. We say that a closed sub-prescheme $Y$ of $X$ is primary in $X$ if the Ideal $\mathfrak{J}$ of ${\mathcal{O}}_X$ that defines $Y$ is primary in ${\mathcal{O}}_X$.

Definition (3.2.5).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. We call an irredundant decomposition of $\mathcal{F}$ a family $({\mathcal{F}}_{\alpha }{)}_{\alpha \in I}$ of ${\mathcal{O}}_X$-Module quotients of $\mathcal{F}$ such that the ${\mathcal{F}}_{\alpha }$ are irredundant, the family $(Supp({\mathcal{F}}_{\alpha }))$ is locally finite, and the canonical homomorphism $\mathcal{F}\to {⨁}_{\alpha }{\mathcal{F}}_{\alpha }$ is injective. We say that such a decomposition is reduced if the sets $Ass({\mathcal{F}}_{\alpha })$ are pairwise distinct, and if there exists no subset $J\ne I$ such that the sub-family $({\mathcal{F}}_{\alpha }{)}_{\alpha \in J}$ is an irredundant decomposition of $\mathcal{F}$.

If $({\mathcal{F}}_{\alpha }{)}_{\alpha \in I}$ is an irredundant decomposition (resp. reduced irredundant decomposition) of $\mathcal{F}$ and if one sets ${\mathcal{F}}_{\alpha }=\mathcal{F}/{\mathcal{G}}_{\alpha }$, we also say that the family $({\mathcal{G}}_{\alpha }{)}_{\alpha \in I}$ of sub-${\mathcal{O}}_X$-Modules of $\mathcal{F}$ is a primary decomposition of 0 in $\mathcal{F}$; we note that the hypothesis of injectivity of the canonical homomorphism $\mathcal{F}\to {⨁}_{\alpha }{\mathcal{F}}_{\alpha }$ is equivalent to the condition ${\bigcap }_{\alpha \in I}{\mathcal{G}}_{\alpha }=0$.

If $({\mathcal{F}}_{\alpha })$ is an irredundant decomposition of $\mathcal{F}$, to say that it is reduced is equivalent to saying that the $Ass({\mathcal{F}}_{\alpha })$ are pairwise distinct and contained in $Ass(\mathcal{F})$; if $Ass({\mathcal{F}}_{\alpha })=x_{\alpha }$ for every $\alpha \in I$, $\alpha \mapsto x_{\alpha }$ is a bijection of $I$ onto $Ass(\mathcal{F})$: these properties are indeed local and therefore result from Bourbaki, Alg. comm., chap. IV, §2, n° 3, prop. 4.

Proposition (3.2.6).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. Then there exists a reduced irredundant decomposition $({\mathcal{F}}^{(x)}{)}_{x\in Ass(\mathcal{F})}$ formed of coherent ${\mathcal{O}}_X$-Modules such that for every $x\in Ass(\mathcal{F})$, one has $Ass({\mathcal{F}}^{(x)})=x$. For every $x\in Ass(\mathcal{F})$ such that $\overline{x}$ is not embedded, ${\mathcal{F}}^{(x)}$ is uniquely determined as the image of the canonical homomorphism $\mathcal{F}\to i_{\ast }(i^{\ast }(\mathcal{F}))$, where $i$ is the canonical morphism $\mathrm{Spec}(k(x))\to X$.

For every $x\in Ass(\mathcal{F})$, let $U$ be an affine open neighbourhood of $x$, with ring $A$, and let $\mathcal{F}|U=M$, where $M$ is an $A$-module of finite type. We know (Bourbaki, Alg. comm., chap. IV, §1, n° 1, prop. 4) that there exists a sub-module $N$ of $M$ such that, if one sets $P=M/N$, one has $Ass(P)=x$ and $Ass(N)=Ass(M)-x$. Let $\mathcal{G}=P$, which is a quasi-coherent ${\mathcal{O}}_U$-Module, and let $j$ be the canonical injection $U\to X$; let $u:j\ast (\mathcal{F})\to \mathcal{G}$ be the surjective homomorphism corresponding to the homomorphism $M\to P$; from this one deduces a homomorphism $j_{\ast }(u):j_{\ast }(j\ast (\mathcal{F}))\to j_{\ast }(\mathcal{G})$, whence by composition a homomorphism

                            ρ_ℱ                j_*(u)
                  v : ℱ ──────→ j_*(j*(ℱ)) ────────→ j_*(𝒢)

of which $u$ is the restriction to $U$; we shall designate by ${\mathcal{F}}^{(x)}$ the image of $\mathcal{F}$ by this homomorphism, which is a coherent ${\mathcal{O}}_X$-Module (I, 6.1.1). One has $Ass(j_{\ast }(\mathcal{G}))=x$ by virtue of (3.1.13), and a fortiori (3.1.7) $Ass({\mathcal{F}}^{(x)})=x$, since ${\mathcal{F}}^{(x)}\ne 0$. Moreover, if

