According to a comment in Dummit and Foote, $\bigwedge^2( \mathbb{Z}[x,y])=0$. Why is this true?
I can't convince myself that monomials such that $x \otimes y$ or $x \otimes 1$ as elements of $\mathbb{Z}[x,y] \otimes_{\mathbb{Z}} \mathbb{Z}[x,y]$ would go to zero once we quotient out by the ideal generated by $p(x,y) \otimes p(x,y)$.
This is only true if you are taking the wedge square of $\mathbb{Z}[x,y]$ as a $\mathbb{Z}[x,y]$-module, not as a $\mathbb{Z}$-module. The wedge square is then defined as the quotient of $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]$ by the $\mathbb{Z}[x,y]$-submodule $K$ generated by elements of the form $p(x,y)\otimes p(x,y)$. Taking $p(x,y)=1$, $1\otimes 1$ is an element of the submodule $K$. But $1\otimes 1$ generates $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]$ as a $\mathbb{Z}[x,y]$-module, since any simple tensor $p(x,y)\otimes q(x,y)$ is equal to the scalar product $(p(x,y)q(x,y))\cdot (1\otimes 1)$. (Indeed, $\mathbb{Z}[x,y]\otimes_{\mathbb{Z}[x,y]}\mathbb{Z}[x,y]\cong\mathbb{Z}[x,y]$ by sending $p(x,y)\otimes q(x,y)$ to $p(x,y)q(x,y)$, and $1\otimes 1$ maps to $1\in\mathbb{Z}[x,y]$ under this isomorphism.)
(Note there is nothing special about $\mathbb{Z}[x,y]$: this argument shows that for any commutative ring $R$, $\bigwedge_R^2(R)=0$.)