Commutativity of "extension" and "taking the radical" of ideals

Question

Let $K$ be a field (not necessarily algebraically closed) and $\overline{K}$ its algebraic closure. By $K[\text{X}]$, I mean $K[X_1,...,X_n]$.

Is it true that the operations of "extension" and "taking the radical" commute for:

$$\mathfrak{a}\subset{K}[\text{X}]\subset\overline{K}[X],$$

i.e., is it true that $\sqrt{\mathfrak{a}\overline{K}[X]}$=$\sqrt{\mathfrak{a}}{\overline{K}[X]}$?

You may also assume $K$ is a perfect field if necessary.

Thoughts: Did not make much progress at all. Not really sure what to do with the algebraic closure. The natural generalisation of this result to arbitrary ring extensions is false.

Background: The motivation for this is very basic and possibly pedantic and is due to wanting to say immediately that an (affine) algebraic set like $(X-Y)^2=1$ (i.e. zeros of $(X-Y)^2-1\in K[\text{X}]$ in $\mathbb{A}^n(\overline{K})$) is "defined over $K$". All definitions are as according to Chapter 1 of Silverman's book on Elliptic Curves wherein the author claims that such statements are clear (c.f. examples 1.3.1-1.3.3). The definition of "defined over $K$" is given in the second "Definition" of Chapter 1.

Answer 1

The question reduces to show that if $\mathfrak a$ is a radical ideal, then its extension is also radical. Equivalently, if $K[X]/\mathfrak a$ is reduced, then $K[X]/\mathfrak a\otimes_K\overline K$ is also reduced. If $K$ is a perfect field (and this is an assumption on page 1 of Silverman's book), then this holds as it is pointed out in this answer and proved here.