# Properties of the exponent function attached to a nonzero prime ideal in a Dedekind domain

I want to prove properties of $v_\mathfrak{p}$, which I have been told is:

"the exponent function attached to a nonzero prime ideal $\mathfrak{p}$ that maps a given nonzero fractional ideal to the exponent to which $\mathfrak{p}$ appears in the factoriation of the ideal".

All I know about this function is that every nonzero fractional ideal $\mathfrak{b}$ of a Dedekind domain can be expressed of the form $$\mathfrak{b}=\prod_\limits\mathfrak{p}\mathfrak{p}^{v_\mathfrak{p}(\mathfrak{b})}$$where the product is over all nonzero prime ideals of the domain and the $v_\mathfrak{p}(\mathfrak{b})$ are integers of which finitely many are nonzero.

I want to show: $\mathfrak{b}\subseteq A \iff v_\mathfrak{p}(\mathfrak{b}) \geq 0 \quad\forall \mathfrak{p},\\ \mathfrak{a} \subseteq \mathfrak{b} \iff v_\mathfrak{p}(\mathfrak{a}) \geq v_\mathfrak{p}(\mathfrak{b}) \quad \forall \mathfrak{p},\\ v_\mathfrak{p}(\mathfrak{ab})=v_\mathfrak{p}(\mathfrak{a})+v_\mathfrak{p}(\mathfrak{b}),\\ v_\mathfrak{p}(\mathfrak{a+b})=\min\{v_\mathfrak{p}(\mathfrak{a}),v_\mathfrak{p}(\mathfrak{b})\},\\ v_\mathfrak{p}(\mathfrak{a \cap b})=\max\{v_\mathfrak{p}(\mathfrak{a}),v_\mathfrak{p}(\mathfrak{b})\}.\\$ (where A is the Dedekind domain)

I have no idea how to show this. First of all, I don't really understand what the function $v_\mathfrak{p}$ means, let alone prove anything with it. Since it is involved in the representation of a nonzero fractional ideal, I assume it has something to do with the way the ideals interact? Also since it is in the exponent of $\mathfrak{p}$, I assume it will do as powers do, thus implying the third property, but I am not sure how to show this.

Any guidance at all would be helpful!

EDIT: I think I can prove the second property from the first, but I have no idea how to prove the first. Also, the 4th and 5th properties seem counter-intuitive, since I would have thought the intersection of two ideals would be made up of the common parts of both, ie. the minimum of the two exponents. Why is it different?

• What do you know about the ideals of a Dedekind domain? In particular, why is $v_p(b)$ well defined? Also, do you know what is the relationship between inclusion and divisibility of ideals? – A.P. Mar 17 '15 at 18:27
• Do it for $\mathbb Z$ first. – lhf Mar 17 '15 at 18:28
• Hint: the first property follows directly from the second, which you can prove by localising at each prime, in turn. – A.P. Mar 17 '15 at 18:42
• Okay, so I think I have an idea for the second and therefore the first one. Since divides $\iff$ contains, $\mathfrak{a} \subseteq \mathfrak{b} \iff \mathfrak{ab}^{-1} \subseteq A$. However, I don't know how to show this happens when $v_\mathfrak{p}(\mathfrak{a})-v_\mathfrak{p}(\mathfrak{b}) \geq 0$. Once that is proved, the first follows since A is the empty product of nonzero prime ideals, ie. $v_\mathfrak{p}(A)=0 \quad \forall \mathfrak{p}$. Is this along the right lines? – AccioHogwarts Mar 17 '15 at 19:04
• I would have gone with a different route: how do inclusions play along after localisation? In particular, what happens of $\mathfrak{a} \subseteq \mathfrak{b}$ after you localise at a prime $\mathfrak{p}$? – A.P. Mar 17 '15 at 21:35

Recall that the injectivity of a map of $A$-modules is a local property. In particular, this means that for any two ideals $\mathfrak{a},\mathfrak{b}$ of $A$ we have $\mathfrak{a} \subseteq \mathfrak{b}$ if and only if $\mathfrak{a}_{\mathfrak{p}} \subseteq \mathfrak{b}_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$.
For example, this means that to prove the second property it is enough to show that for every prime ideal $\mathfrak{p}$ of $A$ we have $\mathfrak{a}_{\mathfrak{p}} \subseteq \mathfrak{b}_{\mathfrak{p}}$ if and only if $v_{\mathfrak{p}}(\mathfrak{a}) \leq v_{\mathfrak{p}}(\mathfrak{b})$. But this is clear, because if $$\mathfrak{a} = \mathfrak{p}_1^{v_{\mathfrak{p}_1}(\mathfrak{a})} \dotsm \mathfrak{p}_k^{v_{\mathfrak{p}_k}(\mathfrak{a})}$$ then, say $$\mathfrak{a}_{\mathfrak{p}_1} = \mathfrak{p}_1^{v_{\mathfrak{p}_1}(\mathfrak{a})}$$ (do you see why?) and we know that if $m,n \geq 0$, then $\mathfrak{p}^n \subseteq \mathfrak{p}^m$ if and only if $n \geq m$ (define $\mathfrak{p}^0 = (1)$).
Can you now prove the other properties, keeping in mind that every non-zero ideal in $A_\mathfrak{p}$ is a power of $\mathfrak{p}$?
As for the question in the edit: observe that $\mathfrak{a} \cap \mathfrak{b}$ is contained in both $\mathfrak{a}$ and $\mathfrak{b}$, so by the second property $v_\mathfrak{p}(\mathfrak{a} \cap \mathfrak{b})$ is greater or equal than $v_\mathfrak{p}(\mathfrak{a})$ and $v_\mathfrak{p}(\mathfrak{b})$ or, in other words, $v_\mathfrak{p}(\mathfrak{a} \cap \mathfrak{b}) \geq \max\{v_\mathfrak{p}(\mathfrak{a}),v_\mathfrak{p}(\mathfrak{b})\}$.