Let $A$ be a Noetherian domain containing an algebraically closed field $k$. Let $x_1,\ldots,x_r\in A$ be irreducible elements generating a radical ideal $I=(x_1,\ldots,x_r)$. Set $B:=A[y_1,\ldots,y_r]$ where $y_i^n = x_i$. I am wondering whether $\sqrt{I\cdot B} = (y_1,\ldots,y_r)$.

Edit: A minor thought. It would, of course, suffice to show that $J:=(y_1,\ldots,y_r)$ is radical: If $f^n \in IB$, then certainly $f^n\in J$, so $f\in\sqrt{J}$. If $J$ was radical, this would mean $\sqrt{IB}\subseteq J$ and "$\supseteq$" is obvious.

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As I already noted, it suffices to check that $(y_1,\ldots,y_r)=J$ is a radical ideal. Let

$\displaystyle f = \sum_{v=(v_1,\ldots,v_r)} \alpha_v \cdot y_1^{v_1} \cdots y_r^{v_r} \in A[y_1,\ldots,y_r]$,      ;       $\alpha_v\in A$

be any element with $f^n\in J$. Any term in the expression $f^n$ will fall into one of three categories:

1. It is an expression $b\cdot y_i$ with $b\in B$ for some $i$, so it is contained in $J$.
2. It is an expression $a\cdot x_i$ with $a\in A$ for some $i$, so it is contained in $I\subseteq J$.
3. It is the $n$-th power of the contant term of $f$.

Hence, we may replace $f$ by its constant term, i.e. $f\in A$ and $f^n\in J$. Thus, $f^n\in J\cap A = I$ and since $I$ is radical, $f\in I=J\cap A$.

Edit (Proof that $J\cap A=I$): Consider $A \hookrightarrow A[T_1,\ldots,T_r] \twoheadrightarrow B$. Clearly, the preimage of $J$ under the projection is $(T_1,\ldots,T_r)+(T_1^n-x_1,\ldots,T_r^n-x_n)$ and the constant term in any polynomial generated by these will be generated by the $x_i$, so its preimage under the inclusion will be $I$.

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Is it true that $J \cap A = I$? $B$ does not consist of independent variables $y_i$ with coefficients in $A$, since $y_i^n = x_i \in A$. – Ted Oct 17 '11 at 2:41
I added that in my answer. – Jesko Hüttenhain Oct 17 '11 at 9:30