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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?

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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.

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  • $\begingroup$ 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 Oct 20 '16 at 12:25

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