Why does a complex linear normalization of a real algebraic surface inherit a real structure?

Question

Could you recommend any references to (some of) the following very basic assertions in algebraic geometry? (It seems unreasonable to reprove them in a research paper.)

(1) Let a surface $X$ in $\mathbb{P}^n$ be the solution set of a system of polynomial equations with real coefficients. Assume that $X$ has a smooth linear normalization $\bar X$ in $\mathbb{P}^N$. Then the complex conjugation on $\mathbb{P}^n$ lifts to an antiholomorphic involution on the linear normalization $\bar X$ (i.e., a normalization map $\bar X\to X$ commutes with the involutions on $\bar X$ and $X$).

EDIT3. The following examples show that this is not completely trivial:

Ex1. Even if $\bar X$ is a solution set of a system of polynomial equations with real coefficients as well, a complex normalization map $\bar X \to X$ may not be given by real polynomials, and complex conjugation in $\mathbb{P}^N$ may not be the required antiholomorphic involution on $\bar X$. E.g., the cone $\bar X=\{x^2-y^2-z^2=0\}$ in $\mathbb{P}^3$ is a normalization of the cone $X=\{x^2+y^2-z^2=0\}$ in $\mathbb{P}^3$ but the map $\bar X\to X$, $(x,y,z)\mapsto (x,iy,z)$ is not given by polynomials with real coefficients. And the map is indeed a complex linear normalization map because the latter is only defined up to composition with a complex isomorphism.

Ex2. The lift of a fixed point may not be a fixed point. E.g., the curve $y^2=x^2(x-1)$ in $\mathbb{P}^2$ has linear normalization $(1:t:t^2:t^3)$ in $\mathbb{P}^3$. A normalization map is $x(t)=t^2+1, y(t)=t^3+t$ (or more accurately $(x_0:x_1:x_2:x_3)\mapsto (x_0:x_0+x_2:x_1+x_3)$). Then the real point $(x,y)=(0,0)$ of the curve is covered by two distinct complex conjugate points $(1:\pm i:-1:\mp i)$ of the linear normalization.

Assertion (1) might seem a tautology, but it has highly nontrivial consequences: e.g., (1)-(3) together with [1, Theorems 5-7 and Proposition 1] imply that a (nonruled) surface in $\mathbb{P}^n$ with a 2-dimensional set of real points parametrized by complex polynomials of degree $2$ has a parametrization by real polynomials of degree $2$ as well. And analogous assertion for higher degree polynomials does not remain true [1, Remark 4] although there is still a nice map $\mathbb{C}P^2\to X$, just not a linear mormalization anymore.

(2) (The closure of) the Veronese surface $(1:u:v:u^2:uv:v^2)$ in $\mathbb{P}^5$ is biregular to $\mathbb{P}^2$.

(3) The only antiholomorphic involution of $\mathbb{P}^2$ up to projective automorphism is the complex conjugation.

And also a question:

(4) Is (the closure of) the ruled surface $(1:u:v:u^2:uv)$ in $\mathbb{P}^4$ smooth? Which 'standard' surface is it isomorphic to? What are the antiholomorphic involutions on (a desingularization of) the surface?

Ideally, a reference to a particular published theorem, which can be just applied `as is' by a nonspecialist, is requested. Notice that (2)-(3) are mentioned in wikipedia and mathoverflow, but without a proof or a reference.

One should remark that there are some references where a version of (1) is stated even without the assumption that $\bar X$ is smooth but these versions cannot be correct just because an `antiholomorphic involution' is undefined for a nonsmooth surface (https://en.wikipedia.org/wiki/Complex_manifold). EDIT: In fact a generalization to locally ringed spaces is used there without mentioning that; thanks to Angelo for clarification.

[1] J. Schicho, The multiple conical surfaces // Contrib. Algeb. Geom. 42:1 (2001), 71-87.

Answer 1

The notion of anti-analytic involution is perfectly well defined for general analytic spaces: it is an involution of locally ringed spaces, that is antilinear with respect to complex scalars. Any projective variety $X \subseteq \mathbb P^n$ defined by equations with real coefficients has an antilinear involution, obtained by restricting the antilinear involution on $\mathbb P^n$.

If you use the language of schemes, very suitable for this type of questions, the point is the following. If $X$ is a projective scheme over $\mathbb R$, the base change $X_{\mathbb C} := \mathop{\rm Spec}{\mathbb C} \times_{\mathop{\rm Spec}{\mathbb R}}X$ has an involution $X_{\mathbb C} \simeq X_{\mathbb C}$, coming from conjugation. The normalization of a projective scheme over $\mathbb R$ is also a projective scheme over $\mathbb R$, and it stays normal after base changing to $\mathbb C$.

But you can avoid thinking about schemes defined over $\mathbb R$. Let $X \subseteq \mathbb P^n$ be a projective variety defined by equations with real coefficients, $\nu\colon \overline X \to X$ the normalization map, $\tau\colon X \simeq X$ the anti-analytic involution. The normalization has a universal property: if $f\colon Y \to X$ is an analytic, or anti-analytic, map from a (reduced) variety $Y$, which does not map any irreducible component of $Y$ into the non-normal locus of $X$, then $f$ lifts uniquely to an analytic, or anti-analytic, map $Y \to \overline X$. This allows you to lift the involution $\tau$ to an anti-analytic involution of $\overline X$.

[Added later]: maybe the scheme-theoretic proof is not so bad, it boils down to a simple fact in commutative algebra. First of all, the existence of the anti-analytic involution is local in the Zariski topology: since the involution is unique, if it exists, local involutions can be glued together. So, we can assume that $X$ is an algebraic subvariety of $\mathbb C^n$, so it corresponds to a finitely generated $\mathbb R$-algebra $A$. Let $\overline A$ be the normalization of $A$; then my claim is that $\overline X$ corresponds to the algebra $A\otimes_{\mathbb R} \mathbb C$. More concretely, if $\overline A = \mathbb R[x_1, \dots, x_m]/(f_1, \dots, f_r)$ (the normalization of a finitely algebra over a field is also finitely generated), then $\overline A\otimes_{\mathbb R} \mathbb C = \mathbb C[x_1, \dots, x_m]/(f_1, \dots, f_r)$. This defines the anti-holomorphic involution: it comes from conjugation in $\mathbb C$.

To check the fact above, call $K$ the field of fractions of $A$; then the field of rational functions of $A\otimes_{\mathbb R} \mathbb C$ is $K\otimes_{\mathbb R} \mathbb C$. This boils down to the fact that an element $f + i g$ (here $f$ and $g$ are in $K$) is integral over $A \otimes_{\mathbb R} \mathbb C$, or, equivalently, over $A$, if and only if $f$ and $g$ are. One direction is standard: if $f$ and $g$ are integral, so is $f + ig$, because $i$ clearly is integral. In the other direction, if $f + ig$ is integral over $A$, so is its conjugate $f + ig$, and so $f = ((f+ig)+(f-ig))/2$ is integral, and analogously for $g$.

Answer 2

Because of the lack of references, the detailed elementary proofs of (2)-(3) and a version of (1) have been written down in arXiv:2002.01355.