Let $R,S$ be two K algebras, where K is a fixed field. Then we can get a new algebra $R \otimes S$, i.e. the tensor product of these two algebras. Suppose the following sequence $$0 \rightarrow R\otimes S \rightarrow X_1 \rightarrow \dots \rightarrow X_n$$ is an exact sequence of $(R \otimes S)$-projective-injective modules $X_i$.

My question is that how to prove these $X_i$ are projective and injective in $R$-modules and $S$-modules, repsectively?


1 Answer 1


I understand your question that if $X$ is an injective $R\otimes S$ module, then it is an injective $R$ module and the same for projective. In this formulation it seems true, at least if we assume the algebras to have units. Let's start with $X$ being projective. Then there exists an $R\otimes S$ module $Y$ such that $X\oplus Y$ is free. As $R\otimes S$ is a free $R$-module (the tensor product being over a field), $X\oplus Y$ is also free as an $R$-module, so $X$ is projective as an $R$ module.

Next assume $X$ is $R\otimes S$-injective. Let $$ 0\to M\to N $$ be an exact sequence of $R$-modules. Then $0\to M\otimes S\to N\otimes S$ is still exact. Now any $R$-module homomorphism $\alpha: M\to X$ induces an $R\otimes S$ module homomorphism $M\otimes S\to X$, which extends to a homomorphism from $N\otimes S$ to $X$. The ensuing map $N\to N\otimes 1\to X$ will be an extension of $\alpha$. So $X$ is injective as $R$ module.

  • $\begingroup$ This is a nice answer. Alternatively, the second part follows from the usual lemma: A functor which has an exact left adjoint preserves injectives. $\endgroup$
    – HeinrichD
    Commented Oct 20, 2016 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.