Showing whether an ideal in $\mathbb{Z}[x,y]$ is prime. The ideal $(1+x^2,1+y^2)$ is prime in $\mathbb{Z}[x,y]$?
I have this:
Analogously to $\mathbb{Z}[x]/(1+x^2)\simeq \mathbb{Z}[i]$, $\mathbb{Z}[x,y]/(1+x^2,1+y^2)\simeq \mathbb{Z}[i]\times \mathbb{Z}[i]$ and $\mathbb{Z}[i]\times \mathbb{Z}[i]$ is not a  integral domain. Therefore $(1+x^2,1+y^2))$ is not prime. This is correct? pd: The ideal $(p)$, $p$ prime is prime in $\mathbb{Z}[x,y]$? I have this: $\mathbb{Z}[x,y]/(p)\simeq (\mathbb{Z}[x]/(p))[y]\simeq (\mathbb{Z}_{p}[x])[y]$ and $\mathbb{Z}_{p}$ is a field then $\mathbb{Z}_{p}[x]$ is a field?

You need a completer argument why $$\mathbb{Z}[x,y]/(1+x^2,1+y^2)\simeq \mathbb{Z}[i]\times \mathbb{Z}[i]$$
I’m not even sure it is true.
I think the quotient ring is $\mathbb{Z}[i]\otimes_{\mathbb Z}\mathbb{Z}[i].$
Then the zero divisors are $$(i\otimes 1 +1\otimes i)(i\otimes 1-1\otimes i)=0.$$

You are correct, though, the ideal is not prime.
We have
$$(x-y)(x+y)\in (1+x^2,1+y^2),$$ but neither $x+y$ nor $x-y$ is in the ideal.
[You need to show $x-y$ and $x+y$ are not in the ideal, of course.]

The quotient can be written as the ring $R$ of all:
$$a+bi+cj+dij$$
where $a,b,c,d\in \mathbb Z,$ and $i^2=j^2=-1$ and $ij=ji.$ You can show this is the quotient by taking $\mathbb Z[x,y]\to R$ with $x\mapsto i,y\mapsto j$ and show this map is onto and has kernel $(1+x^2,1+y^2).$
