Blow-up and base change Consider a complex smooth (projective) surface $X$ and a blow-up $\epsilon:S\longrightarrow X$ at a point $x\in X$. Let $\sigma\in\text{Aut}(\mathbb C)$ be a field automorphism and moreover let 
$$\varphi:=\epsilon\times_{\text{Spec}\mathbb C}\text{id}_{\text{Spec}\mathbb C}:S\times_{\text{Spec}\mathbb C}\text{Spec}\mathbb C\longrightarrow X\times_{\text{Spec}\mathbb C}\text{Spec}\mathbb C$$
be the morphism induced by the base change through the automorphism $\sigma$. Is the morphism $\varphi$ again a blow-up? I think that the answer is yes since the base change induces many isomorphisms of schemes. What do you think? 
Many thanks in advance.
 A: The following result is the Universal Property of Blowing Up as stated in Hartshorne Algebraic Geometry Proposition II.7.14. 

Let $X$ be a noetherian scheme. Let $I$ be a coherent sheaf of ideals,
   and $\pi:S \rightarrow X$ be the blowing up with respect to $I$. If
   $\bar f:\bar Y \rightarrow X$ is any morphism such that $(\bar
 f^{-1}I)O_\bar Y$ is an invertible sheaf of ideals on $\bar Y$, then there
   exists a unique morphism $\bar Y \rightarrow S$  factoring $\bar f$.

Now suppose $f:Y \rightarrow X$ is an isomorphism with inverse map $g$. Let $I$ and $S$ be as in the proposition.  
In your case, $Y=X \times_{Spec\mathbb{C}} Spec{\mathbb{C}}$ and $f$ is the isomorphism $Y$ to $X$ induced by the field automorphism. $I$ is the ideal sheaf of the point $x$.
Let $J$ be the coherent sheaf of ideals $(f^{-1}I)O_Y$ on $Y$, let $\rho:\bar Y \rightarrow Y$ be the blowing up along $J$, and let $\bar f = f \circ \rho $. Then $(\bar f^{-1}I)O_{\bar Y} = (\rho^{-1}J)O_{\bar Y}$ and the right hand side is an invertible sheaf of ideals on $\bar Y$ because of Hartshorne II.7.13 (in plain language, blowing up the $J$ closed subscheme on $Y$ gives you a divisor on $\bar Y$). Thus, by the quoted result, we get a morphism $h:\bar Y \rightarrow S$ which makes a commutative square with corners $\bar Y$, $S$, $Y$, and $X$. This $h$ is an isomorphism because you can run the same argument with $g = f^{-1}$ in place of $f$. 
The only question now is why is $\bar{Y}$ isomorphic to the base change $Y \times_X S$? Because of the commutative square, we have a factorization of $h$ into $\bar Y \rightarrow Y \times_X S \rightarrow S$. The first map in the factorization must be an isomorphism, because the composite $h$ is an isomorphism and the second map is also an isomorphism (isomorphisms are preserved by base change).
In sum, this proves the more general proposition:
Prop: If $S$ is the blow-up of $X$ along $I$, and $Y \rightarrow X$ is an isomorphism then $Y \times_X S$ over $Y$ is canonically the blow up of the inverse image of $I$ on $Y$.
And finally, to bring it back to your original question, the answer is yes. $\varphi$ is the blowup of $X \times_{Spec{\mathbb{C}}} Spec\mathbb{C}$ at the unique preimage $x'$ of $x$.
A: More generally, blowup commutes with flat base change: https://stacks.math.columbia.edu/tag/085S
