Suppose $M$ is a indecomposable module, so that it cannot be written as $M_1\oplus M_2$ for $M_1\neq M$ and $M_2\neq M$, which is finitely generated over a commutative ring $R$. Why is $\text{supp}(M)$ connected in the Zariski topology?

Now I know $\text{supp}(M)=\{\mathfrak{p}\in\text{Spec}(R): M_\mathfrak{p}\neq 0\}$. Since $M$ is finitely generated over $R$, I was able to derive that $\text{supp}(M)=Z(\text{ann}(M))$, with $Z(\mathfrak{a})=\{\mathfrak{p}\in\text{Spec}(R):\mathfrak{p}\supset\mathfrak{a}\}$. For suppose $\mathfrak{p}\in\text{Spec}(R)$ such that $M_\mathfrak{p}\neq 0$. If $aM=0$, $aM_\mathfrak{p}=0$, so $a\notin\mathfrak{p}$ then $M_\mathfrak{p}=0$, and thus $\text{ann}(M)\subset\mathfrak{p}$.

In the other direction, take $\text{ann}(M)\subset\mathfrak{p}$, and suppose $m_1,\dots,m_r$ are generators for $M$ over $R$. If $M_\mathfrak{p}=0$, then there is some $a_i\notin\mathfrak{p}$ such that $a_im_i=0\in M$ for all $i$. Then setting $a=a_1\cdots a_r$, $aM=0$, and $a\in\mathfrak{p}$, contradiction.

So I suppose $\text{supp}(M)=Z(\text{ann}(M))=A\cup B$ where $A$ and $B$ are closed sets in the Zariski topology. To see $\text{supp}(M)$ is connected, I'm trying to conclude one of $A$ or $B$ is empty. Since $M$ is indecomposable, I know it can't be written as $M=M_1\oplus M_2$ for $M_1\neq M$ and $M_2\neq M$. I'm also aware that $\text{supp}(M_1\oplus M_2)=\text{supp}(M_1)\cup\text{supp}(M_2)$.

I thought maybe if it isn't connected, I could possibly show that $M$ is actually decomposable. Can someone please explain the correct idea here? Thanks.

(If possible, I'd appreciate an explanation that uses mostly commutative algebra, if one exists, since I'm unfamiliar with ideas from algebraic geometry.)

  • $\begingroup$ Your idea is probably right... If it's disconnected, maybe cook up the two modules it should be a sum of. Perhaps it would be helpful to use geometric intuition: a module is a quasi-coherent sheaf over the spectrum... if it's support is disconnected, what does that say about the sheaf picture? $\endgroup$ Commented Feb 27, 2012 at 8:29
  • $\begingroup$ @DylanWilson Thanks, I'm afraid I don't know what a sheaf is, or much algebraic-geometry for that matter, so I really can't say. $\endgroup$
    – Buble
    Commented Feb 27, 2012 at 8:55

2 Answers 2


Yes, if a finitely generated module $M$ is indecomposable, its support is connected.

The support of $M$ is $V(Ann(M))$.
Replacing $R$ by $R/Ann(M)$, we may assume that $Ann(M)=0$ and that $Supp(M)=Spec(R)$.
If that space $SuppM=Spec(R)$ were non-connected, it is well-known that there would exist two non-zero idempotents $e,f\in R$ with $ef=0$ and $e+f=1$.
Then we would have a non trivial decomposition $M=eM\oplus fM$. Contradiction.

The fact I mentioned on idempotents is Exercise 2.25 in Eisenbud's Commutative Algebra (the exercise is practically solved by its copious hints).

Second Edit
In order not to give an exercise as sole reference, here is a proof (different from Eisenbud's) that a ring with two non-zero idempotents $e,f\in R$ such that $ef=0$ and $e+f=1$ is a non-trivial product of rings (and thus has non-connected spectrum).

The trick is to realize that the ideal $eR$, although not a subring of $R$, is a ring of its own, with addition and multiplication inherited from $R$, but with unit element $e$. Similarly for $fR$.

Then the morphism $R\to eR\times fR:r\mapsto (er,fr)$ is an isomorphism of rings.
(For surjectivity notice that $ (er_1,fr_2)$ is the image of $er_1+fr_2$.
Injectivity is the trivial observation that $er=fr=0$ implies $r=(e+f)r=0$)

  • $\begingroup$ Dear Georges: I first posted an incomplete answer, which I deleted. Then I wrote the second version, and when I posted it, I noticed that you had just posted yours. So, you gave the first correct answer. It seems to me we're using the same argument, but, since the wording is slightly different, it might be worth leaving mine (assuming, as I hope, it is correct). ($+1$ of course.) $\endgroup$ Commented Feb 27, 2012 at 10:20
  • 1
    $\begingroup$ Dear @Pierre-Yves, what else can I say than "les bons esprits se rencontrent" ? (+1, évidemment). $\endgroup$ Commented Feb 27, 2012 at 10:26
  • $\begingroup$ Thanks both, I thinking I'm beginning to understand this now. $\endgroup$
    – Buble
    Commented Feb 28, 2012 at 21:59

By taking a quotient, we can assume that the annihilator of $M$ is $0$. Then the support of $M$ is the full spectrum $X$.

Suppose $X$ is not connected. By Corollary $2$ to Proposition $15$ Section II.$4.3$ of Bourbaki's Algèbre commutative, there are nontrivial ideals $\mathfrak a$ and $\mathfrak b$ satisfying $$ A=\mathfrak a+\mathfrak b,\quad\mathfrak a\cap\mathfrak b=0. $$ By the Chinese Remainder Theorem, this implies $$ A\simeq\frac{A}{\mathfrak a}\times\frac{A}{\mathfrak b}\quad. $$ Let $M=N\oplus P$ be the corresponding decomposition.

As $\mathfrak b$ is contained in the annihilator of $N$, we have $N\neq M$, and similarly for $P$.


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