9
$\begingroup$

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.

  • 0
    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.2012-08-26

1 Answers 1