Showing whether an ideal in $\mathbb{Z}[x,y]$ is prime.

Question

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?

Answer 1

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 $$(i\otimes 1 +1\otimes i)(i\otimes 1-1\otimes i)=0.$$

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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.]

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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 $(1+x^2,1+y^2).$