Are these two rings isomorphic? And can I use the chinese remainder theorem to prove it? Are these two rings isomorphic?
1) $\frac{k[x,y]}{(x^2+1,y^2+1)}$
2) $\frac{k[x]}{(x^2+1)}\times\frac{k[y]}{(y^2+1)}$
and if they are, does it follow from the Chinese Remainder Theorem? And if yes...how?
 A: Assume that $\text{char} k \neq 2$.
If $x²+1$ is irreducible over $k$ one has that $k[x,y]/(x²+1,y²+1) \simeq k(i)[y]/(y²+1)$.
Since in $k(i)[y]$ one has $(y²+1) = (y+i)(y-i)$ and since $(y+i) - (y-i) = 2i \in k(i)^\times$ it follows by the chinese remainder theorem that $k(i)[y]/(y²+1) \simeq k(i)[y]/(y+i) \times k(i)[y]/(y-i)$ which is in turn isomorphic to $k(i) \times k(i)$, as is the second ring.
So in this case both rings are isomorphic to each other.
If $x²+1 = (x+i)\cdot (x-i)$ is reducible over $k$ one has that the first ring is isomorphic to $(k \times k)[y]/(y²+1)$ which should be isomorphic to $k[y]/(y²+1) \times k[y]/(y²+1)$ - so I think in this case both rings are isomorphic too.
Not sure about $\text{char}k = 2$ though..
A: Just to fill in the gap left by @lush in his answer, the story is quite different in characteristic two.
Let us then consider a field $k$ of characteristic two, and the first named ring $k[x,y]/\langle x^2+1,y^2+1\rangle$; since $x^2+1=(x+1)^2$, you see that our ring is also isomorphic to $k[x,y]/\langle x^2,y^2\rangle$, a local ring in which $x$ and $y$ are nilpotent. The unique maximal ideal is $\langle x,y\rangle$. The $k$-dimension is still four, the basis being $\{1,x,y,xy\}$.
Use the same argument on $k[x]/\langle x^2+1\rangle\times k[y]/\langle y^2+1\rangle\cong k[x]/\langle x^2\rangle\times k[y]/\langle y^2\rangle$, not a local ring, since it has the two maximal ideals $\langle(x,0),(0,1)\rangle$ and $\langle(1,0),(0,y)\rangle$. Alternatively, this ring has the four idempotents $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$, while the first ring has only the two idemopotents $0$ and $1$.
