Proving $\dim A = \dim A\otimes_k K$ by reducing to an irreducible component. Let $A$ be an equidimensional $k$-algebra with $\dim A = n$. Let $K/k$ be an algebraic extension. I just finished a proof that $A\otimes_k K$ is equidimensional with $\dim A\otimes_k K = n$ as well, but I am not comfortable with the first step I made, which was to reduce to the case $A$ is an integral domain.
I noted that since $A$ is equidimensional, we can mod out by any minimal prime $\mathfrak q$ and then we will have $\dim A = \dim A/\mathfrak q$, and $\dim A\otimes_k K=\dim A/\mathfrak q\otimes_k K$. The first equality is obvious, but the second I only made hand-wavingly so. How do I justify it more rigorously?
 A: Faster proof: By Noether normalization, $A$ is finite over some $k[T_1,\dotsc,T_n]$. Hence, $A \otimes_k K$ is finite over $K[T_1,\dotsc,T_n]$. qed.
A: I believe I solved this exercise without using this directed colimit argument to reduce to the finite type case. I'll leave it here in case anyone is interested. 
Let $p: X_K \to X$ be the projection. We know the map $\mathrm{Spec} K \to \mathrm{Spec} k$ is integral, surjective, and flat, hence the same for $p$. This immediately tells us that $\dim(X) = \dim(X_K)$. The hard part is showing pure dimension. Suppose $X = \mathrm{Spec} A$ is pure dimension $n$. Then any irreducible component is o.t.f. $\mathrm{Spec} (A/\mathfrak{p})$ with $\mathfrak{p}$ a minimal prime of $A$. If $E \subset X_K$ is an irreducible component, we know $p(E) \subseteq X$ is irreducible (Tag 0379), closed (exercise 7.3.M, because $p$ is integral), and contained in some irreducible component $F$ of $X$.
Let's look at Vakil's hint for part (a). He suggests we look at an arbitrary irreducible component of $X$, say $F = \mathrm{Spec}(A/\mathfrak{p})$ for $\mathfrak{p}$ a minimal prime. Then if $E \subset X_K$ is an irreducible component mapping into $F$ (all components are of this form for some $E$ so this is OK), it corresponds to a morphism of rings $A \to A' \to A'/\mathfrak{q}$ factoring through $A/\mathfrak{p}$, where $A' = A \otimes_k K$, $\mathfrak{q} \vartriangleleft A'$ is minimal, and $\mathfrak{p} \vartriangleleft A$ is minimal. If we just consider the map $A/\mathfrak{p} \to A'/\mathfrak{q}$, this is integral as quotients are integral and by exercise 7.2.B(a) in Vakil. One can show this map is injective and use exercise 11.1.E to conclude that $\dim(A'/\mathfrak{q}) = \dim(A/\mathfrak{p}) = n$, hence $X_K$ is pure dimension $n$. It might be interesting to note that injectivity of the above map is the same as saying $E$ surjects onto $F$ under $p$.
