How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$?
I know I can mod out the ideal one piece at a time and show $\mathbb C[x]/(x^2+1)$ is not a field since $(x^2+1)$ is not maximal in $\mathbb C[x]$, but is there another way of showing this?
On
The zeros of these polynomials are the complex points $(i,i)$, $(i,-i)$, $(-i,i)$ and $(-i,-i)$. These fall into two orbits under the action of complex conjugation, viz., $(i,i)$ and $(-i,-i)$ which also satisfy $x=y$, and $ (i,-i)$ and $(-i,i)$ which also satisfy $x=-y$. It follows that say, $\left<x^2+1,y^2+1,x-y\right>$ is a proper ideal containing $\left<x^2+1,y^2+1\right>$ as it still has zeroes over the complex numbers (e.g., $(i,i)$) but $\left<x^2+1,y^2+1\right>$ has a zero (e.g., $(i,-i)$) which is not a zero of $\left<x^2+1,y^2+1,x-y\right>$.
Let $I = (x^2+1,y^2+1)$, let $J=(x^2+1,x-y)$, and let $K=(x^2+1,x+y)$.
If $1 \in J$, then $a(x^2+1)+b(x-y)=1$, for some $a,b \in \mathbb{R}[x,y]$. But then, letting $y=x$, we get $a(x,x)(x^2+1)=1$, contradiction, since a nonzero multiple of $x^2+1$ must have degree at least $2$.
If $1 \in K$, then $c(x^2+1)+d(x+y)=1$, for some $c,d \in \mathbb{R}[x,y]$. But then, letting $y=-x$, we get $c(x,-x)(x^2+1)=1$, contradiction, since a nonzero multiple of $x^2+1$ must have degree at least $2$.
Thus, $J$ and $K$ are proper ideals.
Since $x-y \in J$, we get $(x+y)(x-y) \in J$, hence $x^2-y^2\in J$. Then since $y^2+1 = (x^2 + 1) - (x^2 - y^2)$, we get that $y^2 + 1 \in J$. It follows that $I \subseteq J$.
Since $x+y \in K$, we get $(x+y)(x-y) \in K$, hence $x^2-y^2\in K$. Then since $y^2+1 = (x^2 + 1) - (x^2 - y^2)$, we get that $y^2 + 1 \in K$. It follows that $I \subseteq K$.
Suppose $I$ is maximal. Then we must have $J=I$ and $K=I$.
From $J=I$, we get $x-y\in I$, and from $K=I$, we get $x+y\in I$.
From $x-y \in I$ and $x+y \in I$, we get $x \in I$.
But then, from $x \in I$ and $x^2+1\in I$, we get $1 \in I$, contradiction.
It follows that $I$ is not maximal.