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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?

user46225
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  • You should try to find an isomorphism between that module and $k[x]\otimes k[x]$ with its usual structure. – Mariano Suárez-Álvarez Nov 28 '15 at 19:01
  • @MarianoSuárez-Alvarez: following the answer below, I guess the decomposition you say is $k[x]\otimes k[x']$ where $x$ acts on the right factor by multiplying a polynomial in $x'$ by $x'$, is this what you meant? – user46225 Dec 01 '15 at 12:11
  • Possibly related: https://math.stackexchange.com/questions/158445 – Watson Oct 01 '16 at 16:13

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$k[x]\otimes_k k[x]\cong k[x,y]$. 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 $k[x]$.

user26857
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Mohan
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  • I'm afraid I don't understand entirely: why consider $k[x-y,y]$ and how do you deduce freeness? – user46225 Nov 30 '15 at 21:27
  • What is it that you have problems with? Any element $f(x,y)\in k[x,y]$ can be written as $g(x-y,y)$. Expand with respect to $y$ so that it looks like $a_0+a_1y+\cdots+a_ny^n$ where $a_i$s are polynomials in $x-y$. So, it follows that $y^n$ generates $k[x,y]$ as a module over $\phi(k[x])$. They are linearly independent is equally easy. – Mohan Nov 30 '15 at 21:58
  • Thank you, that's clearer. The simple observation that the "action through $\varphi$" amounts to "action of the subring $\varphi(k[x])$ through the given action" is actually quite clarifying. – user46225 Dec 01 '15 at 09:49
  • Shouldn't the basis be ${(x-y)^n: n\geq 0}$? – user46225 Dec 01 '15 at 10:50
  • I mean, the action is "multiply by $x-y$"; it's clear that the resulting module is free if instead of considering $k[x-y][x]$ as you say you consider $k[x][x-y]$, right? i.e. taking the $a_i$ to be polynomials in $x$ and taking $x-y=:x'$ as the indeterminate – user46225 Dec 01 '15 at 12:08