let $R,S$ be associative algebras over $\mathbb{C}$. Let $\mathcal{C} \subseteq$ $R$-Mod be a full abelain subcategory of $R$-Mod which is the category of $R$-modules. Let $B$ and $C$ be, a $(R,S)$-bimodule, a left $S$-module, respectively.

In addition, let $P\in \mathcal{C}$ be a projective object in $\mathcal{C}$. Assume that $P$ is finitely-generated $R$-module.

Note that $\text{Hom}_{R}(P,B)$ and $B\otimes_S C$ are right $S$-modules and left $R$-modules in a natural way.

$\bf{My \ \ Question:}$

Prove that $\text{Hom}_{R}(P,B)\otimes _SC \cong \begin{align} \text{Hom}_{R}(P,B \otimes_S C) \end{align}$ provided $\text{dim}C< \infty$, $\text{dim}B\otimes_S C <\infty$ and $B\in \mathcal{C}$.

In fact, I am not sure whether the condition on finite-dimensional spaces are suitable. It is just to make this equality correct. I would be appreciated if anyone can help me improve or correct this conditions.

My idea is that the map $\Phi:\text{Hom}_{R}(P,B)\otimes _SC \rightarrow \begin{align} \text{Hom}_{R}(P,B \otimes_S C) \end{align}$ given by

$\Phi(f \otimes v)(a) : = f(a)\otimes v$, for all $a\in P, f\in \text{Hom}_{R},(A,B) $ and $v\in C$, is an isomorphism. Thank you in advance!

  • 3
    $\begingroup$ I'm not certain everyone will believe the question belongs here (as opposed to Mathematics.StackExchange), but the easiest way is probably to observe that for $P$ finitely generated projective, we have a canonical natural isomorphism $P^\ast \otimes_R - \cong \hom_R(P, -)$, from which the statement follows by associativity of tensor products of bimodules. You can drop the finiteness assumption on $C$. $\endgroup$
    – Todd Trimble
    Feb 29, 2016 at 13:02
  • $\begingroup$ Thank you very much for your comments. But I am still confused about the isomorphism $P^*\otimes_R - \cong \text{hom}_R(P,-)$ on $\mathcal{C}$ since $\mathcal{C}$ is an arbitrary full abelian subcategory. ($P$ is only projective in $\mathcal{C}$). $\endgroup$
    – tzelin1016
    Feb 29, 2016 at 13:13
  • $\begingroup$ @tzelin1016 Could you clarify what you mean by $P$ being "finitely generated"? Do you mean finitely generated as an $R$-module? Or finitely generated in some sense as an object of $\mathcal{C}$ (in which case could you say exactly what you mean by that)? $\endgroup$ Feb 29, 2016 at 13:30
  • 1
    $\begingroup$ @Jeremy Rickard Thank you very much for your question. Sorry for this unclear point. That is, $P$ is finitely generated over $R$. But I am also interested in any condition that will make this an natural isomorphism. (For example, $\mathcal{C}$ comes from some (not necessarily full) abelian subcategory of other module category (say, modules over algebra $T$) in which the corresponding $P$ is finite generated $T$-module. But I know this don't make sense yet...) Thanks again! $\endgroup$
    – tzelin1016
    Feb 29, 2016 at 13:43
  • $\begingroup$ Crossposted at MSE. $\endgroup$ Mar 2, 2016 at 19:37

1 Answer 1


Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.

As noted in the question, for any $R$-$S$-bimodule $_RM_S$ there is an obvious map $$\Phi_M:\operatorname{Hom}_R(P,M)\otimes_SC\to\operatorname{Hom}_R(P,M\otimes_SC),$$ and it is natural in $M$.

If $M$ is of the form $X\otimes_\mathbb{C}S$ for $_RX$ a left $R$-module, then $\Phi_M$ is an isomorphism, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}V\to \operatorname{Hom}_R(P,X\otimes_\mathbb{C}V)$$ is an isomorphism for any vector space $V$.

The standard projective bimodule presentation $$S\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to S\otimes_\mathbb{C}S\to S\to0$$ of $S$ induces an exact sequence $$\tag{*} B\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to B\otimes_\mathbb{C}S\to B\to0,$$ which is split as an exact sequence of left $R$-modules, so that applying the functor $M\mapsto\operatorname{Hom}_R(P,M)\otimes_S C$ to it preserves exactness.

Also, applying $-\otimes_SC$ to the exact sequence $(*)$ gives an exact sequence $$\tag{**} B\otimes_\mathbb{C}S\otimes_\mathbb{C}C\to B\otimes_\mathbb{C}C\to B\otimes_SC\to0. $$

As a sequence of left $R$-modules, the first term is a (possibly infinite) direct sum $\bigoplus_{i\in I}B$ of copies of $B$ and the second is (since $C$ is finite dimensional) a finite direct sum of copies of $B$.

This is the filtered colimit, over finite subsets $J\subseteq I$, of exact sequences of the form $$\tag{***} \bigoplus_{j\in J}B\to B\otimes_\mathbb{C}C\to U_J\to 0,$$ where the first two terms are in $\mathbb{C}$ since $_RB$ is in $\mathcal{C}$, and since $\mathcal{C}$ is an abelian subcategory, the third term is also in $\mathcal{C}$. Therefore $\operatorname{Hom}_R(P,-)$ preserves exactness of the sequences $(***)$ since $P$ is projective in $\mathcal{C}$. Since $P$ is finitely generated as an $R$-module, $\operatorname{Hom}_R(P,-)$ also preserves filtered colimits, and so applying $\operatorname{Hom}_R(P,-)$ to $(**)$ also gives an exact sequence.

So applying $$\Phi:\operatorname{Hom}_R(P,-)\otimes_SC\to\operatorname{Hom}_R(P,-\otimes_SC)$$ to the sequence $(*)$ gives a map of exact sequences which is an isomorphism on the first two terms and therefore also on the third term.

So $\Phi_B$ is an isomorphism.

This doesn't seem to require $B\otimes_SC$ to be finite dimensional.

  • $\begingroup$ Hmm, I think I may need the fact that $\mathcal{C}$ is closed under coproducts, so that $X\otimes_\mathbb{C}S$ is in $\mathcal{C}$ when $X$ is. I'll think about whether that's really necessary. $\endgroup$ Feb 29, 2016 at 22:23
  • $\begingroup$ This is a very interesting and wonderful proof ! I would appreciate you very much ! I also was wondering if the condition of co-products is needed. $\endgroup$
    – tzelin1016
    Feb 29, 2016 at 23:17
  • $\begingroup$ I have one more question. Does the equation in the second last line need to modified to be the following equation? $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}C\to\operatorname{Hom}_R(P,X\otimes_\mathbb{C}C).$$ Thanks again! $\endgroup$
    – tzelin1016
    Feb 29, 2016 at 23:34
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    $\begingroup$ @tzelin1016 I've edited to give a proof that I think works even if $\mathcal{C}$ is not closed under infinite coproducts. $\endgroup$ Mar 3, 2016 at 11:51

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