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Let $A \subseteq B$ be a finite extension of Dedekind domains such that the extension $K \subseteq L$ of their quotient fields is separable. Let $\mathfrak{p}$ be a maximal ideal of $A$ and let $\mathfrak{q}_1, \dots, \mathfrak{q}_r$ be the primes of $B$ that lie over $\mathfrak{p}$. Let $e_i$ be the ramification index of $\mathfrak{q}_i$ over $\mathfrak{p}$.

Suppose $e_1 \geq 2$ and consider the ideal $I = \mathfrak{q}_1 \mathfrak{q}_2^{e_2} \cdots \mathfrak{q}_r^{e_r}$. I'd like to have a proof of the following containment: $$ \mathfrak{p} \supseteq \mathrm{Tr}_{L/K} (I), $$ where $\mathrm{Tr}_{L/K} \colon L \to K$ is the trace of the field extension $L \supseteq K$. This fact is Exercise 37-(a) at page 95 of Marcus' Number fields.

Thanks to all!

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up vote 5 down vote accepted

If $a\in I$ (or if $a\in\mathfrak{q}_1\mathfrak{q}_2\dots\mathfrak{q}_r$), consider the map $A:B/\mathfrak{p}\to B/\mathfrak{p}$ given by the multiplication by $a$. The map $A$ is nilpotent (as $a^k\in\mathfrak{p}$ for large $k$), so its trace (when we see $A$ as an endomorphism of the vector space $B/\mathfrak{p}$ over $A/\mathfrak{p}$) is $0$. This trace is the reduction mod $\mathfrak{p}$ of $Tr_{L/K}a$.

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Thanks! I deleted my previous comment because I understood your answer. The trace of $a$ over $A/\mathfrak{p}$ is the reduction mod $\mathfrak{p}$ of the trace over $K$ because we can suppose that $B$ is free over $A$: in fact it suffices to replace $A$ with $A_\mathfrak{p}$ and $B$ with $B_\mathfrak{p}$. – Andrea Nov 17 '11 at 23:06

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