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?

  • $\begingroup$ 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? $\endgroup$ – A.P. Mar 17 '15 at 18:27
  • 1
    $\begingroup$ Do it for $\mathbb Z$ first. $\endgroup$ – lhf Mar 17 '15 at 18:28
  • $\begingroup$ Hint: the first property follows directly from the second, which you can prove by localising at each prime, in turn. $\endgroup$ – A.P. Mar 17 '15 at 18:42
  • $\begingroup$ 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? $\endgroup$ – AccioHogwarts Mar 17 '15 at 19:04
  • $\begingroup$ 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 $\mathfrak{p}$? $\endgroup$ – 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, $v_\mathfrak{p}(\mathfrak{a} \cap \mathfrak{b}) \geq \max\{v_\mathfrak{p}(\mathfrak{a}),v_\mathfrak{p}(\mathfrak{b})\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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