$\mathbb{Z}^2$ as a quotient of $\mathbb{Z}[x_1, x_2, \dots, x_n]$ Is $\mathbb{Z}^2$ (as a ring) a quotient of $\mathbb{Z}[x_1, x_2, \dots, x_n]$ for some $n$?
 A: For any ring $R$,
$$R[x] / \langle x(1-x) \rangle \cong R^2$$
The forward maps are $f \mapsto (f(0), f(1))$.
The reverse maps are $(a,b) \mapsto a + (b-a) x$.
A: The ring $\mathbb{Z}[x_1,\dots,x_n]$ is the free commutative ring generated by $x_1,\dots,x_n$.  So for a commutative ring to be a quotient of $\mathbb{Z}[x_1,\dots,x_n]$, it just has to be generated by $n$ elements (since you can then map the $x_i$ to those elements).  If you allow $n$ to vary, this just means you need your ring to be finitely generated.
The ring $\mathbb{Z}^2$ is obviously finitely generated: it is finitely generated just as an abelian group, by the elements $(1,0)$ and $(0,1)$.  So you can get a surjective homomorphism $\mathbb{Z}[x_1,x_2]\to\mathbb{Z}^2$ by sending $x_1$ to $(1,0)$ and $x_2$ to $(0,1)$.
(Less obvious is that $\mathbb{Z}^2$ is actually a quotient of $\mathbb{Z}[x]$, meaning it can be generated by one element.  To see this, note that $(1,0)$ alone already generates $\mathbb{Z}^2$ as a ring, since $(1,1)$ is just the unit of the ring and so $(0,1)=(1,1)-(1,0)$ is in the subring generated by $(1,0)$.)