${\mathcal{N}}^{(x)}=Ker(v)$, one has ${\mathcal{N}}^{(x)}|U=\tilde{N}$, hence $x\notin Ass({\mathcal{N}}^{(x)})$. It follows that the homomorphism $\mathcal{F}\to {⨁}_{x\in Ass(\mathcal{F})}{\mathcal{F}}^{(x)}$ is injective, for its kernel $\mathcal{H}$ is contained in every ${\mathcal{N}}^{(x)}$, hence $Ass(\mathcal{H})$ is contained in the intersection of the $Ass({\mathcal{N}}^{(x)})$, which is empty; consequently (3.1.5), $\mathcal{H}=0$. Taking into account that $Ass(\mathcal{F})$ is locally finite (3.1.6), it is clear that $({\mathcal{F}}^{(x)}{)}_{x\in Ass(\mathcal{F})}$ is a reduced irredundant decomposition of $\mathcal{F}$ verifying the conditions of the statement. The characterization of ${\mathcal{F}}^{(x)}$ when $\overline{x}$ is not embedded follows from Bourbaki, Alg. comm., chap. IV, §2, n° 3, prop. 5, the question being local, and taking account of (I, 1.6.7).

Corollary (3.2.7).

Under the hypotheses of (3.2.6), if $\mathcal{F}$ has no embedded associated prime cycle, there exists only one reduced irredundant decomposition of $\mathcal{F}$.

Corollary (3.2.8).

Let $X$ be a Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. There exists a finite filtration $({\mathcal{F}}_i{)}_{0\le i\le n}$ of $\mathcal{F}$ such that ${\mathcal{F}}_0=\mathcal{F}$, ${\mathcal{F}}_n=0$, formed of coherent ${\mathcal{O}}_X$-Modules and such that the quotients ${\mathcal{F}}_i/{\mathcal{F}}_{i+1}$ are zero or irredundant, and $Ass({\mathcal{F}}_i/{\mathcal{F}}_{i+1})\subset Ass(\mathcal{F})$.

Indeed, $\mathcal{F}$ is isomorphic to a sub-${\mathcal{O}}_X$-Module of a finite direct sum ${⨁}_{j=1}^n{\mathcal{G}}_j$, where the ${\mathcal{G}}_j$ are irredundant and coherent (3.2.6); as every quasi-coherent sub-${\mathcal{O}}_X$-Module of ${\mathcal{G}}_j$ is zero or irredundant (3.1.7), the ${\mathcal{F}}_i=\mathcal{F}\cap ({⨁}_{j=1}^{n-i}{\mathcal{G}}_j)$ answer the question, ${\mathcal{F}}_i/{\mathcal{F}}_{i+1}$ being isomorphic to a coherent sub-${\mathcal{O}}_X$-Module of ${\mathcal{G}}_{n-i}$.

3.3. Relations with flatness

Proposition (3.3.1).

Let $f:X\to Y$ be a morphism, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module and $f$-flat, $\mathcal{G}$ a quasi-coherent ${\mathcal{O}}_Y$-Module. If, for every $y\in Y$, one sets ${\mathcal{F}}_y=\mathcal{F}{\otimes }_{{\mathcal{O}}_Y}k(y)$, one has

(3.3.1.1)                       Ass(ℱ ⊗_{𝒪_Y} 𝒢) ⊃ ⋃_{y ∈ Ass(𝒢)} Ass(ℱ_y)

and the two sides are equal if $Y$ is locally Noetherian.

(Of course, ${\mathcal{F}}_y$ is a sheaf on the fibre $f^{-1}(y)$, and one identifies this fibre with a subspace of $X$ (I, 3.6.1).) The question being local on $X$ and on $Y$, one is reduced to the case where $X$ and $Y$ are affine, and the proposition is then proved in Bourbaki, Alg. comm., chap. IV, §2, n° 6, th. 2.

Corollary (3.3.2).

Let $Y$ be a locally Noetherian prescheme without embedded associated prime cycles, $f:X\to Y$ a morphism, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module and $f$-flat. Then, for every $x\in Ass(\mathcal{F})$, $f(x)$ is a maximal point of $Y$.

It suffices to apply (3.3.1) with $\mathcal{G}={\mathcal{O}}_Y$, since $Ass({\mathcal{O}}_Y)$ is by hypothesis the set of maximal points of $Y$.

Corollary (3.3.3).

Under the hypotheses of (3.3.1), suppose in addition that $X$ and $Y$ are locally Noetherian, $\mathcal{E}$ and $\mathcal{F}$ coherent. Then the following conditions are equivalent:

a) $\mathcal{E}{\otimes }_Y\mathcal{F}$ is without embedded associated prime cycle.

b) For every point $y\in Ass(\mathcal{E})\cap f(Supp(\mathcal{F}))$, $\overline{y}$ is a non-embedded associated prime cycle of $\mathcal{E}$ and ${\mathcal{E}}_y\otimes {\mathcal{F}}_y$ is without embedded associated prime cycle.

