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.

Let N be a submodule of the module M. Suppose M/N and N are semi-simple. Does it follow that M is semi simple?

I think the answer is yes but I am not sure how to prove it. Any help would be appreciated.

share|improve this question
1  
No. In fact this is false even if you assume that $M/N$ and $N$ are simple, and even for a ring as well-behaved as $\mathbb{Z}$. –  Qiaochu Yuan May 15 '12 at 22:21

2 Answers 2

No. Consider the ring $R=\mathbb C[t]/(t^2)$. The regular module $M=R$ contains the submodule $N=(t)$ which is simple, and $M/N$ is also simple, yet $M$ is not semisimple.

In fact, under sensible hypotheses if a ring is such that your statement does hold, then the ring must be semisimple. For example, if the ring is a finite dimensional algebra over a ring (or even an artin algebra)

share|improve this answer

Adding on Qiaochu's comment and trying to come up with an easier, perhaps, counter-example than that given by Mariano: the only simple $\mathbb{Z}-\,$modules (which are the same as abelian groups) are the cyclic groups of prime order, thus: take Klein's viergrup $$\,M:=C_2\times C_2\,\,,\,\,N:= \{1\}\times C_2\,\,,\,\,M/N\cong C_2$$ Both $M\,,\,M/N\,$ are even simple, as mentioned by Qiaochu

share|improve this answer
1  
But the big module in your example is semisimple :) –  Mariano Suárez-Alvarez May 16 '12 at 2:12
1  
If you replace $M$ with $C_4$ and $N$ with the cyclic group of order $2$, though, then you'll be cooking with gas. –  Arturo Magidin May 16 '12 at 2:29

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.