Ring of polynomials as free module

Question

Is it true that $R=k[x,y]$ is a free $R$-module ?

I think that it isn't true. Natural candidate for the base is $\{x^{\alpha}y^{\beta}\}_{\alpha,\beta}$, but :

$x\cdot (xy) +(-y)\cdot x^2 =0$

and $x\neq 0,\ -y \neq 0$.

Answer 1

You have confused $k$-vector space basis as $R$-module basis, hence this issue.

ANy commutative ring is a free module overitself with basis the singleton $\{1\}$, consiting of the multiplicative identity.

Given two elements $a,b\in $R, by commutativity we have $a.b -b.a=0$ and so a basis has to have less than 2 elements.