Dimension of scheme of finite type over a field under base change (Hartshorne Ex. II.3.20)

Question

Consider an integral scheme $X$ of finite type over a field $k$. If $k\subseteq k'$ is a field extension, then the scheme $X' = X\otimes_k k'$ is not necessarily integral. For instance, take $X = \operatorname{Spec} \mathbb{R}[x,y]/(x^2+y^2)$ over $\mathbb R$ and $k' = \mathbb C$. Then $$X' = \operatorname{Spec}\mathbb C[x,y]/(x^2+y^2) = \operatorname{Spec} \mathbb C[x,y]/(x+iy) \cup \operatorname{Spec} \mathbb C[x,y]/(x-iy)$$ has two irreducible components.

Nonetheless, the dimension of each irreducible component of $X'$ is equal to the dimension of $X$. Why?

(This is Ex. II.3.20(f) in Hartshorne's book "Algebraic Geometry".)

Answer 1

1. If $A$ is a finitely generated $k-$algebra, then $\dim(A\otimes_k k')=\dim(A)$.

By Noether's normalization lemma, we may find $a_1,\dots,a_d\in A$, such that $k[x_1,\dots,x_d]\to A,x_i\mapsto a_i$ is injective and finite. So, $d=\dim A$. $k'$ is a flat $k-$module, therefore $k'[x_1,\dots,x_d]\to A\otimes_k k'$ is injective and finite. The first statement follows.

2. Let $\operatorname{Spec} A_i$ be an affine open cover of $X$, then $\operatorname{Spec}(A_i\otimes_kk')$ is an affine open cover of $X'$ and $\dim X'=\sup_i \dim(A_i \otimes_kk')=\sup_i \dim A_i=\dim X$.