# 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 more argument to show the quotient is $\mathbb Z[i]\times\mathbb Z[i].$ I am not even sure if it is true. Jul 30, 2020 at 15:01
• $\Bbb{Z}[x,y]/(p)$ is an integral domain, so $(p)$ is prime, but $(\Bbb{Z}/(p))[x]$ is not a field. Jul 30, 2020 at 16:46

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

• I see. But, my argument $\mathbb{Z}[x,y]/(1+x^1,1+y^2)\simeq \mathbb{Z}[i]\times \mathbb{Z}[i]$ not integral domain then is not prime is correct? Jul 30, 2020 at 15:00
• If you can show that isomorphism, yes, but I’m not sure it is true. Jul 30, 2020 at 15:02
• $x+y$ is not in $(1+x^2,1+y^2)$ by an argument of the grade of $x+y$ right? Jul 30, 2020 at 15:50
• @Subhajit $\Bbb{Z}[x,y]/(1+x^2,1+y^2)\cong \Bbb{Z}[i,y]/(1+y^2)$. By standard properties of the tensor product, $\Bbb{Z}[i,y]/(1+y^2)\cong\Bbb{Z}[i]\otimes_{\Bbb{Z}}\Bbb{Z}[y]/(1+y^2)\cong\Bbb{Z}[i]\otimes_{\Bbb{Z}}\Bbb{Z}[i]$. Jul 30, 2020 at 16:44