Exponents in the factorization of fractional ideals - Dedekind Domains

249 Views Asked by At

I'm reading the public notes by Altman and Kleiman of Commutative Algebra.

I have a couple of doubts strictly connected to each other:

1) page 149. The theorem 24.10 regarding the factorization of ideals in Dedekind domain. Consider $A$ a non zero ideal and $p$ a prime ideal. They say: if $v_p$ denotes the valuation of $R_p$ then [..] and $v_p(A)=min\{v_p(a)|a\in A\}$. Firstly I do not understand what exactly the valuation of $R_p$ is. Secondly, by $v_p(a)$ I suppose they mean $v_p(a/1)$ right? I mean $a$ is an element in $R$ not in $R_p$. So in this way $a/1$ has nonnegative valuation and the minimum exists.

2) Very similarly, consider the theorem 25.14 on page 154. It is about factorization of fractional ideals. Given a fractional Ideal $M$, I do not understand why the minimum $v_p(M)=min\{v_p(x)|x\in M\}$ exists and what $v_p(x)$ means since $x\in M\subseteq Frac(R)$ and $v_p$ has as domain the fraction field of $R_p$.

1

There are 1 best solutions below

10
On BEST ANSWER

Your remarks for 1) are indeed correct: $v_p(A)$ will be nonnegative. (and yes, $a$ is a shortcut for $a/1$)

To find the minimum in 2) it suffices to 'clear out' the denominator of the fractional ideal by multiplying $M$ by a nonzero $r\in R$ where $r$ is chosen such that $rM\subset R$ (This $r$ exists by the usual definition for a fractional ideal): $rM$ is an ideal of $R$ and so by your argument for 1) you find that the valuation $v_p(rM)$ attains a nonnegative minimum. Now simply subtract the valuation of $r/1$ from this minimum to get the desired minimum for $M$.