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Let $A$ be a reduced, noetherian ring and let $M \subseteq \operatorname{Frac}(A)$ be a finitely generated $A$-module (for which then some regular $g \in A$ exists such that $gM \subseteq A$). Further, assume that $M_P \cong A_P$ for every minimal prime ideal $P$ of $A$.

Is it possible that $M$ contains no regular element of $A$?

I would like to see an example or an argument that this is not possible.

The assumption at the minimal primes implies that $M$ is not contained in one of the minimal primes. But is it still possible that $M$ is contained in the union of the minimal primes (which is the set of non-regular elements of $A$)?

windsheaf
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1 Answers1

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Note I'm assuming your ring has identity in what follows, I haven't thought about what happens otherwise.

Let $\{\mathfrak{p}_\alpha\}$ be the set of minimal primes of $R$. Since $R$ is Noetherian, there are finitely many minimal primes.

Let $S$ be the complement of the union of the minimal primes. $S$ is a (saturated) multiplicative set and we can localize at it. Moreover when $R$ is reduced, $S$ is precisely the set of regular elements of $R$.

Consider that when $A$ has finitely many minimal primes, we have

Lemma: $A_S \cong \prod_{i=1}^n A_{\mathfrak{p}_i}$ canonically.

See this MSE question for discussion.

We are assuming that $M$ is fractional with $gM \subset A$, so the above isomorphism and our assumptions about $M_{\mathfrak{p}_i} \cong A_{\mathfrak{p}_i}$

induce a nice pair of isomorphisms

$$gFrac(A) \cong \prod_{i = 1}^{n} (gA)_{\mathfrak{p}_i} \cong \prod_{i = 1}^{n}(gM)_{\mathfrak{p}_i}$$

Look at numerators in the image of any regular element of $gFrac(A)$ in $\prod_{i = 1}^{n}(gM)_{\mathfrak{p}_i}$ (which is regular).

Badam Baplan
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