Is $\langle x^2+1,y\rangle$ maximal or prime in $\Bbb{R}[x,y]$ or $\Bbb{C}[x,y]$

Question

Let $R$ be a ring and $R[x,y]$ be the ring of all polynomials in variables $x$ and $y$ with coefficients in $R$. Then I need to check whether the ideal $\langle x^2+1,y\rangle$ is maximal, prime in $R[x,y]$?

My understanding is that $\langle x^2+1,y\rangle$ is not maximal, as it is a proper subset of the ideals $\langle x^2+1\rangle$ and $\langle y\rangle$.

However I think it is a prime ideal if $R=\Bbb{R}$, and not prime if $R=\Bbb{C}$. Am I right? Then how do I show it?.

Answer 1

$\langle x^2+1, y \rangle$ is not a subset of $\langle y \rangle$ or of $\langle x^2+1 \rangle$. It's a superset of those.

In $\Bbb{C}[x,y]$ it is a proper subset of $\langle x+i, y \rangle$ and hence not maximal. It's also not prime since factoring out $\langle y \rangle$ maps it to the ideal $\langle x^2+1 \rangle$ in $\Bbb{C}[x]$, which is not maximal and therefore not prime since $\Bbb{C}[x]$ is a PID. (And therefore $\Bbb{C}[x,y]/\langle x^2+1,y \rangle \cong \Bbb{C}[x]/\langle x^2+1 \rangle$ which is not an integral domain.)

In $\Bbb{R}[x,y]$ it is both maximal and prime, since $\Bbb{R}[x,y]/\langle x^2+1,y \rangle$ is isomorphic to a field, namely $\Bbb{C}$.

Answer 2

The ideal $(x^2+1, y)$ is maximal in $\mathbb{R}[x, y]$, since $\mathbb{R}[x, y] / (x^2+1, y) \simeq \mathbb{R}[x] / (x^2 +1)$ by the third isomorphism theorem, and $\mathbb{R}[x] / (x^2 +1)$ is $\mathbb{C}.$
On the other hand, it is not a maximal ideal in $\mathbb{C}[x, y]$ (it is not even prime, in fact). This is because $\mathbb{C}[x, y] / (x^2+1, y) \simeq \mathbb{C}[x] / (x^2+1) \simeq \mathbb{C}[x] / (x - i) \times \mathbb{C}[x] / (x + i) \simeq \mathbb{C}^2$.