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Let $A, B$ be $k$-algebras. $M$ is $A$-$B$-bimodule that is finitely generated projective as a left $A$-module and as a right $B$-module, $N, U$ are $B$-$A$-bimodules that is finitely generated projective as a left $B$-module and as a right $A$-module.

As a left $B$-module, $U$ is a direct summand of $B^n$ for some positive integer $n$, hence $M ⊗_B U$ is a direct summand of $M^n$ as a left $A$-module. This shows that $M ⊗_B U$ is fnitely generated projective as a left $A$-module. A similar argument shows that $M ⊗_B U$ is finitely generated projective as a right $A$-module.

I can't see how a similar argument works. I tried and found I cannot tensor with M anymore as $A$ is not a $B$-module. Thank you!

Pedro
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scsnm
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  • Note that $A-B$ is parsed as $A$ minus $B$. If you want hyphens, do not put - between dollar signs. – Pedro Sep 14 '20 at 11:11
  • noted. will keep that in mind – scsnm Sep 14 '20 at 11:12
  • I don't see a question in your post. You say I can't see how a similar argument works. but I cannot understand what you're talking about. – Pedro Sep 14 '20 at 11:17
  • thx! I meant how to use direct summand and tensoring argument to show this since the author used the word similar. – scsnm Sep 14 '20 at 11:20

1 Answers1

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You want to show that $M\otimes_B U$ is finitely generated projective as a right $A$-module. I suppose you can see it is finitely generated, so let us check that the functor $$X\longmapsto\hom_A(M\otimes_B U,X)$$ is exact. By adjunction, this functor is equal to $$X\longmapsto\hom_B(M,\hom_A(U,X))$$ and this is a composition of two functors that are exact, since $M$ and $U$ are projective.

Pedro
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