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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 30, 2020 at 15:01
  • $\begingroup$ $\Bbb{Z}[x,y]/(p)$ is an integral domain, so $(p)$ is prime, but $(\Bbb{Z}/(p))[x]$ is not a field. $\endgroup$
    – Stahl
    Commented Jul 30, 2020 at 16:46

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

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  • $\begingroup$ 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? $\endgroup$
    – eraldcoil
    Commented Jul 30, 2020 at 15:00
  • $\begingroup$ If you can show that isomorphism, yes, but I’m not sure it is true. $\endgroup$ Commented Jul 30, 2020 at 15:02
  • $\begingroup$ $x+y$ is not in $(1+x^2,1+y^2)$ by an argument of the grade of $x+y$ right? $\endgroup$
    – eraldcoil
    Commented Jul 30, 2020 at 15:50
  • $\begingroup$ Can anyone show that isomorphism? Or any hints about this? $\endgroup$
    – A learner
    Commented Jul 30, 2020 at 16:04
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    $\begingroup$ @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, $\Bbb{Z}[i,y]/(1+y^2)\cong\Bbb{Z}[i]\otimes_{\Bbb{Z}}\Bbb{Z}[y]/(1+y^2)\cong\Bbb{Z}[i]\otimes_{\Bbb{Z}}\Bbb{Z}[i]$. $\endgroup$
    – Stahl
    Commented Jul 30, 2020 at 16:44

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