<s>[INST] How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$? Is there a geometric interpretation for this? 
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?
 [/INST] 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.
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