Is $(x^2 + 1, y^2 + 1)$ a prime ideal in $\mathbb{Q}[x, y]$?

Question

At first I was looking for a ring homomorphism from $\mathbb{Q}[x, y]$ to a domain with $(x^2 + 1, y^2 + 1)$ as it's kernel, but I could not find one.

Now I am thinking: maybe $(x + y)(x - y) = x^2 - y^2 \in (x^2 + 1, y^2 + 1)$ while $(x + y), (x - y) \notin (x^2 + 1, y^2 + 1)$, but I'm not sure how to show the second part.

Can anyone give me a hint?

Answer 1

We show for example that $x-y$ is not in the ideal. As a function on the complex numbers, $x^2+1$ vanishes at $i$, and $y^2+1$ vanishes at $-i$. So all members of the ideal vanish when $x=i$ and $y=-i$. But $x-y$ does not.

A similar argument takes care of $x+y$.

Answer 2

$\mathbb{Q}[x,y]/(x^2+1,y^2+1) = \mathbb{Q}[x]/(x^2+1)[y]/(y^2+1) = \mathbb{Q}[x]/(x^2+1)[y]/(y+x)(y-x)$. Now use the Chinese Remainder Theorem. The result will be a product of two fields. Hence, the quotient has zero divisors, i.e. $(x^2+1,y^2+1)$ is not prime.