Suppose a) verified. The hypotheses imply that $\mathcal{E}{\otimes }_Y\mathcal{F}$ is a coherent ${\mathcal{O}}_X$-Module $(0_I,\mathrm{5.3.11}and\mathrm{5.3.5})$; its associated prime cycles are therefore the irreducible components of $Supp(\mathcal{E}{\otimes }_Y\mathcal{F})$ (3.1.4), and for every maximal point $x$ of $Supp(\mathcal{E}{\otimes }_Y\mathcal{F})$, $f(x)=y$ is a maximal point of $Supp(\mathcal{E})$ and $x$ a maximal point of $Supp({\mathcal{F}}_y)$ (2.5.5). Since, by virtue of (3.3.1) and the fact that the relation $y\in f(Supp(\mathcal{F}))$ entails ${\mathcal{F}}_y\ne 0$ (I, 9.1.13), every point of $Ass(\mathcal{E})\cap f(Supp(\mathcal{F}))$ is the image by $f$ of a maximal point of $Supp(\mathcal{E}{\otimes }_Y\mathcal{F})$, we see that condition b) is verified.

Conversely, suppose b) verified, and let us show that if $z$, $z^{\prime }$ are two distinct points of $Ass(\mathcal{E}{\otimes }_Y\mathcal{F})$, neither of them can be adherent to the other. First, if $f(z)=f(z^{\prime })=y$, one has $y\in Ass(\mathcal{E})$ and $z$ and $z^{\prime }$ belong to $Ass({\mathcal{F}}_y)$ by (3.3.1), whence $y\in f(Supp(\mathcal{F}))$; as by hypothesis neither of the two points $z$, $z^{\prime }$ is adherent to the other in $f^{-1}(y)$, neither of them can be adherent to the other in $X$. If $y=f(z)$ and $y^{\prime }=f(z^{\prime })$ are distinct, they belong to $Ass(\mathcal{E})\cap f(Supp(\mathcal{F}))$, hence neither of them can be adherent to the other in $Y$; it follows from the continuity of $f$ that neither of the points $z$, $z^{\prime }$ can be adherent to the other in $X$.

Proposition (3.3.4).

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 and $f$-flat. Then the following conditions are equivalent:

a) $\mathcal{E}{\otimes }_Y\mathcal{F}$ is reduced (3.2.2).

b) For every point $y\in Ass(\mathcal{E})\cap f(Supp(\mathcal{F}))$, $\overline{y}$ is a non-embedded associated prime cycle of $\mathcal{E}$, $long({\mathcal{E}}_y)=1$ and ${\mathcal{F}}_y$ is reduced.

Suppose a) verified. We already know (3.3.3) that for every $y\in Ass(\mathcal{E})\cap f(Supp(\mathcal{F}))$, $\overline{y}$ is a non-embedded associated prime cycle of $\mathcal{E}$ and ${\mathcal{F}}_y$ is without embedded associated prime cycle. Moreover (2.5.5), for every $x\in Ass(\mathcal{E}{\otimes }_Y\mathcal{F})\cap f^{-1}(y)$, one has 1 = long((ℰ ⊗_Y ℱ)_x) = long(ℰ_y) · long((ℱ_y)_x), hence $long({\mathcal{E}}_y)=long(({\mathcal{F}}_y{)}_x)=1$, which proves b).

Conversely, suppose b) verified; we already know that every point $x\in Ass(\mathcal{E}{\otimes }_Y\mathcal{F})$ is a maximal point of $Supp(\mathcal{E}{\otimes }_Y\mathcal{F})$, that $y=f(x)$ is a maximal point of $Supp(\mathcal{E})$ and $x$ a maximal point of $Supp({\mathcal{F}}_y)$ (3.3.1 and 3.3.3); moreover it follows from the hypothesis and from (2.5.5) that $long((\mathcal{E}{\otimes }_Y\mathcal{F}{)}_x)=1$, which proves a).

Corollary (3.3.5).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a flat morphism; if $Y$ is reduced at the points of $f(X)$ and if $f^{-1}(y)$ is a reduced $k(y)$-prescheme for every $y\in f(X)$, then $X$ is reduced.

Since the Nilradical ${\mathcal{N}}_Y$ is coherent, the set of points where $Y$ is reduced is open $(0_I,\mathrm{5.2.2})$, and one may restrict to the case where $Y$ is reduced. It then suffices to apply (3.3.4) to $\mathcal{E}={\mathcal{O}}_Y$ and $\mathcal{F}={\mathcal{O}}_X$.

Proposition (3.3.6).

Let $f:X\to S$, $g:Y\to S$ be two morphisms, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module, $\mathcal{G}$ a quasi-coherent ${\mathcal{O}}_Y$-Module. Suppose that: 1° $\mathcal{G}$ is $g$-flat; 2° $X$ is locally Noetherian, and for every $s\in S$, $g^{-1}(s)$ is locally Noetherian (which will be the case if $Y$ is also locally Noetherian). Let $Z=X{\times }_{S}Y$; for every couple $(x,y)$ such that $x\in X$, $y\in Y$ and

$f(x)=g(y)=s$, let $T_{x,y}$ be the prescheme $\mathrm{Spec}(k(x){\otimes }_{k(s)}k(y))$, and let $I_{x,y}$ be the image of $Ass({\mathcal{O}}_{T_{x,y}})$ by the canonical monomorphism $T_{x,y}\to Z$ (I, 3.4.9). One then has

(3.3.6.1)              Ass(ℱ ⊗_S 𝒢) = ⋃_{x ∈ Ass(ℱ)} ( ⋃_{y ∈ Ass(𝒢_{f(x)})} I_{x,y} )

where for every $s\in S$, ${\mathcal{G}}_s=\mathcal{G}{\otimes }_{{\mathcal{O}}_S}k(s)$.

