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I'm trying to understand Serre's proof (in Local Fields) of the fact that a fractional ideal $\mathfrak{a}$ of a Dedekind domain factors uniquely as $\mathfrak{a} = \prod_{\mathfrak{p}}\mathfrak{p}^{v_{p}(\mathfrak{a})}$. The approach he took was that since $v_{\mathfrak{p}}(\mathfrak{a}) = 0$ for all but finitely many $\mathfrak{p}$, say, $v_{\mathfrak{p_{j}}}(\mathfrak{a}) = e_{j} \neq 0$ for $j = 1,\ldots, k$. Let $\mathfrak{b} = \prod_{j=1}^{k}\mathfrak{p_{j}}^{e_{j}}$. Then show $\mathfrak{a} = \mathfrak{b}$ locally, i.e, $\mathfrak{a}A_{\mathfrak{p}} = \mathfrak{b}A_{\mathfrak{p}}$ for all prime $\mathfrak{p}$; hence, they must be the same. I have been able to fill in most details except for the fact that $\mathfrak{a} = \mathfrak{b}$ locally. If they are integral ideal of $A$, I can prove it, but since they are fractional with possibly negative powers, I could not figure out how to prove that claim. Note that Serre seems to use the following definition for $v_{\mathfrak{p}}(\mathfrak{a})$ for a fractional ideal $\mathfrak{a}$:http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal and I have trouble applying that definition here. Any help is greatly appreciated.

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Dear George: Thank you very much for the answer –  Tin Bui Jan 19 '11 at 17:39
    
for whatever reason you had two different accounts, which is why you were not able to comment on this question earlier. (If you log in with the same credentials every time, then you can comment on any posts you made and any answer to your questions.) I have merged your two accounts into one. –  Willie Wong Jan 19 '11 at 17:42

1 Answer 1

The local ring $A_\mathfrak{p}$ is a DVR and in particular a principal ideal domain. So the fractional ideal $A_{\mathfrak{p}}\mathfrak{a}$ is a free $A_{\mathfrak{p}}$-module (since finitely generated and torsion free) of rank 1, say with basis element $x$ . Any other basis element $x' \in A_{\mathfrak{p}}\mathfrak{a}$ must have the form $x' = ux$ for a unit $u \in A_{\mathfrak{p}}$.

Now, the valuation $v_{\mathfrak{p}}$ is a function $K^\times \to \mathbf{Z}$ where $K$ is the field of fractions of $A$ and also of $A_{\mathfrak{p}}$. So $v_{\mathfrak{p}}(\mathfrak{a}) := v_{\mathfrak{p}}(x)$ is well-defined.

And I believe that you will find that this definition of $v_{\mathfrak{p}}(\mathfrak{a})$ should make clear what is required for your question.

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