TITLE: Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.
QUESTION [14 upvotes]: Disclaimer:  Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic. 
I want to show $\mathbb{Q} \simeq \mathbb{Q}[x,y]/\langle x,y \rangle$ is not projective as a $\mathbb{Q}[x,y]$ module.  I've tried two methods, neither of which gets me to the conclusion. 
I first tried what seems to be sort of standard when proving that something is not projective: show that the lifting of the identity yields a contradiction.  So I let $\pi: \mathbb{Q}[x,y] \to \mathbb{Q}[x,y]/\langle x,y \rangle$ be my surjection given by $f \mapsto \bar{f}$ and the identity map is $id:  \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]/\langle x,y \rangle$. So all I need to show is that a homomorphism $\phi: \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]$ such that $\pi \circ \phi =id$ does not exist.  But if $$\pi(f) = \bar{f} = \overline{a_0+a_{10}x+a_{01}y+a_{11}xy+\cdots+a_{n0}x^n + a_{0n}y^n} = \bar{a_0}$$ then doesn't the map $\bar{a_0} \mapsto a_0$ work?  After all, $$ (\pi\circ \phi)(\bar{a_0}) = \pi(a_0) = \bar{a_0} = id(\bar{a_0}).$$I was concerned at first about this not being well defined, but since every element of a particular coset has the same constant term, it does not depend on choice.  So either I have already made a mistake, or this is just the wrong map from which to derive a contradiction.
The next thing I tried used a different characterization of projective modules: that $P$ is a projective $R$-module iff there is a free module $F$ and an $R$-module $K$ such that $F \simeq K\oplus P$. In our case, this means there is a free module $F$ and a $\mathbb{Q}[x,y]$-module $K$ such that 
$$
\mathbb{Q}[x,y] \oplus \cdots \oplus \mathbb{Q}[x,y] \simeq F \simeq K \oplus \mathbb{Q}[x,y]/\langle x,y \rangle \simeq K \oplus \mathbb{Q}.$$
From here, my concern is that I am waving my hand too much when I say: obviously this cannot be true, since every element of the LHS, which is a tuple of polynomials, cannot be broken up with one chunk in $K$ and the other in $\mathbb{Q}$.  Do agree?  If so, how can I make this argument more rigorous?
One more trouble: nowhere in either of these methods did I explicitly use that the polynomial ring here is only in two variables.  The fact that the question did not use $\mathbb{Q}[x_1,\ldots,x_n]$ instead of $\mathbb{Q}[x,y]$ worries me.

REPLY [10 votes]: A projective module over a domain has no nonzero torsion element, since it is a  submodule of a free module.
  But every element of your module is a torsion element: it is killed by $x$.