Let $A$ be a Dedekind domain and $\mathfrak{a},\mathfrak{b}$ be fractional ideals of $A$. Then we know that $\mathfrak{a}$ and $\mathfrak{b}$ can be decomposed into $\mathfrak{a}=\prod\limits_{\mathfrak{p}}\mathfrak{p}^{v_{\mathfrak{p}}(\mathfrak{a})}$ and $\mathfrak{b}=\prod\limits_{\mathfrak{p}}\mathfrak{p}^{v_{\mathfrak{p}}(\mathfrak{b})}$, where $\mathfrak{p}$ are prime ideals of $A$ and the $v_{\mathfrak{p}}(\mathfrak{a}),v_{\mathfrak{p}}(\mathfrak{b})$ are integers, all but finitely many of which are zero.

Prove that 1) $v_{\mathfrak{p}}(\mathfrak{a}\mathfrak{b})=v_{\mathfrak{p}}(\mathfrak{a})+v_{\mathfrak{p}}(\mathfrak{b})$

2) $v_{\mathfrak{p}}(\mathfrak{a}+\mathfrak{b})=\min\lbrace v_{\mathfrak{p}}(\mathfrak{a}),v_{\mathfrak{p}}(\mathfrak{b}) \rbrace$

3) $v_{\mathfrak{p}}(\mathfrak{a}\cap\mathfrak{b})=\max\lbrace v_{\mathfrak{p}}(\mathfrak{a}),v_{\mathfrak{p}}(\mathfrak{b}) \rbrace$

I can see that this $p$-adic valuation for ideals is supposed to be a generalization of the similar object for integers and prime numbers. But due to multiplication of ideals being slightly more complicated, I haven't been able to prove the first one. I think the last two don't look too hard if I can become comfortable with this thing.

$\mathfrak{a}\mathfrak{b}=\left\lbrace \sum\limits_{i=1}^na_ib_i |a_i \in \mathfrak{a},b_i \in \mathfrak{b} \right\rbrace$. It's not clear to how I can use this to say something about the p-adic valuations of the product of these two ideals.

EDIT: my attempt at 2:

Without loss of generality, say $v_{\mathfrak{p}}(\mathfrak{a}) \leq v_{\mathfrak{p}}(\mathfrak{b})$ $(*)$

I have already proven that $\mathfrak{b} \subseteq \mathfrak{a} \iff v_{\mathfrak{p}}(\mathfrak{a}) \leq v_{\mathfrak{p}}(\mathfrak{b})$ for all $\mathfrak{p}$. Since the chosen $\mathfrak{p}$ in $(*)$ is arbitrary, I conclude that $\mathfrak{b} \subseteq \mathfrak{a}$. Hence $\mathfrak{a}+\mathfrak{b} \subseteq \mathfrak{a}+\mathfrak{a}=\mathfrak{a}$. And clearly $\mathfrak{a} \subseteq \mathfrak{a}+\mathfrak{b}$, so $\mathfrak{a}+\mathfrak{b}=\mathfrak{a}$ $(**)$

2 follows easily from $(**)$.

I am suspicious about this proof, particularly because $(**)$ seems stronger than what I was trying to show in the first place. I suspect I have misused $(*)$.

• Have you done this for $A=\Bbb Z$? – Lubin Mar 16 '15 at 2:42
• I have done this for the integers and regular old prime numbers, but it seems difficult to use exactly the same proofs, especially for 2 and 3. – Ducky Mar 17 '15 at 0:29

The unique factorization of ideals exactly mirrors how this works in $\Bbb Z$, that's the trick. I write $v$ for the valuation, the prime $\mathfrak{p}$ being understood. I also assume $\mathfrak{a}\mathfrak{b}\ne 0$ since that case is easy to see.

• For the first one, you write $\mathfrak{a}=\mathfrak{p}^n\mathfrak{m}, \mathfrak{b}=\mathfrak{p}^m\mathfrak{l}$ with $\mathfrak{m},\mathfrak{l}$ not divisible by $\mathfrak{p}$, so that the first one immediately follows by the uniqueness of the factorization.

• For the second the easiest approach is to note that Dedekind domains are characterized by the fact that all of their localizations at each prime are DVRs. Then by passing to the localization, $A_{\mathfrak{p}}$, we may assume $\mathfrak{a},\mathfrak{b}$ are principal ideals, with $\mathfrak{a}=(\pi^n), \mathfrak{b}=(\pi^m)$ and $\pi$ a local uniformizing parameter. Then assume, WLOG, that $n=\min\{n,m\}>m$ (in the case of equality the result is immediate) and note

$$v(\mathfrak{a}+\mathfrak{b})=v((\pi^n)+(\pi^m))=v(\{r\pi^n+s\pi^m: r,s\in A_{\mathfrak{p}}\})=n$$

The equality coming from the fact $\pi^n$ clearly divides all elements of that ideal, so $v\ge n$ and since $\pi^n$ is in it, $v\le n$.

