# Dual of a projective module

Let $$R$$ be a noncommutative ring with unit, let $$P$$ be a projective left $$R$$-module, and denote $$^{\vee}\!P := \,_R\mathrm{Hom}(P,R)$$. One often sees it written that projectivity implies an isomorphism $$\mathrm{ev}:\,^{\vee}\!P \otimes_R P \to R, ~~ \phi \otimes p \mapsto \phi(p).$$ But I don't see that this is well defined! Consider $$(\phi.r) \otimes p \mapsto \phi.r(p) = (\phi(p))r$$ where we have used the definition of the right $$R$$-module structure of $$^{\vee}\!P$$. Compare this with $$\phi \otimes (r.p) \mapsto \phi(r.p) = r \phi(p),$$ where we have used the fact that $$\phi$$ is a left $$R$$-module map. Now since $$R$$ is not commutative, these two values are not guaranteed to be equal, and the evaluation map is not guaranteed to be well-defined. What is wrong here?

• Sorry for my misunderstanding earlier. Your question, and the answers below, seem to be connected to Exercise 2.20(4), in T.Y. Lam's "Lectures on Modules and Rings". When $P$ is a f.g. projective left $R$-module, then $P^{\ast}\otimes_R P\cong {\rm End}(_RP)$. Feb 21, 2020 at 15:44
• Did you mean to say "morphism" instead of "isomorphism"? Feb 21, 2020 at 20:06

You are right. There is no such a map as the one you are trying to describe.

Here is a map that actually exists. Let $$R$$ and $$S$$ be two noncommutative rings with units, and let $$P$$ be an $$R$$-$$S$$-bimodule. Consider the $$S$$-$$R$$-bimodule $$Q={}_R\mathrm{Hom}(P,R)$$. Then the evaluation is an $$R$$-$$R$$-bimodule map $$\mathrm{ev}\colon P\otimes_S Q\to R, \qquad (p\otimes\phi)\mapsto \phi(p),$$ with the tensor product taken over the ring $$S$$.

In particular, if you do not have an $$R$$-$$S$$-bimodule but only a left $$R$$-module $$P$$, you can take $$S=\mathbb Z$$. Then you get an $$R$$-$$R$$-bimodule map $$\mathrm{ev}\colon P\otimes_{\mathbb Z}Q \to R,$$ with the tensor product taken over the ring integers.

Notice that the $$S$$-$$R$$-bimodule $$Q$$ is (generally speaking) very different from the $$S$$-$$R$$-bimodule $$Q'=\mathrm{Hom}_S(P,S)$$. For the bimodule $$Q'$$, the evaluation is an $$S$$-$$S$$-bimodule map $$\mathrm{ev}\colon Q'\otimes_R P\to S, \qquad (\psi\otimes p)\mapsto \psi(p).$$

• Thanks for the answer. But it seems that $Q \otimes_R P$ is well defined as a $\mathbb{Z}$-module. So is the issue here that $Q = \,_R\mathrm{Hom}$ is really a "right dual" for $P$ and not a "left dual"? (Here I mean right and left dual in rough analogy with the monoidal category sense.) Feb 20, 2020 at 22:52
• I've added a paragraph at the end of the answer which is relevant to your question in the comment. Yes, this is the difference between the left and right duals. Feb 20, 2020 at 22:58
• I think that clears it up - thanks! Feb 20, 2020 at 23:07

Doc, you ain't write no evaluation map. If $$R$$ is commutative, you write the trace map. If $$R$$ is noncom, god knows what you write. The evaluation map, that is an isomorphism for a finitely generative projective generator and a homomorphism of $$R$$-$$R$$-bimodules, in general, is $$p \otimes \phi \mapsto \phi (p), \ P \otimes_{End_RP} P^{\vee} \rightarrow R \, .$$ Use it with care.

In terms of Leonid's answer, you have a canonical $$S:=End_RP$$, lying around.

• Taking $S$ to be the endomorphism ring of $P$ over $R$ is a good idea, but still the evaluation map is not an isomorphism for every finitely generated projective left $R$-module $P$ (not even in the commutative case). Take $P=0$ to see that the evaluation map does not need to be an isomorphism for a finitely generated projective. I guess the evaluation (with $S={}_R\mathrm{Hom}(P,P)^{\mathrm op}$) is an isomorphism when $P$ is a finitely generated projective generator of $R{-}Mod$. Feb 21, 2020 at 12:28
• Right! I will add this condition Feb 21, 2020 at 14:21
• What is a "finitely generative projective generator "? Feb 21, 2020 at 14:30
• A generator, which is finitely generated and projective. See en.wikipedia.org/wiki/Generator_(category_theory) Feb 21, 2020 at 14:32
• A projective left $R$-module $P$ is a generator of $R{-}Mod$ if and only if for every nonzero left $R$-module $M$ there exists a nonzero $R$-module morphism $P\to M$. Equivalently, for every left $R$-module $M$ there exists a surjective left $R$-module morphism onto $M$ from a direct sum of sufficiently many copies of $P$. The general definition of what it means for an object of a category to be a generator is slightly more complicated; see e.g. the link above. Feb 21, 2020 at 15:57