In Bourbaki's Algebra, there is the following exercise ($A$ is an arbitrary ring with unity):

Suppose that $I$ is an $(A,A)$-bimodule which is injective as a left $A$-module and a right $A$-module and that for every (left or right) cyclic $A$-module $E\neq 0$, there exists a non-zero $A$-homomorphism of $E$ into $I$. Show that under these conditions, if $P$ is a left projective $A$-module, $\operatorname{Hom}_A(P,I)$ is a right injective $A$-module.

The following straightforward proof does not use the condition on the cyclic modules, so there is probably something wrong with it, and I hope someone can point out the faulty step.

$\operatorname{Hom}(P,I)$ can be seen as a right $A$-module in a canonical way, using the bimodule structure of $I$.

To show that $\operatorname{Hom}(P,I)$ is injective, it suffices to show that, for any right $A$-module $N$ and a submodule $N'$ of $N$, the homomorphism $\Phi$ of $\operatorname{Hom}(N,\operatorname{Hom}(P,I))$ into $\operatorname{Hom}(N',\operatorname{Hom}(P,I))$ given by the restriction of mappings is surjective.

Since $I$ is injective, the homomorphism of $\operatorname{Hom}(N,I)$ into $\operatorname{Hom}(N',I)$ given by restriction is surjective. Since $P$ is projective, the induced mapping $\Psi$ of $\operatorname{Hom}(P,\operatorname{Hom}(N,I))$ into $\operatorname{Hom}(P,\operatorname{Hom}(N',I))$ is surjective.

But $\operatorname{Hom}(P,\operatorname{Hom}(N,I))$ is canonically isomorphic to $\operatorname{Hom}(N,\operatorname{Hom}(P,I))$, and likewise $\operatorname{Hom}(P,\operatorname{Hom}(N',I))$ to $\operatorname{Hom}(N',\operatorname{Hom}(P,I))$. $\Phi$ and $\Psi$ correspond to each other via these isomorphisms, so $\Phi$ must be surjective as well.

Where is the mistake? If you think there is none, please tell me so.

  • $\begingroup$ There are situations where the criterion on cyclic modules is unnecessary: consider $A=\mathbb{Z}$ and $I=\mathbb{Q}$. Then $\mathrm{Hom}(P,\mathbb{Q})$ is divisible and thus injective as a right $\mathbb{Z}$-module. $\endgroup$
    – KReiser
    Aug 26, 2012 at 19:58

1 Answer 1


I didn't analyze what you wrote very carefully, but I do know the answer to your title question.

Suppose $P$ is an $R$-$S$ bimodule which is flat as a left $R$ module, and $M$ is an injective right $S$ module. Then $\mathrm{Hom}(P_S,M_S)$ is an injective right $R$ module (where the right $R$ module action is given in the usual way: $(f\cdot r)(x)=f(rx)$ ).

This is the "Injective Producing Lemma" which appears (among other places I'm sure) in Lam's Lectures on Modules and Rings pg 61. It's used several times throughout the book.

The proof is really short, and I think essentially says what you have above. Paraphrasing:

We have the standard isomorphism:

$$\mathrm{Hom}(A_R,\mathrm{Hom}(P_S,M_S)_R)\cong \mathrm{Hom}((A\otimes _R P)_S,M_S)$$ Since $_R P$ is flat, $-\otimes_R P$ is exact, and since $M_S$ is injective $\mathrm{Hom}(-,M_S)$ is exact. Thus the composition on the right hand side is exact on $A$, and therefore the left hand side is exact on $A$. Consequently, $\mathrm{Hom}(P_S,M_S)$ is an injective right $R$ module.

  • $\begingroup$ I can't make the transfer to the statement of the question. The problem is that $P$ is not assumed to be a bimodule. $\endgroup$
    – Stefan
    Aug 30, 2012 at 17:26
  • $\begingroup$ @StefanWalter Sorry: I picked different letters and it looked so similar...! So maybe full projectivity of $P$ is necessary here with the bimodule $M$ to get things going... $\endgroup$
    – rschwieb
    Aug 30, 2012 at 17:52

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