$k[x]\otimes k[x]$ as a right $k[x]$-module Let $k$ be a commutative ring. Consider the ring map $\varphi:k[x]\to k[x]\otimes_k k[x]$ given by $\varphi(x)=x\otimes 1-1\otimes x$.
Now consider $k[x]\otimes_k k[x]$ as a right module over itself. Via $\varphi$, it becomes a $k[x]$-module.
Is this module projective?

$k[x]\otimes_k k[x]\cong k[x,y]$. Thus $\phi(x)=x-y$. Then, changing variables, you may assume that $\phi:k[x]\to k[x-y,y]=k[x,y]$ with $\phi(x)=x-y$ and thus $k[x,y]$ is a free module with basis $y^n, n\geq 0$ over $k[x]$.
