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?
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, $v_\mathfrak{p}(\mathfrak{a} \cap \mathfrak{b}) \geq \max\{v_\mathfrak{p}(\mathfrak{a}),v_\mathfrak{p}(\mathfrak{b})\}$.