# In which case $M_1 \times N \cong M_2 \times N \Rightarrow M_1 \cong M_2$ is true?

Usually for modules $M_1,M_2,N$ $$M_1 \times N \cong M_2 \times N \Rightarrow M_1 \cong M_2$$ is wrong. I'm just curious, but are there any cases or additional conditions where it gets true?

James B.

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Hm... if $M_1$, $M_2$ and $N$ are free and finitely generated over a ring with the IBN, it holds true. –  martini Jun 14 '12 at 17:11
This is a very hard question: For example, let $R=k$x_1,\ldots,x_n$$. If $P$ is projective module of finite rank and $P\oplus R^n\cong R^{n+m}$, then $P\cong R^m\ \ldots$ But this is the Serre's Conjecture! –  H. Kabayakawa Jun 14 '12 at 20:50

A standard result in this direction is the Krull-Schmidt Theorem:

Theorem (Krull-Schmidt for modules) Let $E$ be a nonzero module that has both ACC and DCC on submodules (that is, $E$ is both artinian and noetherian). Then $E$ is a direct sum of (finitely many) indecomposable modules, and the direct summands are unique up to a permutation and isomorphism.

In particular:

Corollary. If $M_1$, $M_2$, and $N$ are both noetherian and artinian, and $M_1\times N \cong M_2\times N$, then $M_1\cong M_2$.

Proof. Both $M_1\times N$ and $M_2\times N$ satisfy the hypothesis of the Krull-Schmidt Theorem; decompose $M_1$, $M_2$, and $N$ into a direct sum of indecomposable modules. The uniqueness clause of Krull-Schmidt yields the isomorphism of $M_1$ with $M_2$. $\Box$

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I recommend A Crash Course on stable range, cancellation, substitution and exchange by T.Y. Lam, 2004, which is a pretty good guide to the phenomenon.

Update: Another question surfaced which contains a sufficient condition for cancellation. The condition is: $\hom(M_1,N)=\hom(M_2,N)=\{0\}$.

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