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This question comes from Lorenzini's book "An Invitation to Arithmetic Geometry." It's question 26b on page 129. We're given that $A$ is a Dedekind domain with field of fractions $K$, and that $L=K(\sqrt[n]{a})$ for some nonzero element $a\in K$. Under the assumption that $(a) = I^n$ for some ideal $I$ of $A$, I want to show that $B/A$ is unramified except possibly at the primes dividing n. In part a of this question, it's assumed that $gcd(n, char(K))=1$, where I showed $B/A$ is unramified implies that $(a) = I^n$. I've carried over this assumption into part b. There is a hint for this question to work locally.

If $f(x) = x^n-a$ and $g(x)$ is the minimal polynomial of $\sqrt[n]{a}$ over $A$, then as $gcd(n, char(K))=1$ we find that $g(x)$ is separable, so $L/K$ is a separable extension and $B$ is indeed a Dedekind domain.

The first thing I want to show is that given any prime ideal $P$ of $A$ such that $n \notin P$, the residue field extension $B/M$ is separable over $A/P$ for the maximal ideals $M$ of $B$ dividing $P$.

Given any element $\bar{\alpha} \in B/M$, I then consider $\alpha \in B$ with $\alpha \equiv \bar{\alpha} \pmod{M}$. Because $[L:K] = deg(g(x))$, the degree of the minimal polynomial of $\alpha$ over $A$ should divide $deg(g(x))$. Must the degree of $g(x)$ divide n? If that is the case, I can say that, $\pmod{P}$, the derivative of the minimal polynomial of $\alpha$ is nonzero, from which I can conclude that the residue field extension is separable. I'm aware that if $f(x)$ is reducible, then for some prime dividing n we have $a \in K^p$ or $a \in -4K^4$ if $4|n$ - would that be relevant here?

I also must show that $PB_M = MB_M$ for all such maximal ideals $M$ dividing $P$. I can't see where to use that $(a) = I^n$. I had thought to consider that should a prime ramify in $B$ that does not contain n that this would contradict $(a) = I^n$, but this approach did not prove successful.

Could someone point me in the right direction? I've been stuck on this question for the past day, and I'm very unsure on how to proceed!

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The thing that I immediately thought of would be to show that if $\mathfrak{p}\nmid n$, then $\mathfrak{p}$ is coprime to the conductor of $A[\sqrt[n]{a}]$. Then, show that for all primes $\mathfrak{p}\nmid n$ you have that $x^n-a$ factors $\mod\mathfrak{p}$ with the $e_i$'s all $1$, and separable. – Alex Youcis Sep 29 '13 at 22:06
I'm not really familiar with conductors. Is there another way to view this? – Garnet Sep 29 '13 at 23:20
Sorry, conductors are a new concept to me. My understanding of them comes entirely from this discussion:… If $P$ is coprime to the conductor, then when we localize the conductor at $P$, we just get $B_P$, right? Would this imply that $B/A[\sqrt[n]{a}] = 0$? That would let me apply all the theory I know with $A[\sqrt[n]{a}]$, which would certainly be nice. – Garnet Sep 29 '13 at 23:51

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