TITLE: $A$ regular, $k'/k$ transcendental. How to prove that $A \otimes_k k'$ is regular?
QUESTION [8 upvotes]: Let $k$ be a field and $k'$ a purely transcendental extension of $k$. 
Let now $A$ be an integral finitely generated $k$-algebra.

How to prove that if $A$ is regular then $A \otimes_k k'$ is also
  regular?

Thank you!

REPLY [3 votes]: We use a result of Grothendieck which says:

If $k'$ and $K$ are extension fields of $k$ such that
  either $k'$ or $K$ is finitely generated over $k$ and if $k'$ is separable over $k$, then $k'\otimes_kK$ is regular.

and the Theorem 33.2(i) from Matsumura, CRT, which asserts:

If $A\to B$ is a faithfully flat and regular morphism of Noetherian rings, then $A$ is regular iff $B$ is regular.

Now notice that the morphism $k\to k'$ is regular (use the Grothendieck's result), then change the base via the finite type morphism $k\to A$ and get that $A\to k'\otimes_kA$ is regular. Apply the theorem from Matsumura and you are done.