A question about whether an ideal is maximal or not

Question

Edited New question : my earlier asked question was assosiated with this question:Consider the ideal $I=(x^2+1,y)$ in the polynomial ring $\mathbb{C}[x,y]$.Then which of the following is true

I already knew this question exists on MSE but here it's not answered by question asker clearly why I is not maximal and Answering username Gone didn't answered about maximal ideal.

So, I should have asked in comments but user: Gone has left website and user which asked the question was last seen in Mid of 2017 . So, the chances that question asker will reply my comment are very bleak. So, I am asking again here.

Due to above mentioned reasons please don't close this question.

I am trying assignment question for my abstract algebra course and I am unable to solve this particular problem.

I is an ideal defined by $I=(x^2+1 , y)$ in polynomial ring $\mathbb{C} [x,y] $ . Then is Prime? Is I maximal?

I showed that I is not prime by proving that $x-i$ doesn't belongs to $I$. But I am unable to think how can I prove it maximal.

Can someone please give hints.

Answer 1

To check whether $I$ is maximal or prime, simply consider the quotient as for a commutative ring with unity say $R$, an ideal $J$ is maximal / prime $\iff$ the quotient ring $R/J$ is a field / an integral domain.

Now in our case, $$\frac{\Bbb C[x,y]}{I}\cong \frac{\Bbb C[x]}{(x^2+1)}$$

Since, $x^2+1 $ splits as $x^2+1=(x+i)(x-i)$ in $\Bbb C[x]$ and $x+i+x-i=2i$ thus, the principal ideals $(x+i)$ and $(x-i)$ are co-maximal, thus by CRT, $$\frac{\Bbb C[x]}{(x^2+1)} \cong \frac{\Bbb C[x]}{(x+i)} \times \frac{\Bbb C[x]}{(x-i)} \cong \Bbb C \times \Bbb C$$ which is not even an integral domain, thus also not a field. Hence by the fact we mentioned it follows that $I$ is neither a Prime ideal nor a Maximal ideal.