# $X⊗_A$ − sends finite dimensional module to a projective module

$$A$$ 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

$$M$$ 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.

• HI Qiaochu, thank you. question: why $hom_k(M,-)$ is exact, $M$ is arbitrary here, right?
– ssu
Oct 3, 2020 at 2:50
• @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. Oct 3, 2020 at 2:56
• clear now. thank you!
– ssu
Oct 3, 2020 at 3:00