Showing $A[\theta] \subseteq B \subseteq \frac{1}{d} A[\theta]$, where $A$ is a Dedekind domain

Question

Suppose we have $A$ a Dedekind domain, $K$ its field of fractions. Let $L$ be a field extension of $K$ of degree $n$ and also that it is separable. Let $B$ be the integral closure of $A$ in $L$. Suppose $\theta \in B$ and $L = K(\theta)$.

How can we show that $A[\theta] \subseteq B \subseteq \frac{1}{d} A[\theta]$ for some $d$ in $A$? I can do this if $L/K$ is Galois but not sure otherwise. I would greatly appreciate any hint/answers! Thank you very much!!

score 1 · Accepted Answer · answered Feb 19 '16 at 22:11

1

If $b\in B$ then $b\in L$, so $b\in K[\theta]$ (since $\theta$ is algebraic over $K$). Then there is $a\in A$, $a\ne0$ such that $ab\in A[\theta]$. As you already found out, $B$ is a finitely generated $A$-module. For $b_1,\dots,b_n\in B$, a generating system, there are non-zero elements $a_1,\dots,a_n\in A$ such that $a_ib_i\in A[\theta]$. Set $d=a_1\cdots a_n$. Then $dB\subseteq A[\theta]$.

answered Feb 19 '16 at 22:11

user26857

53,190

I haven't quite figure out that $B$ is a finitely generated $A$-module yet. In fact, I thought I could use this fact to show that $B$ is a finitely generated $A$-module... – Johnny T. Feb 19 '16 at 22:24
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See here for the usual proof. – user26857 Feb 19 '16 at 22:32
I see. Thank you very much! – Johnny T. Feb 20 '16 at 18:31

Showing $A[\theta] \subseteq B \subseteq \frac{1}{d} A[\theta]$, where $A$ is a Dedekind domain

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