TITLE: Is the ideal $ (X^{2}+1,Y^{2}+1) $ prime in $ \mathbb{Q}[X,Y] $?
QUESTION [5 upvotes]: I am not very confident that what I have done is right although I can't see any flaw in it. So here it goes: 
Since $ (Y^{2}+1) \subseteq  (X^{2}+1,Y^{2}+1) $, by the third isomorphism theorem for rings we have that $$ \frac{\mathbb{Q}[X,Y]}{(X^{2}+1,Y^{2}+1)}  \simeq \frac{(\mathbb{Q}[X])[Y]\big/(Y^{2}+1)}{(X^{2}+1,Y^{2}+1)\big/(Y^{2}+1)} $$ 
Hence $$ \frac{\mathbb{Q}[X,Y]}{(X^{2}+1,Y^{2}+1)}  \simeq \frac{(\mathbb{Q}[i])[X]}{(X^{2}+1)}= \frac{(\mathbb{Q}[i])[X]}{(X+i)(X-i)} $$ which by the Chinese Remainder Theorem is isomorphic to $ \mathbb{Q}[i] \times \mathbb{Q}[i] $ and since clearly this is not a domain, we conclude that $ (X^{2}+1,Y^{2}+1) $ does not belong to the spectrum of $ \mathbb{Q}[X,Y] $. 
I would be thankful for any comments and hints in case my solution is wrong. Thank you!

REPLY [5 votes]: Very well done, I think you can skip the CRT part, since $X \pm i$ is a nontrivial zero-divisor, which is easily seen from $\frac{(\mathbb{Q}[i])[X]}{(X+i)(X-i)}$.