Let $p:Z\to X$, $q:Z\to Y$ be the canonical projections, so that one has the commutative diagram

                                X ←─── Z
                                       p
                                ↓ f    ↓ q
                                S ←─── Y
                                   g

Set ${\mathcal{G}}^{\prime }=q\ast (\mathcal{G})$, so that $\mathcal{F}{\otimes }_S\mathcal{G}=\mathcal{F}{\otimes }_X{\mathcal{G}}^{\prime }$; as ${\mathcal{G}}^{\prime }$ is $p$-flat (2.1.4), it follows from (3.3.1) that one has

(3.3.6.2)                       Ass(ℱ ⊗_X 𝒢') = ⋃_{x ∈ Ass(ℱ)} Ass(𝒢'_x)

with ${\mathcal{G}}_x^{\prime }={\mathcal{G}}^{\prime }{\otimes }_{{\mathcal{O}}_X}k(x)$. If $s=f(x)$, one has ${\mathcal{G}}_s=\mathcal{G}{\otimes }_{{\mathcal{O}}_S}k(s)$, and $p^{-1}(x)=g^{-1}(s){\otimes }_{k(s)}k(x)$; moreover, since the field $k(x)$ is a flat $k(s)$-module, the morphism $p^{-1}(x)\to g^{-1}(s)$ is flat (2.1.4); applying (3.3.1) to this morphism, it comes

(3.3.6.3)                       Ass(𝒢'_x) = ⋃_{y ∈ Ass(𝒢_s)} Ass(𝒪_{T_{x,y}})

whence the proposition.

We note that if, in the statement, one suppresses hypothesis 2°, one may still conclude, by virtue of (3.3.1), the relation

(3.3.6.4)             Ass(ℱ ⊗_S 𝒢) ⊃ ⋃_{x ∈ Ass(ℱ)} ( ⋃_{y ∈ Ass(𝒢_{f(x)})} I_{x,y} ).

Corollary (3.3.7).

Under the hypotheses of (3.3.6), suppose in addition that $S$ is locally Noetherian and that $f(Ass(\mathcal{F}))\subset Ass({\mathcal{O}}_S)$. Then one has

(3.3.7.1)                             Ass(ℱ ⊗_S 𝒢) = ⋃_{(x,y) ∈ C} I_{x,y}

where $C$ is the set of couples $(x,y)$ such that $x\in Ass(\mathcal{F})$, $y\in Ass(\mathcal{G})$ and $f(x)=g(y)$.

Since $\mathcal{G}$ is $g$-flat, it follows indeed from (3.3.1) that the relation "$s\in Ass({\mathcal{O}}_S)$ and $y\in Ass({\mathcal{G}}_s)$" is equivalent to $y\in Ass(\mathcal{G})$: the conclusion follows from (3.3.6.1).

Remarks (3.3.8).

We shall see later (4.2.2) that under the hypotheses of (3.3.6), $T_{x,y}$ is a prescheme without embedded associated prime cycle; it will follow that if $\mathcal{F}$ and the ${\mathcal{G}}_s$ are without embedded associated prime cycle, the same is true of $\mathcal{F}{\otimes }_S\mathcal{G}$.

Corollary (3.3.9).

Under the conditions of (3.3.7), one has

(3.3.9.1)                             q(Ass(ℱ ⊗_S 𝒢)) ⊂ Ass(𝒢)

(where $q:X{\times }_{S}Y\to Y$ is the canonical projection).

Indeed, if $(x,y)\in Z$, one has $q(I_{x,y})=y\subset Ass(\mathcal{G})$.

3.4. Properties of the sheaves $\mathcal{F}/t\mathcal{F}$

Proposition (3.4.1).

Let $X$ be a locally Noetherian prescheme, $t$ a section of ${\mathcal{O}}_X$ over $X$, $Y$ the closed sub-prescheme of $X$ defined by the Ideal $t{\mathcal{O}}_X$ of ${\mathcal{O}}_X$. Let $\mathcal{F}$ be a coherent ${\mathcal{O}}_X$-Module, $S$ the reduced closed sub-prescheme of $X$ having $Supp(\mathcal{F})$ as underlying space, $(S_i)$ the family of reduced closed sub-preschemes of $X$ having as underlying spaces the irreducible components of $S$; we designate by $s_i$ the generic point of $S_i$. Finally, let $Z$ be an irreducible component of $Supp(\mathcal{F}/t\mathcal{F})=S\cap Y$, $z$ its generic point.

(i) For every $i$ such that $Z\subset S_i$, $Z$ is an irreducible component of $S_i\cap Y$.

(ii) If $Z$ is not equal to any of the $S_i$, one has

$$(\mathrm{3.4.1.1})long((\mathcal{F}/t\mathcal{F}{)}_z)\ge \sum _{i}long({\mathcal{F}}_{s_i})$$

where the sum on the right-hand side is extended to all $i$ such that $Z\subset S_i$.

