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!

• This is not true. For example, let $A = k[x,z]$, $I = (xz)$ and $x = x$ and choose any $n> 1$. Then $B= k[y,z]$ and $IB = (y^n z)$ which is clearly not radical. – naf Oct 20 '11 at 15:38
• Ah. Yea. I can make a much smarter assumption here, see my edit. – Jesko Hüttenhain Oct 20 '11 at 16:02
• Actually, that already helped me. If we assume that no minimal prime over $I$ contains $x$, then no minimal prime over $IB'$ should contain $x$ either. – Jesko Hüttenhain Oct 20 '11 at 16:06

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...

• That is great. Thanks a lot! Since you never needed $A$ to be regular, that's even better. – Jesko Hüttenhain Oct 22 '11 at 6:14
• Yes $A$ can be any commutative ring (even not necessarily Noetherian, because $A\to B$ is of finite presentation). But the condition $n$ invertible in $A$ is important. Otherwise, if $A$ has positive characteristic $p$ dividing $n$, then $B$ would not be reduced when $x=z^p$ for some $z\in A$. – Qing Liu Oct 22 '11 at 20:44