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

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?.

• Why do you think it is prime for $\mathbb{R}$ and not for $\mathbb{C}$? In addition, think about whether $y \in \langle x^2 + 1 \rangle$. Do you have theorems relating maximal/prime ideals to field/integral domain quotients? If so, those are the best approach for this question. Apr 18, 2018 at 4:11
• Because $x^2+1$ is irreducible in the first case, but reducible in the latter case.
– QED
Apr 18, 2018 at 4:13
• That's a good observation - in the case of $\mathbb{C}$, there's an easy way to show that the ideal is not prime now, using the reduction you mention. Apr 18, 2018 at 4:16

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 $$\mathbb{C}[x, y] / (x^2+1, y) \simeq \mathbb{C}[x] / (x^2+1) \simeq \mathbb{C}[x] / (x - i) \times \mathbb{C}[x] / (x + i) \simeq \mathbb{C}^2$$.
$$\langle x^2+1, y \rangle$$ 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 $$\Bbb{C}$$.