7
$\begingroup$

Suppose that $A$ is a commutative noetherian Jacobson ring and $M$ is an $A$-module. Suppose in addition that $M$ is $\mathfrak{m}$-adically separated for every maximal ideal $\mathfrak{m}$, and that $M/\mathfrak{m}M$ is $A/\mathfrak{m}$-finite for every $\mathfrak{m}$. Does it follow that $M$ is $A$-finite?

If I imposed no separatedness conditions, then $\mathbf{Q}$ as a $\mathbf{Z}$-module would be a counterexample. If I did not require $A$ to be Jacobson, or if I only imposed a separatedness condition at one maximal ideal, then $k[\![t]\!]$ as a $k[t]$- or $k[t]_{(t)}$-module would be a counterexample.

If it helps, I'm happy to assume that the $\mathfrak{m}$-adic completions $M_{\mathfrak{m}}^\wedge$ of $M$ are all finite projective of the same rank. The rings I have in mind are polynomial rings over fields, or (strict) affinoid algebras over discretely valued fields, so I'm also happy to make additional assumptions (such as excellence) on $A$.

$\endgroup$
1
  • $\begingroup$ Just to clarify: even for a Noetherian, Jacobson ring $A$, most finite $A$-modules are not $\mathfrak{m}$-adically separated, e.g., $\mathbb{Z}/2\mathbb{Z}$ is not $p$-adically separated except for $p=2$. Krull proved that for every Noetherian ring $A$, for every finitely generated $A$-module $M$, and for every prime ideal $\mathfrak{p}$ of $A$, the localized module $M_{\mathfrak{p}}$ is $\mathfrak{p}$-adically separated. $\endgroup$ Jun 23, 2014 at 15:48

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.