Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $M$ is a left $\mathbb{C}[t] \langle \partial _t \rangle$-module ( a left module over the Weyl algebra), then $\mathrm{Hom}(M,\mathbb{C}[t] \langle \partial _t \rangle)$ is equipped with a structure of right $\mathbb{C}[t] \langle \partial _t \rangle$-module (by $(\varphi . P)(m) = \varphi(m)P$). If we replace the Weyl algebra by a ring of differential operators $k\langle \partial _t \rangle$ (when $k$ is a (skew)field) and consider a left module $N$ then, Is $\mathrm{Hom}(N,k\langle \partial _t \rangle)$ a right module with the action defined above? In what other contexts this "duality" left-right could be considered?

Thanks in advance

share|improve this question
add comment

1 Answer

If $A$ is an algebra, $M$ is a left $A$-module, and $B$ is an $A$-bimodule (e.g. $A$ itself, like in your example), then $Hom_A(M,B)$ is a right $A$-module, with the action given, as you suggest, by $$(\phi.a)(m)=\phi(m).a.$$ Does that help?

A remark along the same lines: $B\otimes_A M$ is again a left $A$-module, with the action given by $$a.(b\otimes m)=(a.b)\otimes m.$$ This construction, in a more general situation of $B$ being an $A$-$A'$-bimodule for another algebra $A'$, shows up when you want to show that the categories $A-mod$ and $A'-mod$ are equaivalent (that is, the algebras $A$ and $A'$ are Morita equivalent).

share|improve this answer
    
Thanks a lot Vladimir. Do you have any reference for this result? (It is just for the sake of completeness) –  Polarbear Jan 3 '12 at 13:00
    
Sure! Jacobson, Basic algebra II, Sec. 3.8, especially Prop. 3.5. (With $S$ being the ground field $k$.) –  Vladimir Dotsenko Jan 3 '12 at 15:01
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.