5
$\begingroup$

Let $R$ be a commutative ring with $1$ and let $S,T$ be $R$-modules such that $S$ is finitely generated and such that $S \cong S \oplus T$. Must $T=0$?

This is certainly true if $R$ is a PID, but what if $R$ is just a commutative ring with $1$? (if $R$ is a PID then the fact that $S \cong S \oplus T$ and $S$ is finitely generated implies $T$ is finitely generated as well, and then the result follows from the cancellation law for modules over a PID).

2 Answers 2

7

Yes. Let $\mathfrak{m}$ be any maximal ideal of $R.$ Then

$S/{\mathfrak{m}S} \cong S\otimes_{R} R/\mathfrak{m} \cong (S\oplus T) \otimes_{R} R/\mathfrak{m} \cong (S\otimes_{R} R/\mathfrak{m})\oplus (T \otimes_{R} R/\mathfrak{m}) \cong S/{\mathfrak{m}S} \oplus T/{\mathfrak{m}T}.$

It follows

$dim_{R/\mathfrak{m}}(T/\mathfrak{m}T) = 0.$

Hence,

$T_{\mathfrak{m}}/\mathfrak{m}T_{\mathfrak{m}} = T/\mathfrak{m}T = 0.$

Therefore by Nakayama's lemma

$T_{\mathfrak{m}} = 0.$

And as this is true for all maximal ideals of $R,$ it must be the case that $T = 0.$

  • 0
    Dear @jspecter, why $R/m$ is a flat $R$-module? one can have $m \not = m^2.$2012-12-18
4

Here's a proof not using localization.

Suppose there is a maximal ideal $M$ of $R$ which contains the annihilator of $T$. Then $(R/M) \otimes_R T$ is a nonzero vector space over $R/M$. Moreover, since,

$(R/M) \otimes_R S \cong [(R/M) \otimes_R S] \oplus [(R/M) \otimes_R T],$

counting dimensions over $R/M$ shows that $(R/M) \otimes_R T=0$, for every maximal ideal $M$. Thus no maximal ideal of $R$ contains the annihilator of $T$, which implies of course that the annihilator of $T$ is all of $R$, i.e. $T=0$.

  • 0
    I guess you are right! It's just so intuitive that it's nonzero. Thanks for the heads up. I'll leave my answer here for general edification. :)2011-07-28