If $A$ is a commutative ring, $I \subset A$ an ideal and $f:A \rightarrow B$ a ring homomorphism, then the extension of $I$, $I^e = \langle f(a): a \in I \rangle$ does not commute with the radical, I mean, $\sqrt{I^e} \neq (\sqrt{I})^e$ in general.


I'm struggling with the following problem: I have two fields, $K$ and $L$, where $L/K$ is an extension of fields. I am considering the two Laurent polynomial rings: $A=K[x\_1^{\pm 1},\dots,x\_n^{\pm 1}]$ and $B=L[x\_1^{\pm 1},\dots,x\_n^{\pm 1}]$, and $f:A \rightarrow B$ is the trivial injection.


I am wondering if, given an ideal $I \subset A$, it is true or not that $\sqrt{I^e} =(\sqrt{I})^e$. I know that the inclusion $(\sqrt{I})^e \subset \sqrt{I^e}$ is general.


I have noted that, for this homomorphism $f$, it is true that $I^{ec} = I$ for each ideal $I \subset A$. Taking radicals, $\sqrt{I^{ec}} = \sqrt{I}$. As contraction of ideals commute with radical, $(\sqrt{I^e})^c = \sqrt{I}$. If I now take the extension, then $$(\sqrt{I^e})^{ce} = (\sqrt{I})^e.$$


On the other hand, $\sqrt{I} = (\sqrt{I})^{ec} = ((\sqrt{I})^e)^c$. So we have $\left( (\sqrt{I})^e \right)^c = (\sqrt{I^e})^c$. Therefore, the ideals $\sqrt{I^e}$ and $(\sqrt{I})^e$ have the same contraction. But I'm stucked here again.


I don't know how to continue. And I don't know even if the statement $\sqrt{I^e} = (\sqrt{I})^e$ is true or not in this case, but I couldn't find any counterexample.


Any help would be really appreciated, thanks in advance.


Pd: I'm new here, I hope this question is appropiate.



Extension of the radical and radical of the extension of an ideal
Let $K$ be a field of characteristic $p>0$ and let $a\in K$ an element without $p$th root. Let $L=K(b)$, where $b$ is the $p$th root. Let $I$ be the ideal in $K[x^{\pm 1}]$ generated by $x^p-a$. Then $I$ is reduced, but its extension is not.


