$\divideontimes$ Refer to the Janusz, Algebraic number fields, Chap II- Example 2 or Nuekirch, Algebraic number theory, p.69.
Let $\mathcal{o}$ be a Dedekind domain wiht fraction field $K$. Let $\mathfrak{p} \neq 0$ be a nonzero prime ideal in $\mathcal{o}$. For any nonzero $x$ in $\mathcal{o}$ let $v_{\mathfrak{p}}(x)$ denote the power to which $\mathfrak{p}$ appears in the factorization of $(x)$. We may then wrie
$$ (x) = \prod \mathfrak{p}^{v_{\mathfrak{p}(x)}}, \ \mathfrak{p} \text{ runs through the primes. }$$
Then I don't know how can we define corresponding valuation $v_{\mathfrak{p}} : K^{*}\to \mathbb{Z}$ clearly. How can we lift the function $v_{\mathfrak{p}}$ on $\mathcal{o}$ to $K^{*}$?
And assoicated question : Let $L|K$ be a ( finite ) Galois extension of number fields. $\mathfrak{P}_1, \mathfrak{P}_j \subseteq \mathcal{O}_{L}$ be prime ideals of $L$ such that $\sigma \mathfrak{P}_1 = \mathfrak{P}_j$ for some $\sigma \in G(L|K)$. Then the corresponding valuations satisfy $v_{\mathfrak{P}_j} \circ \sigma = v_{\mathfrak{P}_1}$ ?
It is sufficient to show the relation only on $\mathcal{O}_L$? Assume this. Fix $x\in \mathcal{O}_L$. Then
$$ \mathfrak{P}_j^{v_{\mathfrak{P}_j}(\sigma(x))} \prod_{\mathfrak{P}_j \neq \mathfrak{P} } \mathfrak{P}^{v_{\mathfrak{P}}(\sigma(x))} = \prod_{\mathfrak{P}}\mathfrak{P}^{v_{\mathfrak{P}}(\sigma(x))} = ( \sigma(x) ) \stackrel{?}{=} \sigma((x)) = \sigma(\prod_{\mathfrak{P}}\mathfrak{P}^{v_{\mathfrak{P}}(x)}) = \sigma ( \mathfrak{P}_1^{v_{\mathfrak{P}_1}(x)} \prod_{\mathfrak{P}_1 \neq \mathfrak{P}}\mathfrak{P}^{v_{\mathfrak{P}}(x)}) \stackrel{?}{=} \sigma(\mathfrak{P}_1)^{v_{\mathfrak{P}_1}(x)} \prod_{\mathfrak{P}_1 \neq \mathfrak{P}}(\sigma(\mathfrak{P}))^{v_{\mathfrak{P}}(x)} = \mathfrak{P}_j^{v_{\mathfrak{P}_1}(x)}\prod_{\mathfrak{P}_1 \neq \mathfrak{P}}(\sigma(\mathfrak{P}))^{v_{\mathfrak{P}}(x)} . \tag{1} $$
Note that $\{ \mathfrak{P} \in \operatorname{Spec}(\mathcal{O}_L) : \mathfrak{P}_j \neq \mathfrak{P} \} = \{ \sigma(\mathfrak{P}) : \mathfrak{P} \in \operatorname{Spec}(\mathcal{O}_L), \mathfrak{P}_1 \neq \mathfrak{P} \}$ ( Check ). So for far left and far right in $(1)$, in $\prod_{\mathfrak{P}_j \neq \mathfrak{P} } \mathfrak{P}^{v_{\mathfrak{P}}(\sigma(x))}$ and $\prod_{\mathfrak{P}_1 \neq \mathfrak{P}}(\sigma(\mathfrak{P}))^{v_{\mathfrak{P}}(x)}$, there is no $\mathfrak{P}_j$-term. So by comparing exponentials, $v_{\mathfrak{P}_j}(\sigma(x)) = v_{\mathfrak{P}_1}(x)$. And it remains to prove our claim for arbitrary $x\in K^{*}$. For this, proof of the case $x\in \mathcal{O}_L$ is sufficient? Can anyone help?
