What are the closed points of $\mathbb{A}_{\mathbb{R}}^2 = \operatorname{Spec}(\mathbb{R}[x,y])$? I am trying to find all the closed points of $\mathbb{A}_{\mathbb{R}}^2$.
After a quick google research, I found that $\mathbb{A}_{\mathbb{R}}^2 = \operatorname{Spec}(\mathbb{R}[x,y])$ and then all we need to find is the maximal ideals of $\mathbb{R}[x,y]$.
However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.
And more than that, how do I find the maximal ideals of $\mathbb{R}[x,y]$? Seems hard.
Any help would be appreciated.
Thanks!
 A: Here is a more direct and canonical approach than the first one I posted:
First, an important fact: if $R=k[X,Y]$, and $\mathfrak{m} \subset R$ is a maximal ideal, then the field $R/\mathfrak{m}$ is a finite extension of $k$.  This is known as Zariski's Lemma, and it's the "hard" part of the Nullstellensatz, though there is at least one very short proof for general $k$, and a one-line proof when $k=\mathbb{R}$.
Now, we can make quick work of this problem.  Since $R/\mathfrak{m}$ is a finite extension of $k$, we can identify $R/\mathfrak{m}$ with a subfield of $K$, an algebraic closure of $k$. (When $k=\mathbb{R}$, we have $K=\mathbb{C}$, which you might know as the fundamental theorem of algebra).
Identify the cosets $X + \mathfrak{m},Y+\mathfrak{m}$ with $a,b\in K$, let $p(X)$ be the minimal polynomial for $a$ over $k$, and let $q(X,Y)$ be the minimal polynomial for $b$ over $k[a]$, considered as a polynomial in $Y$ with coefficients in $k[1,X,\ldots , X^{(\deg p) - 1}]$.  Clearly, $p(a)=q(a,b)=0$, so $p,q\in\ker(R\to R/\mathfrak{m})=\mathfrak{m}$
I claim that $\mathfrak{m} = (p,q)$.  Indeed, if $f\in R$, then there is a unique $g\in f+(p,q)$ with $\deg_X g<\deg p$ and $Y$ degree $\deg_Y g<\deg q$ (use the polynomial division algorithm twice).
If $f\in\mathfrak{m}$, then $g\in\mathfrak{m}$, so $g(a,b)=0$.  By the definition of the minimal polynomial, $g(a,Y)=0$.  By the same definition applied to the coefficients of $g\in (k[X])[Y]$, we have $g(X,Y)=0$.  In other words, $f\in (p,q)$.

When $k=\mathbb{R}$, we can be very concrete:


*

*If $a,b\in\mathbb{R}$, then $p=X-a$, $q=Y-b$.

*If $a\in\mathbb{R}$ but $b\notin\mathbb{R}$, then $p=X-a$, $q=(Y-b)(Y-\overline{b})$

*If $a\notin\mathbb{R}$, then $1$ and $a$ are linearly independent over $\mathbb{R}$, so we can pick $c,d\in\mathbb{R}$ with $ca+d=b$.  Then $p=(X-a)(X-\overline{a})$, $q=Y-(cX+d)$.



My original suggested approach:


*

*Classify the maximal ideals of $\mathbb{C}[X,Y]$ using the Nullstellensatz.

*Show that every maximal ideal of $\mathbb{R}[X,Y]$ is the restriction of a maximal ideal of $\mathbb{C}[X,Y]$.

*Describe when different maximal ideals of $\mathbb{C}[X,Y]$ restrict to the same maximal ideal of $\mathbb{R}[X,Y]$. (Hint: the number of maximal ideals above a given one is either $1$, or $2$.)


This proof is somehow more "geometric" than the one above, more natural but requiring more difficult tools.  Here is how it goes:
The inclusion $\mathbb{R}[X,Y]\hookrightarrow \mathbb{C}[X,Y]$ induces a morphism $f:\mathbb{A}^2_\mathbb{C}\to \mathbb{A}^2_\mathbb{R}$, which is surjective because it is the base change to $\mathbb{A}^2_\mathbb{Z}$ of the surjective morphism $\operatorname{Spec}\mathbb{C}\to\operatorname{Spec}\mathbb{R}$.
If $\mathfrak{m}\in\mathbb{A}^2_\mathbb{R}$ is a closed point with residue field $\kappa$, then the fiber $f^{-1}(\mathfrak{m})$ can be identified with the fiber product $\mathbb{A}^2_\mathbb{C} \times_{\mathbb{A}^2_\mathbb{R}} \operatorname{Spec} \kappa \cong \operatorname{Spec} (\mathbb{C}[X,Y]\otimes_{\mathbb{R}[X,Y]} \kappa) \cong \operatorname{Spec}(\kappa)$ or $\operatorname{Spec}(\kappa\times \kappa)$, depending on whether $\kappa=\mathbb{R}$ or $\kappa=\mathbb{C}$.
So each fiber has either one or two points, corresponding to a real point $(a,b)$ , or a pair of conjugate points $\{(c,d),(\overline{c},\overline{d})\}$.
A: Re: 

However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.

The points of a scheme are*, by definition, prime ideals of a ring.  If this is new to you, you're probably not ready to be looking at this question until you go back and learn what a scheme is.
Once you understand the question, Slade's answer shows you a good way of working the problem.
*Or, if you want to be incredibly pedantic, correspond to the ideals of a ring, since a scheme is a locally ringed space that happens to be locally isomorphic to the spectrum of a ring.
