most recent 30 from http://mathoverflow.net 2013-05-21T18:00:12Z http://mathoverflow.net/feeds/question/52198 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal T.B. 2011-01-16T00:56:25Z 2011-01-16T01:22:00Z

<p>I am reading Local fields and see Serre using $v_{\mathfrak{p}}(\mathfrak{a})$ where $\mathfrak{a}$ is a fractional ideal of the Dedekind domain $A$ and $v_{\mathfrak{p}}$ is the valuation associated to the discrete valuation ring $A_{\mathfrak{p}}$. Serre did not really define this in the book, I looked it up on the web and found the following definition: since $\mathfrak{a}$ is a fractional ideal, there exists $d\in A$ such that $d\mathfrak{a}\subset A$. Define $v_{\mathfrak{p}}(\mathfrak{a})$ as $v_{\mathfrak{p}}(d\mathfrak{a}) - v_{\mathfrak{p}}(d)$ where $v_{\mathfrak{p}}(d\mathfrak{a})$ is defined to be the power $k$ that gives $d\mathfrak{a}A_{\mathfrak{p}} = (\mathfrak{p}A_{\mathfrak{p}})^{k}$. This work fine in proving the theorems following this idea in Serre's book. However, I could not prove that this is well-defined since $d$ is certainly not unique. Clarification is appreciated. Thanks.</p>

http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal/52199#52199 zeb 2011-01-16T01:10:30Z 2011-01-16T01:10:30Z

<p>Fractional ideals in Dedekind domains have unique factorization into (positive or negative) powers of prime ideals. To find $v_p(a)$, see what power of $p$ occurs in the factorization of $a$.</p> <p>Now check that the strange definition of valuation you got off the web is equivalent to this definition (and thus independent of $d$).</p>

http://mathoverflow.net/questions/52198/what-is-the-definition-of-the-valuation-of-a-fractional-ideal/52200#52200 Pete L. Clark 2011-01-16T01:17:31Z 2011-01-16T01:17:31Z

<p>Let $d'$ be any other nonzero element of $R$ such that $d' \mathfrak{a} \subset R$. Then</p> <p>$v_{\mathfrak{p}}(d' \mathfrak{a}) + v_{\mathfrak{p}}(d) = v_{\mathfrak{p}}(dd' \mathfrak{a}) = v_{\mathfrak{p}}(d') + v_{\mathfrak{p}}(d\mathfrak{a})$. </p> <p>So </p> <p>$v_{\mathfrak{p}}(d'\mathfrak{a}) - v_{\mathfrak{p}}(d') = v_{\mathfrak{p}}(d \mathfrak{a}) - v_{\mathfrak{p}}(d)$.</p> <p>(In general, Serre's <em>Local Fields</em> routinely leaves computations like this to the reader, so you should probably get practice working them out for yourself now, before the material becomes more difficult.)</p>