Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Let $d'$ be any other nonzero element of $R$ such that $d' \mathfrak{a} \subset R$. Then

$v_{\mathfrak{p}}(d' \mathfrak{a}) + v_{\mathfrak{p}}(d) = v_{\mathfrak{p}}(dd' \mathfrak{a}) = v_{\mathfrak{p}}(d') + v_{\mathfrak{p}}(d\mathfrak{a})$.

So

$v_{\mathfrak{p}}(d'\mathfrak{a}) - v_{\mathfrak{p}}(d') = v_{\mathfrak{p}}(d \mathfrak{a}) - v_{\mathfrak{p}}(d)$.

(In general, Serre's Local Fields 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.)

share|improve this answer
add comment

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 $a$.

Now check that the strange definition of valuation you got off the web is equivalent to this definition (and thus independent of $d$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.