TITLE: Is $(x^2 + 1, y^2 + 1)$ a prime ideal in $\mathbb{Q}[x, y]$?
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: $\mathbb{Q}[x,y]/(x^2+1,y^2+1) = \mathbb{Q}[x]/(x^2+1)[y]/(y^2+1) = \mathbb{Q}[x]/(x^2+1)[y]/(y+x)(y-x)$. 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.