Posted by Jason Polak on 19. September 2017 · Write a comment · Categories: homological-algebra, modules

Consider a field $k$. Define an action of $k[x,y]$ on $k[x]$ by $f*g = f(x,x)g(x)$ for all $f\in k[x,y]$ and $g\in k[x]$. In other words, the action is: multiply $f$ and $g$ and then replace every occurrence of $y$ by $x$.

Is $k[x]$ a projective $k[x,y]$-module? Consider first the map $k[x,y]\to k[x]$ given by $f\mapsto f(x,x)$. It’s easy to check that this map is in fact a $k[x,y]$-module homomorphism. It would be tempting to try and split this map with the inclusion map $k[x]\to k[x,y]$. But this doesn’t work: this inclusion is not a $k[x,y]$-module homomorphism.

In fact, the $k[x,y]$-module homomorphism $k[x,y]\to k[x]$ given by $f\mapsto f(x,x)$ cannot split simply because there are no nonzero $k[x,y]$-module homomorphisms $k[x]\to k[x,y]$. Therefore, $k[x]$ is not projective as a $k[x,y]$-module, using the module structure we gave it.

Here are two more ways to see this:

  1. Through the notion of separability: by definition, $k[x]$ being a projective $k[x,y]\cong k[x]\otimes_k k[x]$-module under the structure that we have defined means that $k[x]$ is a separable $k$-algebra. However, all separable $k$-algebras are finite-dimensional as vector spaces over $k$, whereas $k[x]$ is infinite-dimensional.
  2. Through Seshradi’s theorem: this theorem says that every finitely-generated projective module over $k[x,y]$ is actually free. Therefore, we just have to show that $k[x]$ is not free because $k[x]$ is certainly finitely-generated as a $k[x,y]$-module. But $(x^2y – xy^2)$ annihilates all elements of $k[x]$, which cannot happen in a free module.

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