(iii) Suppose that $Z$ is equal to none of the $S_i$. For the two sides of (3.4.1.1) to be equal, it is necessary and sufficient that the two following conditions be satisfied:

α) $t_z$ is ${\mathcal{F}}_z$-regular (0, 15.1.4).

β) For every $i$ such that $Z\subset S_i$, the canonical image of the germ $t_z$ in ${\mathcal{O}}_{S_i,z}$ generates the maximal ideal of this ring (which entails that ${\mathcal{O}}_{S_i,z}$ is a discrete valuation ring and the image of $t_z$ a uniformizer of this ring).

(i) If $j:Y\to X$ is the canonical injection, one has $\mathcal{F}/t\mathcal{F}=\mathcal{F}{\otimes }_{{\mathcal{O}}_X}{\mathcal{O}}_Y=j\ast (\mathcal{F})$, hence $Supp(\mathcal{F}/t\mathcal{F})=j^{-1}(S)=S\cap Y$ (I, 9.1.13), whence the assertion.

(ii) and (iii). As the $s_i$ such that $Z\subset S_i$ are those belonging to $\mathrm{Spec}({\mathcal{O}}_z)$, we may, in order to prove (ii) and (iii), replace $X$ by $\mathrm{Spec}({\mathcal{O}}_z)$; and if $M={\mathcal{F}}_z$, we may therefore suppose that $\mathcal{F}=\tilde{M}$, whence ${\mathcal{F}}_{s_i}=M_{{\mathfrak{p}}_i}$, where we designate by ${\mathfrak{p}}_i$ the minimal ideals of ${\mathcal{O}}_z$. Moreover, as $M$ is an ${\mathcal{O}}_z$-module of finite type, one has $S=S_{red}^{\prime }$, with $S^{\prime }=\mathrm{Spec}({\mathcal{O}}_z/\mathfrak{a})$, where $\mathfrak{a}$ is the annihilator of $M$ in ${\mathcal{O}}_z$ $(0_I,\mathrm{1.7.4})$, and the two sides of (3.4.1.1) keep the same values, whether one considers $M$ as an ${\mathcal{O}}_z$-module or as an $({\mathcal{O}}_z/\mathfrak{a})$-module; one therefore sees that one can finally replace $X$ by $S^{\prime }=\mathrm{Spec}(A)$, where $A$ is a Noetherian local ring, $M$ being a faithful $A$-module; since $Z$ is closed in $S$, the hypothesis that $Z\ne S_i$ for every $i$ means that $s_i\notin Z$, hence that $\dim (A)>0$; finally, to say that $z$ is the generic point of $Z$, an irreducible component of $S\cap V(t)$, means that $A/tA$ is of dimension 0 (in other words, is a local Artinian ring). One is therefore reduced to proving the following statement:

Lemma (3.4.1.2).

Let $A$ be a Noetherian local ring of dimension > 0, ${\mathfrak{p}}_i$ the minimal prime ideals of $A$, $\mathfrak{m}$ its maximal ideal, $t$ an element of $\mathfrak{m}$ such that $A/tA$ is Artinian. Then, for every $A$-module of finite type $M$, one has

$$(\mathrm{3.4.1.3})long(M/tM)\ge \sum _{i}long(M_{{\mathfrak{p}}_i});$$

moreover, for the two sides of (3.4.1.3) to be equal, it is necessary and sufficient that the following conditions be satisfied:

α) $t$ is $M$-regular;

β) for every $i$ such that $M_{{\mathfrak{p}}_i}\ne 0$, the image of $t$ in $A/{\mathfrak{p}}_i$ generates the maximal ideal of this ring (which entails that $A/{\mathfrak{p}}_i$ is a discrete valuation ring).

As $A$ is not of dimension 0 and $A/tA$ is Artinian, one has necessarily $\dim (A)=1$ (0, 16.3.4) and $t\notin {\mathfrak{p}}_i$ for every $i$: the principal ideal (t) is therefore an ideal of definition of $A$, and hence contains a power of its maximal ideal $\mathfrak{m}$. Let $N$ be the submodule of elements of $M$ annihilated by a power of $t$ (or by a power of $\mathfrak{m}$, which amounts to the same thing as we have just seen); if one sets $P=M/N$, $t$ is $P$-regular, since the relation $tx\in N$ for an $x\in M$ entails $t^k(tx)=0$ for some integer $k$, hence $x\in N$. This being so, one has the

Lemma (3.4.1.4).

Let $A$ be a ring,

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

an exact sequence of $A$-modules. If $t\in A$ is M''-regular, the sequence

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

is exact.

Since $M/tM=M{\otimes }_A(A/tA)$, it suffices to prove exactness at $M^{\prime }/t{M}^{\prime }$; now, if the image $x\in M$ of an element $x^{\prime }$ of $M^{\prime }$ is such that $x=ty$ with $y\in M$, one deduces, for the images x'', y'' of x, y in M'', $x^{\prime \prime }=t{y}^{\prime \prime }$; but as $x^{\prime \prime }=0$, the hypothesis entails $y^{\prime \prime }=0$, hence $y$ is the image of an element $y^{\prime }\in M^{\prime }$, and the relation $x=ty$ entails $x^{\prime }=t{y}^{\prime }$ since $M^{\prime }\to M$ is injective.

