# Find primitive element such that conductor is relatively prime to an ideal (exercise from Neukirch)

This is an exercise from Neukirch, "Algebraic Number Theory", Ch I, Sec 8, Exercise 2, pg 52. It really has me stumped.

Suppose $A$ is a Dedekind domain, $K$ its field of fractions, $L$ a finite, separable extension of $K$, and $B$ the integral closure of $A$ in $L$ (so $B$ is also Dedekind).

Now suppose $J$ is any integral ideal of $B$. The exercise asks to show that there exists a $\theta \in B$ such that $\theta$ is a primitive element (i.e. $L=K(\theta)$) and such that $J$ is relatively prime to the conductor $\mathfrak f = \{\alpha \in B|\alpha B \subseteq A[\theta] \}$ of $A[\theta]$.

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It appears that the claim in the exercise is false. I think this is a counterexample:

Let $A=\mathbb Z_{(2)}$, i.e. $\mathbb Z$ localized at the prime $(2)$. Let $K=\mathbb Q$ and let $L=\mathbb Q(\zeta)$, where $\zeta^3-\zeta^2-2\zeta-8=0$. Let $B$ be the integral closure of $A$ in $L$, so $B=S^{-1}\mathcal O_L$, where $S=\mathbb Z \smallsetminus 2\mathbb Z$. Let $J=2B$.

$A$ is a DVR with $2A$ its only prime, which factors as $2B=P_1^{e_1}\cdots P_r^{e_r}$, and every prime of $B$ is one of the $P_i$. Thus, the conductor $\mathfrak f$ is relatively prime to $J=2B$ iff $\mathfrak f=B.$ So we must find a $\theta \in B$ such that $B=A[\theta]$. Without loss of generality, we can assume that $\theta \in \mathcal O_L$.

But this is impossible. Indeed, it can be shown that $2$ divides $[\mathcal O_L:\mathbb Z[\theta]]$ for every $\theta$ in $\mathcal O_L$ (as noted in Keith Conrad's notes "Factoring after Dedekind", pg. 5-6). Choose $b \in \mathcal O_L$ such that $b\mod \mathbb Z[\theta]$ is an element of even order in $\mathcal O_L/\mathbb Z[\theta]$. Then for any odd integer $k$, $kb\notin \mathbb Z[\theta]$, so $b \notin A[\theta]$. Thus $B \not= A[\theta]$.

If anyone finds a flaw in this counterexample, or thinks the claim in the exercise should be correct, please let me know! Thanks.

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Were you able to verify with others that the claim in this problem is false? I'm having a really hard time proving this claim (as well as the one that follows it in Neukirch's book). Thanks! –  user17360 Oct 9 '11 at 18:57
More generally, if $A$ is a DVR with prime $\pi$ and we let $J = \pi{B}$ then the only nonzero ideal in $B$ that's relatively prime to $J$ is $B$, so if $\theta$ exists then $B = A[\theta]$. Writing $h(x)$ for the minimal poly. of $\theta$ over $K$ (equiv., over $A$) we'd have $B \cong A[x]/(h(x))$. Set $n = [L:K]$. If $\pi$ splits completely in $B$ then (since $B = A[\theta]$) $h(x) \bmod \pi$ has $n$ distinct (monic) linear factors. If $A/\pi$ has size less than $n$ there aren't enough factors. So having $\#(A/\pi) < [L:K]$ and $\pi$ splitting completely gives a counterexample. –  KCd Oct 9 '11 at 19:50
@George: what is the next claim in Neukirch's book? I don't have it in front of me. –  KCd Oct 9 '11 at 19:51
I wonder if this question is the solution to the problem raised at mathoverflow.net/questions/48759/… –  KCd Oct 9 '11 at 19:59
@George: Concerning the following claim (Exer 3), Neukirch probably intended it to be proven using Exer 2. To prove it without using that Exercise (whose claim seems incorrect), you can see Lang's Algebraic Number Theory pg 40, although his proof uses completions, etc. which aren't developed until Ch 2 of Neukirch. –  John M Nov 15 '11 at 6:23

The claim is indeed false, as John's counterexample shows.

It is, however, true if $J$ is assumed to be a prime ideal of $B$ - at least in the number field case. A proof can be found in Hecke's "Vorlesungen über die Theorie der Algebraischen Zahlen" (1923), Hilfssatz c on page 135 in the second edition (Chelsea Publishing Co, 1970, available from the AMS).

The proof uses the finiteness of the residue class field, so it does not generalize directly to the case of general Dedekind domains.

Oh, just found it in Serre's "Local Fields", Ch. III, exercise 2 on p. 59. It holds in the general Dedekind domain case if the residue extension $A/J\cap A \subset B/J$ is separable.

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