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This question have two parts.

The first one:
Let A be a k-algebra that is projective as k-module, where k is a commutative ring. why is $A\otimes_kA$ projective as $A^{ev}=A\otimes_kA^{op}$-module?

I guess it's must be easy, but I can't find a justification.

A counterexample that show how $A$ could be NOT projective as $A^{ev}$-module despite being projective as left and right $A$-module.

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  • I think that $A\otimes_k A$ with the canonical $A^{ev}$-module structure is just the ring $A^{ev}$ itself; and hence, it is free over $A^{ev}$. – Pavel Aug 01 '18 at 08:07

1 Answers1

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I'm not sure about the first but here's a counterexample for the second: Take $A = k[x]$. Note this is commutative so the $op$ construction doesn't matter. $A \otimes A \simeq k[x, y]$ (where $y$ is just the $x$ from the right tensor factor) and the $k[x, y]$-module structure on $k[x]$ is to let both $x$ and $y$ act as $x$.

Now to see that $k[x]$ isn't a projective $k[x, y]$-module show that there is no injection $k[x] \hookrightarrow k[x, y]^n$ into a free module. Do this by asking where $1 \in k[x]$ is sent to and note that whatever $f \in k[x, y]^n$ this is, equivariance of the injection implies that $xf = yf$, or $(x - y)f = 0$, so $f = 0$.

Jim
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