# Discussion of the curve x^2+y^2+1 over $\mathbb{R}$

I am currently studying von Szamuelys "Galois Groups and Fundamental Groups" and in the chapter 4.3 he discusses the curve $x^2+y^2+1$ over $\mathbb{R}$ via the base change to $\mathbb{C}$.

Since I'm new to algebraic geometry I am having a hard time understanding a few of his statements.

Before he discusses this example he defines a integral algebraic curve as the ringed space $(X=Spec(A),\mathcal{O}_X)$ where $A$ is an integral domain of an arbitrary field $k$ with trancendence degree 1. (In this case every prime ideal is maxiaml). And $\mathcal{O}_X$ is a sheaf of rings on $X$ (equipped with the Zarski Topology) via:

$$$\mathcal{O}_X(U):=\bigcap_{P \in U}A_p$$$ for any open U.

The closed points of a curve are those maximal ideals that do not contain an ideal I.

(I am not sure if this is a common definition, so I repeated it here)

When discussing the curve $f=(x^2+y^+2+1)$ he considers the $\mathbb{R}$-curve $X=Spec(\mathbb{R}[x,y]/(f))$

It is not clear to me how this works with the definition since $A$ was supposed to be an integral domain of a field, but even though $\mathbb{R}[x,y]/(f)$ is a integral domain, it is not part of $\mathbb{R}$. Is my understanding of the definition wrong?

We now conclud that the closed points of $X$ correspond to the maximal ideals of $\mathbb{R}[x,y]$ not containing $(f)$. Why? shouldent those be the open sets since we factor out f?

For each such point we now have $\mathbb{R}[x,y]/M \simeq \mathbb{C}$ because $\mathbb{C}$ is the only nontrivial finite extension of R and X has no points over R. (this is clear)

With the base change morphism $X_C \to X$ there are two closed points lying above each closed point of X, (yes) because $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathbb{C} \oplus \mathbb{C}.$ If we make Gal (C|R) act on the tensor product via its action on the second term, then on the right-hand side the resulting action interchanges the components. (why would we even consider this?)

1. I think you should write $\mathbb{R}[x,y]/(f)$ rather than $\backslash$, since I think the former is the common notation for quotients.
2. For any integral domain you can take its field of fractions. Here if you take the field of fractions of $\mathbb{R}[x,y]/(f)$ you will get a field of transcendence degree $1$ over $\mathbb{R}$. Geometrically the transcendence degree corresponds to the dimension of your variety. In this case you have a curve so it is $1$.
3. The closed points of $\mathrm{Spec}(A)$ are its maximal ideals. In this case the maximal ideals of $\mathbb{R}[x,y]/(x^2+y^2+1)$ are the maximal ideals of $\mathbb{R}[x,y]$ containing the ideal $(x^2+y^2+1)$.
4. Note that the Galois action sends $z\mapsto \bar{z}$. If you identify $\mathbb{C}$ with $\mathbb{R}[x]/(x^2+1)$, then your tensor product there is given by $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[x]/(x^2+1) \cong \mathbb{C}[x]/(x+i) \oplus \mathbb{C}[x]/(x-i) \cong \mathbb{C} \oplus \mathbb{C}$. Explicitly, this maps the element $a+bx \mapsto (a-bi,a+bi)$. Your Galois action acts by mapping $x\mapsto -x$, and you can see that under the identification this corresponds to interchanging the two components.