I am trying to solve Robin Hartshorne's Exercise 2.4.7, and it reads:
Let $X$ be a separated shceme of finite type over $\mathbb{C}$, let $\sigma$ be a semilinear involution on $X$, which satisfies $\sigma^2=\mathrm{id}$ and the following diagram commutes $$ X \to X\\ \downarrow\qquad\downarrow\\ \mathbb{C}\to \mathbb{C} $$ where $\mathbb{C}\to \mathbb{C}$ is a complex conjugation. Suppose for any two point $x_1,x_2\in X$ there is an open affine subset containing both of them. Show that thre is a unique separated scheme $X_0$ of finite type over $\mathbb{R}$, such that $X_0\times_\mathbb{R} \mathbb{C}\cong X$, and such that this isomophism identifies the given involution of $X$ with the one on $X_0 \times_\mathbb{R} \mathbb{C}$.
So my strategy was this: First, define a topological space $X^\sigma=\{x\in X: \sigma(x)=x\}$ with the subspace topology. Give a scheme structure $\mathcal{O}_{X^\sigma}$ as following: $$ \mathcal{O}_{X^\sigma}(U\cap X^\sigma)=\mathcal{O}_X(U)^\sigma, $$ where $\mathcal{O}_X(U)^\sigma=\{a\in \mathcal{O}_X(U): \sigma_U (a)=a\}$. This is certainly a sheaf since $\sigma$ is a sheaf morphism, and moreover, this is a scheme of finite type over $\mathbb{R}$, since $$ X=\bigcup_{i=1}^n U_i $$ where $U_i = \mathrm{Spec} A_i$, $A_i$'s are finitely generated $\mathbb{C}$-algebra. Hence, $A_i = \mathbb{C}[x_1,\ldots,x_{n_i}]/I$, where $\sigma$ is given by the conjugation. Then we have $A_i^\sigma = \mathbb{R}[x_1,\ldots,x_{n_i}]/I\cap \mathbb{R}[x_1,\ldots,x_{n_i}]$. Thus, $$ X^\sigma = \bigcup_{i=1}^n U_i \cap X^\sigma = \bigcup_{i=1}^n \mathrm{Spec} A_i^\sigma $$ Hence it is covered by a finite affine cover, which is a finitely generated $\mathbb{R}$-algebra. Moreover, since each $A_i^\sigma$ is closed in $A_i$ (it is an integral extension), it is also separated.
I'm not sure this proof is right. I found out that I did not use the condition that for any $x_1, x_2$, there is an affine open subset containing two points. I think there must be a mistake. I guess that $A_i^\sigma$ is not always of the form $\mathbb{R}[x_1,\ldots,x_n]/I\cap \mathbb{R}[x_1,\ldots,x_n]$, since $\sigma$ might act on $x_i$'s with some permutation of order $2$ maybe? Can anyone give some hints to correct the proof?
Thanks in advance.