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I am in a weird situation with problem II 4.7 in Hartshorne's Algebraic Geometry. I can do parts b, c,d and e and I understand what to do in part a. but I am stuck on technical details.

Here is the set up. We have a variety over $ \mathbb{C}$ and a degree $2$ automorphism $ \sigma : X \to X$ which conjugates coefficients in $ \mathbb{C}$. We want to produce a variety $ X_0$ over $ \mathbb{R}$ which base changes to $X$. Let $G = \mathbb{Z} / 2 $. Then we have a homomorphism $ G \to {\rm Aut}(X)$ which sends $1 $ to $ \sigma$. The scheme $ X_0$ should be $ X / G$. The fact that $X$ is separated and that any two points are contained in an affine subscheme guarantee that $X/G$ exists as a scheme. Also it is obvious that $X/G$ is finite type over $ \mathbb{R}$.

Problem 1: How does one go about proving that such a quotient is separated?

Problem 2: How do you prove that $X_0$ base changes to $X$?

If the equations of $X$ all have coefficients in $ \mathbb{R}$ and $ \sigma$ is just conjugation, then problem 2 is easy. Here is an example of when $ \sigma$ is not just complex conjugation

Example: Define $ \sigma : \mathbb{A}^2_{\mathbb{C}} \to \mathbb{A}^2_{\mathbb{C}}$ to be the morphism whose pullback is $ a X^p Y^p \mapsto \overline{a} Y^p X^q $. In this case $ \mathbb{A}^2 / G $ is the spectrum of $ \mathbb{R}[X+Y,XY,i(X-Y)]$.

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I don't quite have the time this afternoon to develop this as fully as I'd like, but here is an idea that I believe should work.

I think it is possible to describe the functor $X\mapsto X_0$ very explicitly on affine schemes. This allows us to easily calculate the base change $(X_0)_\mathbb{C}$, and it should also let us prove separatedness, by looking at the morphism $X_0 \to X_0 \times_\mathbb{R} X_0$, and showing that it has closed image by explicitly writing down the ideal sheaf in terms of the corresponding one for $X \to X \times_\mathbb{C} X$.

On an affine patch, say $X = \operatorname{Spec} A$, where $A=\mathbb{C}[X_1,\ldots,X_n]/I$, and we have an involution $\sigma^* : A\to A$ which conjugates $\mathbb{C}$. Let $B$ denote the subring of elements $a\in A$ with $a=\sigma^* a$. Then our $X_0$ should just be $\operatorname{Spec} B$.

There are certain elements that we know are in $B$, namely $X_i + \sigma^* X_i$, and $i(X_i - \sigma^* X_i)$. I believe that it should be possible to use standard techniques of symmetric functions to show that $B$ is generated over $\mathbb{R}$ by these $2n$ elements. (Note, for example, that $X_i (\sigma^* X_i) = ((X_i+\sigma^* X_i)^2 + (i(X_i-\sigma^* X_i))^2)/4$.)

So we can write $B\cong \mathbb{R}[U_1,V_1,\ldots,U_n,V_n]/J$, where $J$ has some nice description in terms of $I$.

If I find the time, I'll come back and see if I can fill in more details. But I hope that it's at least somewhat helpful as is.

Andrew Dudzik
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  • do you mean to say that if one did not assume that $X$ is separated over $\mathbb C$, we may still find such an $X_0$ by patching real affine schemes? I thought that one should use the fact that $X$ is separated to find $X_0$. I'm ignoring the fact that we also want to show $X_0$ is separated over $\mathbb R$. – quantum May 01 '20 at 16:28
  • Actually I believe that $B$ is generated by no more than $n$ elements Because all those $a=\sigma^* a$ are in $\mathbb R[U_1,\dots,U_n]$ where $U_i = X_i+ \sigma^* X_i$: We may assume that the patches are reduced and indeed $a=\sigma^a$ implies that $a+\sigma^a = 2a$ and thus $a$ is in this polynomial ring. – quantum May 01 '20 at 18:38