$A$ 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


1 Answer 1


$M$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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