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Let $R$ be the coordinate ring of an affine complex variety (i.e. finitely generated, commutative, reduced $\mathbb{C}$ algebra) and $M$ be an $R$ module.

Let $s\in M$ be an element, such that $s\in \mathfrak{m}M$ for every maximal ideal $\mathfrak{m}$. Does this imply $s=0$?

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1 Answer 1

up vote 9 down vote accepted

Not in general, no. For example, if $R = \mathbb C[T]$ and $M$ is the field of fractions of $R$, namely $\mathbb C(T)$, then (a) every maximal ideal of $R$ is principal; (b) every element of $M$ is divisible by every non-zero element of $R$. Putting (a) and (b) together we find that $M = \mathfrak m M$ for every maximal ideal $\mathfrak m$ of $R$, but certainly $M \neq 0.$

Here is a finitely generated example: again take $R = \mathbb C[T]$, and take $M = \mathbb C[T]/(T^2).$ Then $s = T \bmod T^2 \in \mathfrak m M$ for every $\mathfrak m$, because $\mathfrak m M = M$ if $\mathfrak m$ is a maximal ideal other than $(T)$, and this is clear from the choice of $s$ if $\mathfrak m = (T)$.

The answer is yes if $M$ is finitely generated and torsion free. For let $S$ be the total quotient ring of $R$ (i.e. the product of functions fields $K(X)$ for each irreducible component $X$ of the variety attached to $R$).
Then $M$ embeds into $S\otimes_R M$ (this is the torsion free condition), which in turn embeds into a finite product of copies of $S$ (since it is finite type over $S$, which is just a product of fields).

Clearing denominators, we find that in fact $M$ then embeds into $R^n$ for some $n$. Thus it suffices to prove the result for $M = R^n$, and hence for $R$, in which case it follows from the Nullstellensatz, together with the fact that $R$ is reduced.

Finally, note that for any finitely generated $R$-module, if $M = \mathfrak m M$ for all $\mathfrak m$ then $M = 0$ (since Nakayama then implies that $M_{\mathfrak m} = 0$ for all $\mathfrak m$). Thus if $M$ is non-zero it can't be that every section lies in $\mathfrak m M$ for all $\mathfrak m$.

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Thanks, Matt E, especially for the $C[t]/t^2$ example (I knew about the other example already and just forgot to mention M should be finitely generated)! Also your proof for the torsion free case is nice, I gladly accept your answer. –  Jan Sep 1 '10 at 18:52

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