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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?

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  • $\begingroup$ 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 $P^{\ast}\otimes_R P\cong {\rm End}(_RP)$. $\endgroup$
    – Pace Nielsen
    Commented Feb 21, 2020 at 15:44
  • $\begingroup$ Did you mean to say "morphism" instead of "isomorphism"? $\endgroup$
    – Michael Bächtold
    Commented Feb 21, 2020 at 20:06

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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 $$ \mathrm{ev}\colon Q'\otimes_R P\to S, \qquad (\psi\otimes p)\mapsto \psi(p). $$

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  • $\begingroup$ 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.) $\endgroup$
    – Fofi Konstantopoulou
    Commented Feb 20, 2020 at 22:52
  • $\begingroup$ 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. $\endgroup$
    – Leonid Positselski
    Commented Feb 20, 2020 at 22:58
  • $\begingroup$ I think that clears it up - thanks! $\endgroup$
    – Fofi Konstantopoulou
    Commented Feb 20, 2020 at 23:07
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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.

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    $\begingroup$ 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 $R{-}Mod$. $\endgroup$
    – Leonid Positselski
    Commented Feb 21, 2020 at 12:28
  • $\begingroup$ Right! I will add this condition $\endgroup$
    – Bugs Bunny
    Commented Feb 21, 2020 at 14:21
  • $\begingroup$ What is a "finitely generative projective generator "? $\endgroup$
    – Fofi Konstantopoulou
    Commented Feb 21, 2020 at 14:30
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    $\begingroup$ A generator, which is finitely generated and projective. See en.wikipedia.org/wiki/Generator_(category_theory) $\endgroup$
    – Bugs Bunny
    Commented Feb 21, 2020 at 14:32
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    $\begingroup$ 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. $\endgroup$
    – Leonid Positselski
    Commented Feb 21, 2020 at 15:57

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