$\newcommand{\Supp}{\operatorname{Supp}}$ $\newcommand{\Spec}{\operatorname{Spec}}$ $\newcommand{\Ann}{\operatorname{Ann}}$ Let $A$ be a commutative ring with 1. Let $M$ be a finitely-generated $A$-module. Then we define the support of $M$ as follows: $$\Supp(M) = \mathcal{V}(\Ann(M)) \subset \Spec(A) \ . $$ That is, a prime ideal $\mathfrak{p}$ is in $\Supp(M)$ if and only if $\Ann(M) \subset \mathfrak{p}$.

Lemma. It is well known that if $$0 \to M' \to M \to M'' \to 0$$ is exact sequence of finitely-generated $A$-modules, then $$\Supp(M) = \Supp(M') \cup \Supp(M'') \ .$$

See also http://en.wikipedia.org/wiki/Support_of_a_module for more information.


Now for the problem: suppose $M$ is a finitely-generated $A$-module such that $\Supp(M) \subset F \cup F'$ where $F,F'$ are closed subsets of $\Spec(A)$. Then I want to show there is a submodule $L$ such that $\Supp(L) \subset F$ and $N = M/L$ satisfies that $\Supp(N) \subset F'$. Next, if $F \cap F' = \emptyset$ then I want to show that $M \cong L \oplus N$.

My attempt

My idea was to pick $L$ such that $\Ann(L)$ is the ideal of elements of $A$ vanishing on $F$. Then by the Nullstellensatz $\Supp(L) \subset F$. Now, by the lemma I want to use the exact sequence $$ 0 \to L \to M \to M/L \to 0 $$ to deduce that $\Supp(M/L) \subset F'$. We have $$ F \cup F' \supset \Supp(M) = \Supp(L) \cup \Supp(M/L)$$ and even if $\Supp(L)=F$ I can't deduce from that $\Supp(M/L) \subset F'$. If there was an equality $\Supp(M) = F \cup F'$ then the desired result would follow.


1 Answer 1


I will suppose $A$ is noetherian. I am not sure that the result holds otherwise.

We can replace $A$ with $A/\mathrm{Ann}M$ and $M$ with $M/\mathrm{Ann}(M)M$. So the support of $M$ is $\mathrm{Spec}(A)$. Let $F=V(I), F'=V(J)$. Then $\mathrm{Spec}(A)=V(I)\cup V(J)=V(I\cap J)$. This means $I\cap J$ is in the nilradical of $A$. As $A$ is noetherian, there exists $n\ge 1$ such that $I^nJ^n=0$. Replacing $I, J$ respectively by $I^n, J^n$ (this doesn't change the sets $F, F'$), we can suppose $IJ=0$.

Let $L=IM$. Then $JL=0$. So $J\subseteq \mathrm{Ann}(L)$ and the support of $L$ is contained $V(J)=F'$. Obviously $M/L$ has support contained in $V(I)=F$.

In case $V(I)\cap V(J)=\emptyset$, then $V(I+J)=V(I)\cap V(J)=\emptyset$. So $I+J=A$ and $M=IM+JM$. Let us show $M=IM\oplus JM$. Let $e\in I, e'\in J$ such that $1=e+e'$. Let $x\in IM\cap JM$. Then $ex\in eJM\subseteq IJM=0$ and similarly $e'x=0$. So $x=ex+e'x=0$. Therefore $M=IM\oplus JM$ and $JM\simeq M/IM=M/L$.

  • $\begingroup$ Thank you. By passing to Supp$(M)$ = Spec$(A)$ you reduced to the case Supp$(M) = F \cup F'$ and not just inclusion, right? $\endgroup$
    – ano
    Jan 2, 2013 at 11:44
  • $\begingroup$ Yes, this is for simplification. Otherwise we have to deal with $\mathrm{Ann}(M)$. But this doesn't really matter. $\endgroup$
    – user18119
    Jan 2, 2013 at 12:18
  • $\begingroup$ But in the case Supp$(M) \subset F \cup F'$ can't be a situation when Supp$(M) \subset F$ alone? $\endgroup$
    – ano
    Jan 2, 2013 at 13:32
  • $\begingroup$ I don't understand the question. Yes this can happen. What I did it write $\mathrm{Supp}(M)=(F\cap \mathrm{Supp}(M)) \cup (F'\cap \mathrm{Supp}(M))$. $\endgroup$
    – user18119
    Jan 2, 2013 at 13:38
  • 1
    $\begingroup$ Answer: in that case we take $L=M$ and then $N = M/L=M/M =0$ and $\mathrm{Supp}(M/L)=\mathrm{Supp}(0)=\emptyset$ and the goal in the problem is achieved. $\endgroup$
    – ano
    Jan 2, 2013 at 15:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .