0
$\begingroup$

If $G$ is a finitely generated non-zero module over the non-trivial commutative Noetherian ring $R$ then is it possible that for all maximal ideal $M$ of $R$ we have $MG=G$ ?

If $R$ is semi-local then by Nakayama's lemma the answer is no (although we don't need to use the fact that $R$ is Noetherian). What about arbitrary Noetherian ring $R$ ?

$\endgroup$
1
  • 2
    $\begingroup$ Doesn't Nakayama work in this case? That is, why do you need semi-locality? You localize at a maximal ideal m in $R$, and note that by Nakayama, as G is finitely generated, $m G_m =G_m$ , and thus $G_m=0$ for all maximal ideals. This implies that $G=0$ . (I might be wrong, but I want to know if I am!) $\endgroup$
    – Dedalus
    Jun 25, 2012 at 19:18

1 Answer 1

Reset to default
2
$\begingroup$

I'll post this as an answer instead of a comment.

I believe that Nakayama's lemma works in this case too and I don't see why you need semi-locality. Localize at a maximal ideal $m$ in R, we then have that if $mG_m=G_m$, then $G_m=0$ by Nakayama's lemma. But if a module satisfies $G_m=0$ for all maximal ideals of R, then $G=0$.

$\endgroup$
1
  • $\begingroup$ This is correct, and indeed neither semilocality nor Noetherianness of $R$ is needed $\endgroup$
    – zcn
    Jan 27, 2014 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.