This lemma established, one derives from it the relation

(3.4.1.5)                       long(M/tM) = long(N/tN) + long(P/tP).

On the other hand, for every $i$, one has $N_{{\mathfrak{p}}_i}=0$ since $t\notin {\mathfrak{p}}_i$, hence $M_{{\mathfrak{p}}_i}=P_{{\mathfrak{p}}_i}$; to prove (3.4.1.3), it suffices to do so by replacing $M$ by $P$; on the other hand, if the two sides of (3.4.1.3) are equal, it follows from the same inequality for $P$ and from (3.4.1.5) that one necessarily has $long(N/tN)=0$, hence $N/tN=0$ and finally $N=0$, by Nakayama's lemma, $N$ being of finite type; now, $N=0$ means that $t$ is $M$-regular. One may therefore reduce to the case where $M=P$, that is, suppose already that $t$ is $M$-regular. Note that this entails $\mathfrak{m}\notin Ass(M)$, since $t$ cannot annihilate an element $\ne 0$ of $M$. As $A$ is of dimension 1, one therefore has necessarily $Ass(M)\subset {\bigcup }_i{\mathfrak{p}}_i$.

Let us then proceed by induction on $n={\sum }_{i}long(M_{{\mathfrak{p}}_i})$. If $n=0$, one has necessarily $M_{{\mathfrak{p}}_i}=0$ for every $i$, hence $M=0$ since none of the ${\mathfrak{p}}_i$ belongs to $Ass(M)$; the two sides of (3.4.1.3) are then zero, and assertion β) of (3.4.1.2) is trivial. If $n>0$, the reasoning at the beginning of the proof of (3.4.1) allows us to suppose moreover that the $A$-module $M$ is faithful: this entails $M_{{\mathfrak{p}}_i}\ne 0$ for every $i$ (Bourbaki, Alg. comm., chap. II, §2, n° 2, cor. 2 of prop. 4), and consequently $Ass(M)={\bigcup }_i{\mathfrak{p}}_i$.

Suppose first $n=1$; there is then only a single minimal prime ideal $\mathfrak{p}$ of $A$,

and to say that $M_{\mathfrak{p}}$ is of length 1 means that $M_{\mathfrak{p}}$ is isomorphic to the residue field $k=A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$ as an $A_{\mathfrak{p}}$-module. Consequently $M_{\mathfrak{p}}$ is annihilated by $\mathfrak{p}{A}_{\mathfrak{p}}$, hence $\mathfrak{p}$ is the annihilator of $M$ (Bourbaki, Alg. comm., chap. II, §2, n° 4, formula (9)), which entails $\mathfrak{p}=0$ since $M$ is supposed faithful; the ring $A$ is therefore integral. This being so, the hypothesis $M\ne 0$ entails $M/tM\ne 0$ by Nakayama's lemma, and consequently $long(M/tM)\ge 1$, which proves (3.4.1.3) in this case. Moreover, if $long(M/tM)=1$, $M$ is necessarily monogenic (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 2 of prop. 4), hence isomorphic to a quotient $A/\mathfrak{b}$; since it is faithful, one necessarily has $\mathfrak{b}=0$ and $M$ is isomorphic to $A$; as $long(A/tA)=1$, tA is necessarily equal to the maximal ideal $\mathfrak{m}$, and as $A$ is a Noetherian integral local ring, this proves that $A$ is a discrete valuation ring (Bourbaki, Alg. comm., chap. VI, §3, n° 6, prop. 9), of which $t$ is the uniformizer. Conversely, if $A$ is a discrete valuation ring, $t$ its uniformizer, $long(M_{\mathfrak{p}})=1$ and if $t$ is $M$-regular, then $M$ is torsion-free, hence isomorphic to a sub-module of $A$ ($M$ being of finite type), and consequently isomorphic to $A$ itself, whence $long(M/tM)=long(A/tA)=1$.

Suppose now $n\ge 2$; there then exists an exact sequence

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

with $M^{\prime }\ne 0$, $M^{\prime \prime }\ne 0$ and Ass(M) = Ass(M') ∪ Ass(M''); indeed, if $Ass(M)$ is not reduced to a single element, this follows from Bourbaki, Alg. comm., chap. IV, §1, n° 1, prop. 4; if on the contrary $Ass(M)$ is reduced to a single prime ideal, this latter is necessarily the unique minimal prime ideal $\mathfrak{p}$ of $A$; the hypothesis then entails $long(M_{\mathfrak{p}})\ge 2$ and it suffices to take for $M^{\prime }$ the inverse image of a submodule of $M_{\mathfrak{p}}$ distinct from 0 and from $M_{\mathfrak{p}}$. As $t$ is $M$-regular, $t$ does not belong to any of the prime ideals of $Ass(M)$ (Bourbaki, Alg. comm., chap. IV, §1, n° 1, cor. 2 of prop. 2), hence, for the same reason, $t$ is $M^{\prime }$-regular and M''-regular. This last property entails by (3.4.1.4) 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; as it is moreover the case for the sequence

$$0\to M_{{\mathfrak{p}}_i}^{\prime }\to M_{{\mathfrak{p}}_i}\to M_{{\mathfrak{p}}_i}^{\prime \prime }\to 0$$

