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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).

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up vote 7 down vote accepted

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

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


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

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 $T = 0.$

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thank you, this looks really nice. Can you please explain the line: $$T_{\mathfrak{m}}/\mathfrak{m}T_{\mathfrak{m}} = T/\mathfrak{m}T$$, is it because in general $S^{-1}(IM)=S^{-1}(I)S^{-1}(M)$ where $S$ is a multiplicative set, $I$ an ideal and $M$ an $A$-module? so apply this to $$I=\mathfrak{m}$$, then $\mathfrak{m}_{\mathfrak{m}}$ is the unique maximal ideal of $A_{\mathfrak{m}}$? is that the reasoning behind? – user6495 Jul 28 '11 at 17:24
Dear @jspecter, why $R/m$ is a flat $R$-module? one can have $m \not = m^2.$ – Ehsan M. Kermani Dec 18 '12 at 3:35

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. $T=0$.

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This is the same proof. Note $T_M = 0$ if and only if $Ann_R(T) \not \subset M.$ – jspecter Jul 28 '11 at 17:24
Yes, in essence it's the same proof. But there's no need for localization, or Nakayama's lemma. – Bruno Joyal Jul 28 '11 at 17:28
A=R, I think. How do you know T/MT is nonzero? It sounds like Nakayama to me. – Jack Schmidt Jul 28 '11 at 17:33
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. :) – Bruno Joyal Jul 28 '11 at 17:46

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