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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

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1 Answer 1

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).

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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

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