for every $i$, one has

                  long(M/tM)     = long(M'/tM') + long(M''/tM'')
                  long(M_{𝔭_i})  = long(M'_{𝔭_i}) + long(M''_{𝔭_i})

and the induction hypothesis therefore entails the inequality (3.4.1.3). Moreover the two sides cannot be equal unless the analogous inequalities for $M^{\prime }$ and M'' are also equalities. By virtue of the induction hypothesis, this is equivalent to property β) for the ${\mathfrak{p}}_i$ such that $M_{{\mathfrak{p}}_i}^{\prime }\ne 0$ or $M_{{\mathfrak{p}}_i}^{\prime \prime }\ne 0$; but these ideals are precisely those for which $M_{{\mathfrak{p}}_i}\ne 0$. Q.E.D.

Corollary (3.4.2).

Under the general hypotheses of (3.4.1), suppose that $Z$ is not equal to any of the $S_i$ and that $long((\mathcal{F}/t\mathcal{F}{)}_z)=1$. Then there exists only one of the $S_i$ containing $Z$, and for this value of $i$, one has $long({\mathcal{F}}_{s_i})=1$; moreover ${\mathcal{O}}_{S_i,z}$ is a discrete valuation ring of which $t_z$ is a uniformizer, and $t_z$ is ${\mathcal{F}}_z$-regular.

This results from (3.4.1), the two sides of (3.4.1.1) then being equal.

Proposition (3.4.3).

Let $X$ be a locally Noetherian prescheme, $t$ a section of ${\mathcal{O}}_X$ over $X$, $Y$ the closed sub-prescheme of $X$ defined by the Ideal $t{\mathcal{O}}_X$ of ${\mathcal{O}}_X$. Let $\mathcal{F}$ be a coherent ${\mathcal{O}}_X$-Module, $T$ an associated prime cycle of $\mathcal{F}$, $T^{\prime }$ an irreducible component of $T\cap Y$, $x$ the generic point of $T^{\prime }$. Suppose that $t_x$ is ${\mathcal{F}}_x$-regular; then one has $x\in Ass(\mathcal{F}/t\mathcal{F})$.

As in the proof of (3.4.1), we can reduce to the case where $X=\mathrm{Spec}({\mathcal{O}}_x)$; the proposition is then (taking into account (3.1.2)) a consequence of (0, 16.4.6.3).

Proposition (3.4.4).

Let $X$ be a locally Noetherian prescheme, $t$ a section of ${\mathcal{O}}_X$ over $X$, $Y$ the closed sub-prescheme of $X$ defined by the Ideal $t{\mathcal{O}}_X$ of ${\mathcal{O}}_X$. Let $\mathcal{F}$ be a coherent ${\mathcal{O}}_X$-Module, $(S_i)$ the family of irreducible components of $Supp(\mathcal{F})$. Let $y$ be a point of $Y$ such that $t_y$ is ${\mathcal{F}}_y$-regular and no embedded associated prime cycle of $\mathcal{F}/t\mathcal{F}$ contains $y$. Then the irreducible components of $Supp(\mathcal{F}/t\mathcal{F})$ that contain $y$ are exactly the irreducible components of the $S_i\cap Y$ that contain $y$, and the associated prime cycles of $\mathcal{F}$ containing $y$ are non-embedded.

Let us first prove the last assertion. Let $T\supset T_1$ be two associated prime cycles of $\mathcal{F}$ containing $y$; if $x$ is the generic point of an irreducible component of $T\cap Y$ containing $y$, $x$ is a generization of $y$, hence contained in every neighbourhood of $y$, and the hypothesis that $t_y$ is ${\mathcal{F}}_y$-regular entails that $t_x$ is ${\mathcal{F}}_x$-regular (0, 15.2.4), hence, by virtue of (3.4.3), one has $x\in Ass(\mathcal{F}/t\mathcal{F})$. Let $x_1$ be the generic point of an irreducible component of $T_1\cap Y$ containing $y$, and let $x$ be the generic point of an irreducible component of $T\cap Y$ containing $x_1$; it follows from what precedes that $x_1$ and $x$ both belong to $Ass(\mathcal{F}/t\mathcal{F})$, and as $x_1\in \overline{x}$, the hypothesis entails that $x_1=x$. Let us again denote by $T$ and T_1 the integral closed sub-preschemes of $X$ having $T$ and T_1 as underlying spaces respectively, and set $A={\mathcal{O}}_{T,x}$, $A_1={\mathcal{O}}_{T_1,x}$; one has therefore $A_1=A/\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal of $A$. By the definition of $x$ and $x_1$, $A/tA$ and $A_1/t{A}_1$ are Artinian rings; on the other hand, we saw above that $t_x$ is ${\mathcal{F}}_x$-regular, hence $x$ cannot belong to $Ass(\mathcal{F})$, and consequently A_1 is not Artinian. One therefore has $\dim A=\dim A_1=1$; but this entails $\mathfrak{p}=0$ and $A=A_1$ (0, 16.1.2.2); as $\mathrm{Spec}(A)$ and $\mathrm{Spec}(A_1)$ are respectively dense in $T$ and T_1, one has indeed $T=T_1$.

