I am trying to solve Ex 11.1.G from Ravi Vakil's FOAG. It says if $X$ is an affine scheme over $k$, a field and $K|_k$ is an algebraic field extension, then $X$ is of pure dimension $n$ iff $X_K:=X\times_k K$ is of pure dimension $n$.
Here's my attempt at a solution. Let $X=\operatorname{Spec}A$. Then $X_K=\operatorname{Spec}(A\otimes_k K)$. The canonical map $X_K \rightarrow X$, in this case corresponds to the homomorphism of $k$-algebras namely $A\rightarrow A\otimes_k K$; $a\mapsto a\otimes 1$
Since $K|_k$ is an integral extension, so is $A\otimes_k K|_A$. Thus the Going-Up theorem holds true.
Also since $K$ is a free $k$-module, $A\otimes_k K$ is a free $A$-module and faithfully flat. Since $A\hookrightarrow A\otimes_k K$ is a flat ring extension, the Going-Down theorem holds true.
Now say $P$ be a minimal prime ideal of $A\otimes_k K$. Then $p:=P \cap A$ is a minimal prime ideal of $A$ by Going-Down theorem. Conversely if $p$ is a minimal prime of $A$ and $P$ lies over $p$ (since it is an integral extension) then $P$ is a minimal prime ideal of $A\otimes _k K$ by Incomparability.
Assume $X$ is of pure dimension $n$. Then for any minimal prime $p$ of $A$, we have $\operatorname{coht}p=\dim X$. If $P$ is a minimal prime of $A\otimes _k K$, then $\operatorname{coht} P =\operatorname{coht} P\cap A=\dim X $ by Going-Up Theorem and Incomparability. Thus $X_K$ is of pure dimension $n$ as well.
Now assume $X_K$ is of pure dimension $n$. Then if $p$ is a minimal prime ideal of $A$, then $p=P\cap A$ where $P$ is a minimal prime of $A\otimes_k K$. Again by Going-Up and Incomparability, we get $\dim X_K=\operatorname{coht}P=\operatorname{coht} P\cap A $. Thus $X$ is of pure dimension $n$.
I hope this is a correct solution. I am pretty sure there exists alternative solution along the lines of Vakil's hints or preferably without using flatness since Vakil does not develop it upto this point but I simply can't get it. To be more precise, Vakil has first asked to reduce the problem to $A$ being an integral domain. I am stuck with this part without using the Going-Down theorem for flat ring extensions. Any help/insight/suggestion will be most welcome.
Does this work for the reduction? If $Q\subset A':=A\otimes_k K$ is a minimal prime, then $PA'\subset Q$ for some minimal prime $P\subset A.$ Thus it suffices to check that $A'/PA'$ is pure of dimension $n$. But $$A'/PA' = (A\otimes_k K)\otimes_A (A/P)= (A\otimes_A A/P)\otimes_k K = A/P\otimes_k K $$ So we've reduced to $A/P\subset A/P\otimes_k K$, as desired.
The only thing that seems slightly sketchy is the second equality in the centered line...
Edit: If you downvote, please let me know what's wrong with the argument, as I am interested in the correct reduction.