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.

I need to prove that a surjective homomorphism of finite $\mathbb{F}_p[\Delta]$-modules $$A \twoheadrightarrow B$$ results in $B$ being a subrepresenation of $A$ of the group $\Delta$ of order prime to $p$. I, however, don't know if this really leads to a subrepresentation, since I've always thought, that a subrepresentation should look like $B \subset A$ in module language. Thanks for your help, Tom

share|improve this question
1  
Exactly how is the original exercise worded? And, is $\Delta$ any group? –  Berci Apr 17 '13 at 16:43
    
a subrepresentation should look like $B\subseteq A$ in module language I agree with that. In any case, $B$ would be a quotient representation of $A$... –  rschwieb Apr 17 '13 at 17:01
1  
A surjection means $B$ is a quotient of $A$. Now $\Delta$ is prime to $p$, so representations are semisimple, so subrep and quotient rep is the same thing. –  Aaron Apr 17 '13 at 19:47
    
Thank you for your hint! Are submodules and quotient modules the same in the semi-simple case because every short exact sequence splits? –  BIS HD Apr 17 '13 at 21:05
2  
The $k[G]$ modules are semisimple, so all submodules are direct summands. –  tharris Apr 17 '13 at 22:29

1 Answer 1

up vote 0 down vote accepted

As tharris and Aaron pointed out, for a field $k$ (e.g. $\mathbb{F}_p$) whose characteristic does not divide the group order $|\Delta|$, Maschke's theorem asserts that any $k[\Delta]$-module is semisimple, which is per definition equivalent to the fact that every submodule is a direct summand. Hence any quotient module must be a submodule.

Thanks to you, guys!

share|improve this answer

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.