The $S_i$ containing $y$ are therefore all the associated prime cycles of $\mathcal{F}$ containing $y$; if $x$ is the generic point of an irreducible component of $S_i\cap Y$ containing $y$, one again deduces from (0, 15.2.4) that $t_x$ is ${\mathcal{F}}_x$-regular, hence, by (3.4.3), that $x\in Ass(\mathcal{F}/t\mathcal{F})$; this proves the first assertion of (3.4.4).

Proposition (3.4.5).

Let $X$ be a locally Noetherian prescheme, $t$ a section of ${\mathcal{O}}_X$ over $X$, $Y$ the closed sub-prescheme of $X$ defined by the Ideal $t{\mathcal{O}}_X$ of ${\mathcal{O}}_X$. Let $\mathcal{F}$

be a coherent ${\mathcal{O}}_X$-Module, $y$ a point of $Y$; suppose that $t_y$ is ${\mathcal{F}}_y$-regular and that $\mathcal{F}/t\mathcal{F}$ is integral at the point $y$ (3.2.4). Then $\mathcal{F}$ is integral at the point $y$.

Taking into account (3.4.4), it suffices to prove that $y$ is contained in a single irreducible component of $Supp(\mathcal{F})$, and that if $s$ is the generic point of this component, one has $long({\mathcal{F}}_s)=1$. Now, by hypothesis, $y$ belongs to only a single irreducible component of $Supp(\mathcal{F}/t\mathcal{F})$, and if $z$ is the generic point of this component, one has $long((\mathcal{F}/t\mathcal{F}{)}_z)=1$; the conclusion therefore follows from (3.4.2).

Proposition (3.4.6).

The hypotheses being those of (3.4.1), let $x$ be a point of $Y$. Suppose that $Y$ contains none of the $S_i$ containing $x$, and that $\mathcal{F}/t\mathcal{F}$ is reduced at the point $x$ (3.2.2). Then $t_x$ is ${\mathcal{F}}_x$-regular and $\mathcal{F}$ is reduced at the point $x$. Moreover, if $z$ is the generic point of an irreducible component of $Supp(\mathcal{F}/t\mathcal{F})$ containing $x$, $z$ is contained in a single one of the $S_i$, and ${\mathcal{O}}_{S_i,z}$ is a discrete valuation ring of which $t_z$ is a uniformizer.

The fact that $t_x$ is ${\mathcal{F}}_x$-regular results from the following lemma applied to the ring ${\mathcal{O}}_x$:

Lemma (3.4.6.1).

Let $A$ be a Noetherian ring, $M$ an $A$-module of finite type, ${\mathfrak{p}}_i$ the minimal elements of $Supp(M)$, $t$ an element of $A$. Suppose that $t$ belongs to none of the ${\mathfrak{p}}_i$ and that $M/tM$ is a reduced $A$-module (3.2.2). Then $t$ is $M$-regular.

Every prime ideal $\mathfrak{p}\in Supp(M)$ contains one of the ${\mathfrak{p}}_i$; as $t$ belongs to none of the ${\mathfrak{p}}_i$, the homothety of ratio $t$ in $M_{\mathfrak{p}}$ is not nilpotent (Bourbaki, Alg. comm., chap. IV, §1, n° 4, cor. of prop. 9). Let us designate by $N$ the submodule of $M$ formed of elements annihilated by a power of $t$, and set $P=M/N$; we shall show that $N=0$. Since $t$ is $P$-regular, one has an exact sequence (3.4.1.4)

$$0\to N/tN\to M/tM\to P/tP\to 0.$$

As $N$ is of finite type, it is annihilated by a power of $t$, and it therefore suffices to show that $N/tN=0$. As $N/tN$ is a submodule of $M/tM$, it suffices to prove that $(N/tN{)}_{\mathfrak{p}}=0$ for every $\mathfrak{p}\in Ass(M/tM)$, or again that the homomorphism $u_{\mathfrak{p}}:(M/tM{)}_{\mathfrak{p}}\to (P/tP{)}_{\mathfrak{p}}$ is bijective for every $\mathfrak{p}\in Ass(M/tM)$. Now one has $(P/tP{)}_{\mathfrak{p}}\ne 0$; indeed, as $\mathfrak{p}\in Supp(M/tM)=Supp(M)\cap V(t)$ $(0_I,\mathrm{1.7.5})$, the image of $t$ in $A_{\mathfrak{p}}$ is contained in the maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$, hence the hypothesis $P_{\mathfrak{p}}=t{P}_{\mathfrak{p}}$ would entail $P_{\mathfrak{p}}=0$ by Nakayama's lemma; one would therefore have $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ and the homothety of ratio $t$ in $M_{\mathfrak{p}}$ would be nilpotent; but this contradicts the remark made at the beginning, since $\mathfrak{p}\in Supp(M)$. This being so, the hypothesis that $M/tM$ is reduced entails $long((M/tM{)}_{\mathfrak{p}})=1$, and as $(P/tP{)}_{\mathfrak{p}}\ne 0$, $u_{\mathfrak{p}}$ is necessarily bijective, which proves the lemma.