Let $k$ be a field and $X$ an irreducible $k$-scheme of finite type. Then is $X_\overline{k}$ equidimensional?
I prove it as follow: In affine case, say $X = \operatorname{Spec}k[T_1 , \cdots , T_n]/ I $, the projection $ X_\overline{k} \to X$ corresponds to $ \phi : k[T_1 , \cdots , T_n]/ I \to \overline{k}[T_1 , \cdots , T_n]/ I $, an injective integral morphism. So for every prime ideal $\mathfrak{p}$ of $\overline{k}[T_1 , \cdots , T_n]/ I$, the dimension of $\operatorname{V}(\mathfrak{p})$ and of $\operatorname{V}(\phi ^{-1}(\mathfrak{p} ))$ are same. So, since irreducible components correspond to minimal prime ideals and for a minimal prime ideal $\mathfrak{p}$ its contraction $\phi^{-1}(\mathfrak{p})$ is also minimal, every irreducible component of $X_\overline{k}$ is of dimension $= \operatorname{dim}X$.
Is this correct?
If it's correct, I have some problems. First, in this proof I use the only property that $X$ is equidimension.
Next, this (exercise 3.2.10 of Liu's book Algebraic Geometry and Arithmetic Curves) is stated as following:
Let $k$ be a field and $K/k$ a finite Galois extension with group $G$. Let $X$ be an algebraic varity over $k$.
(a) Let $L$ be an extension of $k$. Show that $G$ acts transitively on $\operatorname{Spec}L \otimes_k K$.
(b) Let us suppose $X$ is irreducible. Show that $G$ acts transitively on the irreducible components of $X_K$. Deduce from this that the irreducible components of $X_\overline{k}$ have the same dimension.
(c) Let us suppose $X$ is connected. Show that $G$ acts transitively on the connected components of $X_K$.
(An algebraic variety over $k$ is defined by a $k$-scheme of finite type in this book.)
But in my proof, I do not use neither (a) nor the first part of (b).
So I have no confidence.