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 $R$ ?

  • 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


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 $G=0$.

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

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