Geometrical interpretation of $\operatorname{Spec}(\mathbb{R}[x,y]/(x^2+y^2))$

Question

$\def\Spec{\operatorname{Spec}}$If am not mistaken, the prime spectrum of $\Spec(\mathbb{R}[x,y]/(x^2+y^2))$ consists of the points $(\overline{x},\overline{y})$, $(\overline{x}-a)$, $(\overline{y}-b)$, where $a,b\in\mathbb{R}\setminus\{0\}$ (I think one can prove this by using the description of $\Spec k[x,y]$ explained here).

However, I wonder: why is this spectrum? Does it bear some geometric interpretation? I heard that its points are to be interpreted as “two lines that intersect at the origin.” I guess reflecting somehow the fact that over the complex numbers we can factor $x^2+y^2=(x+iy)(x-iy)$, and that $\Spec(\mathbb{C}[x,y]/(x^2+y^2))$ is effectively the classical complex affine variety equal to $V(x+iy)\cup V(x-iy)\subset\mathbb{A}_\mathbb{C}^2\ (=\mathbb{C}^2)$.

My questions are:

1. Is this interpretation of $\Spec(\mathbb{R}[x,y]/(x^2+y^2))$ correct?
2. Is any of this pointing out to any general phenomenon/a going on behind the scenes? If I'm not being precise is because I don't know what I'm looking for. I guess I would be happy with (a) knowing about some relation between a $k$-variety (in the scheme-theoretic sense) and its base change to $\overline{k}$ (whenever the later is still a variety), (b) some explanation or indications of literature about how schemes extend geometry to polynomials that don't have solutions or don't have that many solutions over non-algebraically closed fields (I am not talking about the usage of nilpotents in modern algebraic geometry, everything I would like to find out more about is in the reduced case).

(The most general definition of “scheme-theoretic $k$-(pre)variety” I know is that of a reduced (separated) $k$-scheme of finite type.)

Answer 1

There are further ideals, even maximal ideals.

The maximal ideals must have quotient fields that are extensions of $\mathbb R.$ Consider the cases when the quotient field is $\mathbb R$ or $\mathbb C.$ So the image of $f:\mathbb R[x,y]\to\mathbb C$ must send $x,y$ to elements of $\mathbb C$ such that $f(x)=\pm i f(y).$

You've covered the cases when one of $f(x)$ or $f(y)$ is real.

So, given $f(x)=u+vi$ with $uv\neq 0,$ we can take the ideal: $$I_{u,v}=\left\langle \overline x^2-2u \overline x+(u^2+v^2), \overline y^2-2v \overline y +(u^2+v^2)\right\rangle$$

Adding the two generators and multiplying by $-\frac12,$ we can also write this ideal as:

$$I_{u,v}=\left\langle\overline x^2-2u \overline x+(u^2+v^2), u \overline x +v\overline y -(u^2+v^2)\right\rangle$$

Note that when $u\neq 0, v=0,$ this ideal becomes:

$$I_{u,0}=\langle (\overline x-u)^2,u(\overline x-u)\rangle =\langle \overline x-u\rangle,$$ and similarly. $I_{0,v}=\langle \overline y-v\rangle,$ so this pattern contains your cases, too.

You also left out the obvious zero ideal, $\langle 0\rangle,$ which is prime but not maximal.

It should be possible to prove these are all the primes.

Idea for how to start: First, consider the maps $\mathbb R[x],\mathbb R[y]\to \mathbb R[x,y].$ The image of these maps in the quotient by your prime ideal must be integral domains, so the ideals in $\mathbb R[x]$ and $\mathbb R[y]$ must be either zero or generated by prime polynomials. The prime polynomials are the linear polynomials and the quadratic polynomials with complex roots.

If $p(x)\in \mathbb R[x]$ is a prime in the ideal, then show $p(iy)p(-iy)$ is in $\mathbb R[y]$ and must be in the ideal, too. If $p$ is linear and not $x,$ then $p(iy)p(-iy)$ is prime, but if $p$ is quadratic, then $p(iy)p(-iy)$ must factor.

So the prime ideals are:

$$I_{u,v},(u,v)\neq 0,\\ \langle \overline x,\overline y\rangle,\\ \langle 0\rangle $$