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$A$ a finite-dimensional $k$-algebra, $X$ projective $A ⊗_k A^{op}$-mdoule. The projectivity of $X$ as a bimodule implies that $X⊗_A$ − sends finite dimensional module to a projective module.

I am not quite familiar with bimodule projectivity. Why the tensor is projective? Thank you

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$M$ is projective iff $\text{Hom}(M, -)$ is exact, so we want to know why $\text{Hom}_A(X \otimes_A M, -)$ is exact. The tensor-hom adjunction gives

$$\text{Hom}_A(X \otimes_A M, -) \cong \text{Hom}_{A \otimes_k A^{op}}(X, \text{Hom}_k(M, -))$$

where, if $N$ is an $A$-module, $\text{Hom}_k(M, N)$ acquires an $A \otimes_k A^{op}$-bimodule structure by composing and precomposing by the actions of $A$ on $N$ and $M$ respectively. This means $\text{Hom}_A(X \otimes_A M, -)$ is a composition of two exact functors, namely $\text{Hom}_k(M, -)$ (I assume $k$ is a field here) and $\text{Hom}_{A \otimes_k A^{op}}(X, -)$ (by projectivity).

The following more explicit argument may also be helpful. Projective modules are retracts of free modules. If $X = \bigoplus_i A \otimes_k A^{op}$ is a free bimodule then $X \otimes_A (-) \cong \bigoplus_i A \otimes_k (-)$ is a free $A$-module (again we need to assume here that $k$ is a field). Then taking a retract of $X$ means taking a retract of this free $A$-module, which will be projective.

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  • $\begingroup$ HI Qiaochu, thank you. question: why $hom_k(M,-)$ is exact, $M$ is arbitrary here, right? $\endgroup$
    – ssu
    Oct 3, 2020 at 2:50
  • $\begingroup$ @steven: this is why we need the assumption that $k$ is a field. Then $M$, as a $k$-module, is free, so in particular projective. Without that assumption what we get is that $X \otimes_A (-)$ sends projective $k$-modules to projective $A$-modules. $\endgroup$ Oct 3, 2020 at 2:56
  • $\begingroup$ clear now. thank you! $\endgroup$
    – ssu
    Oct 3, 2020 at 3:00

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