# Projective module, polynomial ring

Let $k$ be a field and consider $k[x,y]$. $k$ is then a $k[x,y]-$module if we let $k[x,y]$ act by $f(x,y)\cdot k = f(0,0)\cdot k$.

Is $k$ projective? I have tried to construct an isomorphism between $k\oplus (x,y) \cong k[x,y]$

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The subtle point is that we have an isomorphism $k\oplus (x,y) \cong k[x,y]$ as $k$-vector spaces but not as $k[x,y]$- modules. – Georges Elencwajg Jan 5 '14 at 11:40

No. Consider the usual $k[x,y]$-module structure on $k[x,y]$. The canonical projection $k[x,y] \to k$ (which sends $1 \mapsto 1$) of $k[x,y]$-modules does not have a section.
In particular, there's no nonzero map of $k[x,y]$-modules $\varphi : k \to k[x,y]$ whatsoever. For $x\varphi(1) = \varphi(x \cdot 1) = \varphi(0) = 0$, so $\varphi(1) = 0$.
A projective module $M$ over a domain $D$ has no torsion: for $d\in D$ and $m\in M$ the equality $dm=0$ implies $d=0$ or $m=0$.
[This is because $M$ is a submodule of a free module and free modules have no torsion over a domain]
Since $M=k$ has torsion over $D=k[x,y]$ (since $x\cdot 1=0$), the module $k$ is not projective.