Simple form of a ring What is a simple form of this ring: $$\mathbb{Z}[\sqrt{2}][x]/(5,x^2+1),$$
I know that $\mathbb{Z}[\sqrt{2}][x]=\mathbb{Z}[x,y]/(y^2-2)$. Probably, I should use second theorem of isomorphism, but I don't see how.
 A: Hint: Use the fact that you can mod out in any order, and use the isomorphism $\mathbb{Z}[\sqrt{2}][x]/(5,x^2 + 1)\cong\mathbb{Z}[x,y]/(5,x^2 + 1,y^2 - 2)$. I'll start you off:
\begin{align*}
\mathbb{Z}[\sqrt{2}][x]/(5,x^2 + 1)&\cong\mathbb{Z}[x,y]/(5,x^2 + 1,y^2 - 2)\\
&\cong\mathbb{F}_5[x,y]/(x^2 + 1,y^2 - 2)
\end{align*}
Now look at these polynomials. What can you say about the quotient of $\mathbb{F}_5[x,y]$ by either? Another hint: you'll want to use the Chinese remainder theorem eventually.
A: You can write it as $$\mathbf F_5[x,y]/(x^2+1,y^2-2).$$ 
As $y^2-2$ has no root in $\mathbf F_5$, it is irreducible over $\mathbf F_5$ and $\mathbf F_5[y]/(y^2-2)$ is a field, isomorphic to $\mathbf F_{25}$.
As for $x^2+1$, it has  roots in  $\mathbf F_5$ ($\pm 2$), so the quotient $\mathbf F_5[x,y]/(x^2+1,y^2-2)$ is isomorphic to $\mathbf F_{25}[x]/((x-2)(x+2))\simeq\mathbf F_{25}[x]/(x-2)\times\mathbf F_{25}[x]/(x+2)$  by the Chinese Remainder Theorem, and the latter is isomorphic to $\mathbf F_{25}\times\mathbf F_{25} $.
A: Hint:
Well you know that:
$\mathbb{Z}[x] / (x^2+1) = \mathbb{Z}[i]$
$\mathbb{Z} / (5) = \mathbb{Z}_5$
