$x^2=y^2=(x+y)^2 = 0$ but $xy \neq 0$?

I'm looking for an example of a commutative ring with two elements $x,y$ such that $x^2=y^2=(x+y)^2=0$, and $xy \neq 0$. Obviously, $2$ must be a zero divisor in such a ring. I don't think products of $\mathbb{Z}/n\mathbb{Z}$ will do. Anyone has an idea?

-
Since we're in a commutative ring $\,R\,$ , $\,(x+y)^2=x^2+2xy+y^2\,$ , so your conditions imply that it must be $\,char(R) = 2\,$ . Perhaps this can help to see why Georges's nice answer takes such a ring. –  DonAntonio Dec 16 '12 at 10:55

An example of what you require is $\mathbb F_2[x,y]\stackrel {def}{=}\mathbb F_2[X,Y]/(X^2,Y^2)$
What might not be completely evident is that $xy\neq 0$.
This means that in $\mathbb F_2[X,Y]$ the monomial $XY$ is not in the ideal $(X^2, Y^2)$ or that an equation $$XY=X^2f(X,Y)+Y^2g(X, Y)\quad \text{with }f(X,Y),g(X,Y)\in \mathbb F_2[X,Y]$$ is impossible.
But this is clear: taking the degree 2 components of both sides yields $XY=X^2f(0,0)+Y^2g(0,0)$, an obvious impossibility.

-
Thank you very much. –  timofei Dec 16 '12 at 10:21
It's a pleasure. And welcome to this site! –  Georges Elencwajg Dec 16 '12 at 11:10
+1 Impressive: The key point is "not completely evident", but "clear" per an "obvious impossibility". ;) –  Hagen von Eitzen Dec 16 '12 at 11:13

Let $E_2=\begin{pmatrix}1&1\\1&1\end{pmatrix}$. Consider the ring of matrices over $\mathbb{F}_2$ generated by $$0,\ I_4,\ x=\begin{pmatrix}E_2&0\\0&E_2\end{pmatrix},\ y=\begin{pmatrix}I_2&I_2\\I_2&I_2\end{pmatrix}$$ under usual matrix addition and multiplication.

-