0
$\begingroup$

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 $R_p$.

  • 0
    Do you know that $R_p$ is a discrete valuation and thus you have a group homomorphism $v_p:K-\{0\}\to\mathbb{Z}$ with $R_p=\{x\in K-\{0\}|v_p(x)\geq 0\}\cup\{0\}$? – 2017-01-13
  • 0
    Thank you for the comment. Yes I know, but I do not know how the homomorphism $v_p$ is defined in this particular case – 2017-01-13
  • 0
    What is special about this particular case? – 2017-01-13
  • 0
    What I meant is that I do not know how the homomorphism is defined explicitly. It picks $a$ in $K^*$ and send it in $\mathbb Z$ in which way? Or we don't care about it in order to answer my questions? Do you know how to answer them? – 2017-01-13
  • 2
    $R_p$ is a discrete valuation ring means, it is a local pid and let $\pi$ be a generator of the maximal ideal. Then any element in $a \in K^*=K-\{0\}$ can be uniquely written as $u\pi^n$ where $u\in R_p$ is a unit and $n\in\mathbb{Z}$. Thus you get a map $v_p(a)=n$ and this is independent of the choice of $\pi$. An element $a\in K^*$ is in $R_p$ if and only if $n\geq 0$. – 2017-01-14
  • 0
    ok got it thank you. However I do not know how to use it to prove that $v_p(M)$ exists. Do you have the answer? – 2017-01-14
  • 0
    The problem is that every element $x\in M$ is also in $K$ so $v_p(x)$ could be negative and so it does not seem clear to me that the minimum exists. – 2017-01-14
  • 1
    Use the fact that $M$ is finitely generated and thus $\pi^nM\subset R_p$ for some $n$, so $v_p(x)\geq -n$ for all $x\in M$. – 2017-01-14
  • 0
    Got it Thank you very much! – 2017-01-15

1 Answers 1

1

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 $M$.

  • 0
    I understand your arguments but it seems to me that what you computed is different from what I wanted. For example, you even did not mention the set $\{v_p(x)|x\in M\}$. – 2017-01-14
  • 0
    I did implicitly: $v_p(M)=v_p(rM)-v_p(r)$ – 2017-01-14
  • 0
    Actually you should prove that $min\{v_p(x)|x\in M\}=v_p(rM)-v_p(r)$. This is not obvious in my opinion. You described a procedure to compute the RHS but you did not prove the above equality – 2017-01-14
  • 0
    yes, because I believe this is not hard so I left the proof for you :-) – 2017-01-14
  • 0
    Let $x\in M$ be such that $v_p(x)$ is the minimum on the left. Let $y\in M$ such that $v_p(ry)=v_p(rM)$. Then: 1)$v_p(x)=v_p(rx/r)=v_p(rx)-v_p(r)\ge v_p(rM)-v_p(r)$ 2) $v_p(rM)=v_p(ry)=v_p(r)+v_p(y)\ge v_p(r)+v_p(x)$. From this we have $v_p(ry)-v_p(r)\ge v_p(x)$ Is it correct in your opinion? – 2017-01-15
  • 0
    Not exactly - you can't assume $v_p(M)$ already exists. I would slightly rephrase it as follows. Let $n=v_p(rM)$ (which we know to exist by 1). Then for any $y\in M$ we have $v_p(y)=v_p(ry)-v_p(r)\ge n- v_p(r)$, thus $v_p(M)$ exists. – 2017-01-16
  • 0
    Actually I assumed that $v_p(M)$ exists by the comments of Mohan above. Considering that, the other my steps should work, right? – 2017-01-16
  • 0
    Once you establish that the minimum already exists you're basically done. There's no point in your reasoning above once you've gone through the steps of Mohan. (Note btw that your last inequality should be $\le$, apart from that I see no trouble) – 2017-01-16
  • 0
    I can't see why it should be $\le$. I simply subtracted $v_p(r)$ from both sides. – 2017-01-17
  • 0
    Sorry I misread, it's ok – 2017-01-17