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 $A$ be a $k$ finite dimensional algebra and let $M$ be a simple finite dimensional right $A$-module. Why is the dual of $M$, i.e $\operatorname{Hom}_{k}(M,k)$ a simple left $A$-module?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Let $M$ be any finite dimensional $A$-module.

Suppose $N$ is a submodule of $\hom_k(M,k)$. Then $N^\perp=\{m\in M:\forall\phi\in N,\phi(m)=0\}$ is a submodule of $M$, and this establishes a bijection $$N\in\operatorname{Sub(hom_k(M,k))}\mapsto N^\perp\in\operatorname{Sub} M$$ between the set of submodules of $M$ and those of $\hom_k(M,k)$.

Now, a module is simple iff it has exactly two submodules, so this bijection tells us that the dual of a finite dimensional simple module is simple.

share|improve this answer
    
it is clearly surjective since giving a submodule of $M$ then take the set of all linear map which vanish on that submodule. Why is it injective though? –  user10 Feb 4 '13 at 22:17
    
This statement should remind of a similar one relating subspaces of a finite dimensional vector space with those of the dual space. The arguments are exactly the same in both cases —this should be explained in pretty much any good linear algebra textbook. –  Mariano Suárez-Alvarez Feb 4 '13 at 22:39
    
(And, in fact, injectivity in our case follows immediately from the injectivity in the case of vector spaces, as my map is just a restriction of that one) –  Mariano Suárez-Alvarez Feb 4 '13 at 22:42
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.