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!



Extension of radical ideal after adjunction of roots
The answer is yes if $n$ is invertible in $A$. That $A$ contains a field doesn't matter in my proof. 


What you want is $B/IB$ is reduced. We have $B/IB=(A/I)[T]/(T^n-\bar{x})$. So after replacing $A$ by $A/I$, we can suppose $I=0$, $A$ is reduced and $x$ doesn't belong to any minimal prime ideal of $A$, and we have to show $B:=A[T]/(T^n-x)$ is reduced. This is equivalent to $X:=\mathrm{Spec}(B)$ is reduced at generic points and has no embedded points (properties R$\_0$ and S$\_1$). 


Denote by $X=\mathrm{Spec}(B)$ and by $f : X\to Y$ the morphism corresponding to $A\to B$. It is flat because it is free. 


Note that $f$ is étale above generic points of $Y$ (here we use the fact that $n$ is invertible in $A$ and the element $x$ is non zero at the generic points of $Y$). Let $\xi$ be a generic point of $X$. As $f$ is flat, using going-down properties (Matsumura, Commutative Algebra, 5.A and 5.D), we see that $f$ takes the generic points of $X$ to generic points of $Y$. So $X$ satisfies R$\_0$. 


It remains to show the property $S\_1$. Suppose $\dim O\_{X,p}\ge 1$. Then $\dim O\_{Y,q}\ge \dim O\_{X,p}\ge 1$ because $A\to B$ is finite. There exists a non-invertible regular element in $O\_{Y,q}$ ($Y$ is reduced). As $f$ is flat, the image of this element is still regular in $O\_{X,p}$, so the latter has positive depth and $X$ satisfies S$\_1$. 


**Update** It is not necessary to use the going-down property to compare the dimensions. As $f$ is flat and finite, we have 
$$ \dim O\_{X, p}=\dim O\_{Y, q}+\dim O\_{X\_q, p}=\dim O\_{Y,q}.$$


There are probably more diret proofs... 