• For the third, we again localize, and note since the valuation is discrete, we have that $k\le \ell\iff (\pi^k)\supseteq (\pi^\ell)$ with equality of numbers iff we have equality of ideals. But then the result immediately follows, since if $k\le\ell$ we have $(\pi^k)\cap(\pi^\ell)=(\pi^\ell)$.

Edit

The messier way that only involves the unique factorization characterization goes like this: first get a common denominator, i.e. write

$$\mathfrak{a}=\mathfrak{p}^n\prod_{i=1}^r\mathfrak{q}_i^{e_i},\, \mathfrak{b}=\mathfrak{p}^n\prod_{i=1}^r\mathfrak{q}_i^{f_i}$$

with some of the exponents being allowed to be zero, and as always we assume all the $\mathfrak{q}_i$ are distinct from one another and $\mathfrak{p}$. Then for peace of mind, clear out the denominators aside from $\mathfrak{p}$, that is to say multiply $\mathfrak{a}+\mathfrak{b}$ by

$$\prod_{i=1}^r\mathfrak{q}_i^{N_i}$$

with $N_i$ so that $N_i+e_i, N_i+f_i\ge 0$ so that we may assume $\mathfrak{a},\mathfrak{b}$ are integral ideals times a power of $\mathfrak{p}$. We note that we may further assume that $(N+e_i)(N+f_i)=0$, i.e either the exponent of $\mathfrak{q}_i$ in $\mathfrak{a}$ is $0$ or the corresponding exponent in $\mathfrak{b}$ is $0$. This modified version does not affect the valuation because the $\mathfrak{q}_i$ are all coprime to $\mathfrak{p}$. Then

$$\mathfrak{a}+\mathfrak{b}=\mathfrak{p}^n\left(\prod_{i=1}^r\mathfrak{q}_i^{N_i-e_i}+\mathfrak{p}^{m-n}\prod_{i=1}^r\mathfrak{q}_i^{N_i-f_i}\right).$$

By assumption the summands within the parentheses are co-prime integer ideals, hence the inside of the sum is merely $A$, so that $v(\mathfrak{a}+\mathfrak{b})=n$ as desired.

For the intersection you do similarly, using the Chinese Remainder theorem after clearing denominators and assuming $(N_i+e_i)(N_i+f_i)=0$, then you have

$$\mathfrak{a}\cap\mathfrak{b}=\mathfrak{p}^n\left(\prod_{i=1}^r\mathfrak{q}_i^{N_i-e_i}\cap\mathfrak{p}^{m-n}\prod_{i=1}^r\mathfrak{q}_i^{N_i-f_i}\right)=\mathfrak{p}^n\cdot\prod_{i=1}^r\mathfrak{q}_i^{2N_i+e_i+f_i}\cdot\mathfrak{p}^{m-n}=\mathfrak{p}^m\cdot\mathfrak{j}$$

with $\mathfrak{j}$ coprime to $\mathfrak{p}$ and this has valuation $m$ as desired.

• Thank you for the help. Unfortunately I'm not familiar with localizations - the only characterization I have seen for Dedekind domains is that each nonzero fractional ideal has a prime ideal factorization. I am editing my question now with a proposed solution, but I think there's an error there. – Ducky Mar 17 '15 at 0:30
• @Ducky I have written out the more cumbersome proof with that characterization. Hopefully that helps clear things up (and also shows off how much easier the other proof is for whenever you learn that machinery). – Adam Hughes Mar 17 '15 at 0:48
• Thank you for taking the time to give a different proof. Just a small nitpick - in the factorization of $\mathfrak{b}$ in the beginning, is it meant to be $\mathfrak{p}^m$ rather than $\mathfrak{p}^n$? – Ducky Mar 17 '15 at 0:57
• @Ducky Yes, I've fixed the typo. Do let me know if you have further questions, or if you think this is an acceptable and understandable answer! :-) – Adam Hughes Mar 17 '15 at 1:05
• "By assumption the summands within the parentheses are co-prime integer ideals, hence the inside of the sum is merely $A$". I am not sure that I understand how you reached this conclusion – Ducky Mar 17 '15 at 1:18