Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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
add comment

2 Answers

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 $M_2$. $\Box$

share|improve this answer
add comment

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: $\hom(M_1,N)=\hom(M_2,N)=\{0\}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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