I want to apologize in advance if this is blatantly trivial, but I already posted on math.stackexchange.com and got no answer at all.

Let $A$ be a Noetherian domain containing an algebraically closed field $\Bbbk$. If you want, you can also assume that $A$ is local and regular. Let $I\subseteq A$ be a radical ideal and $x\in A$. (**edit**) Assume that no minimal prime over $I$ contains $x$. Let $n\in\mathbb{N}$ and $B:=A[y]$ for $y^n=x$. If it becomes necessary, we can require $n$ not to divide $\mathrm{char}(\Bbbk)$ or even $\mathrm{char}(\Bbbk)=0$.

**My question is**, is $IB$ radical?

My feeling says it is, and my attempt at proof looks like this: Let $B'=A[Y]$ with $Y$ an indeterminant, so $\pi: B' \twoheadrightarrow B'/(Y^n-x) = B$.

Claim.$IB'$ is radical.

Proof.Should be easy to see from the geometric intruition: If $X$ is reduced, then so is $X\times_\Bbbk\mathbb{A}^1_\Bbbk$. A more elementary proof would be the following: We first note that for any polynomial $f\in B'$, we have $f\in IB'$ if and only if all coefficients of $f$ are elements of $I$. To show that $f^m\in IB'$ implies $f\in IB'$, we perform induction on the degree of $f$: For $\deg(f)=0$, the statement is clear since $I$ was assumed to be radical. Then, we write $f=a+Yg$ with $a\in A$ the constant term of $f$ and $\deg(g)=\deg(f)-1$. Then, $$f^m = \sum_{k=0}^m \binom{m}{k} \cdot a^k \cdot (Yg)^{m-k}$$ since $f^m\in IB'$, we know $a^m\in IB'$, so $a\in IB'$ since we already handled degree $0$. From the above expression, it therefore also follows that $(Yg)^m\in IB'$ and hence, $g^m\in IB'$. By induction hypothesis, $g\in IB'$ and thus, $f=a+Yg\in IB'$.

Hence, the statement is true if we can prove the following claim:

Claim.If $P$ is a minimal prime over $IB'$, then $P+(Y^n-x)$ is also a prime ideal.

Then, the minimal primes over $IB$ will be the quotitens of these ideals, showing that $\sqrt{IB}$ is $\pi(IB'+(Y^n-x))=\pi(IB')=IB$. Trying to prove this, based on the fact that $x\notin I$, I am stuck.

Thanks a bunch in advance for any help!