For Galois extension of number fields, $\mathfrak{P}_j, \mathfrak{P}_k | \mathfrak{p}$ $\Rightarrow$ $L_{\mathfrak{P}_j} \cong L_{\mathfrak{P}_k}$?

Question

Let $L|K$ be a (finite) algebraic Galois extension of number fields. Let $\mathfrak{p} \subseteq\mathcal{O}_K$ be a prime ideal of its ring of integers. Let $\mathfrak{P}_j , \mathfrak{P}_k \subseteq \mathcal{O}_L$ be prime ideals lying above $\mathfrak{p}$. Note that they are conjugate; i.e., there exists $\sigma \in G(L|K)$ such that $\sigma \mathfrak{P}_j = \mathfrak{P}_k$ ($\because$ Neukirch, Algebraic number theory, I-(9.1) Proposition).

Then my question is, in this case, is there a canonical (?) isomorphism $\psi : L_{\mathfrak{P}_j} \to L_{\mathfrak{P}_k}$ fixing $K_{\mathfrak{p}}$ (completions)? I think it seems to work, but I can't seem to find any proper proof.

Answer 1

Let me try to clarify a bit: $\sigma$ already induces an isomorphism as you can see because if we take $\eta\in L$ we can say that: $\sigma : L_{{P_j}}\rightarrow L_{P_k} $ such that $\sigma(\eta)=\eta$ which therefore induces to $\sigma(P_j)=P_k$. Therefore, the map $\psi := \sigma|_{L_{P_j}} : L_{P_j}\rightarrow L_{P_k}$ with $K_p$ is fixed because $K_p$ is the identity map such that $\sigma$ is restricted to $K$ with $\sigma\in G(L|K)$. Ultimately, $\psi:L_{P_j}\rightarrow L_{P_k}$ is an isomorphism fixing to $K_p$ as $\sigma = \psi$.

Addressing the further progression:

So, if you consider $P_1=P_j$ we can consider the following completions of the evaluation rings: $$ \sigma : \widehat{\mathcal{O}(L,vp_1)} \rightarrow \widehat{\mathcal{O}(L,vp_j)} $$ Since the given isomorphism of $\sigma$ the previous expression can be further expanded as: $$ \sigma : \widehat{\mathcal{O}(L,\hat{v}p_1)} \rightarrow \widehat{\mathcal{O}(L,\hat{v}p_j)} $$ Thefore, the expression also fixes $$ \widehat{\mathcal{O}(K,v_p)} = \mathcal{O}(K_p,\hat{v}_p) $$

Now, by taking the fraction fields of the completed valuation rings we will have: $\text{Frac}(\sigma) : L_{P_1}\rightarrow L_{P_j}$ which will fix $K_p$ as $K_p = \text{Frac}(\mathcal{O}(K_p,\hat{v}_p))$ which will ultimately lead us to $v_{P_j}∘\sigma=v_{P_1}$

I hope this answered your question, let me know for any further analysis. Have a nice day!

Answer 2

For notational simplicity, assume $\sigma \mathfrak{P}_1 = \mathfrak{P}_j$. Note that $v_{\mathfrak{P}_j} \circ \sigma = v_{\mathfrak{P}_1}$ ( $ \because $ Definition of the corresponding valuation from nonzero prime ideals and associated question ).

Note that the map between valuation rings $ \sigma : \mathcal{O}_{(L,v_{\mathfrak{P}_1})}=\mathcal{O}_{(L, v_{\mathfrak{P}_j}\circ \sigma}) \to \mathcal{O}_{(L,v_{\mathfrak{P}_j)}}$ , $x \mapsto \sigma(x)$ is a well-defined isomorphism fixing the valuation ring $\mathcal{O}_{(K,v_{\mathfrak{p}})} \subseteq K $ ( $\because$ $\sigma \in G(L|K)$ ). Consider the promoted isomorphism between completions : $$ \hat{\sigma} : \mathcal{O}_{(L_{\mathfrak{P}_1}, \hat{v}_{\mathfrak{P}_1})} \stackrel{!}{=} \hat{\mathcal{O}}_{(L,v_{\mathfrak{P}_1})} \xrightarrow{\cong} \hat{\mathcal{O}}_{(L,v_{\mathfrak{P}_j})} \stackrel{!}{=} \mathcal{O}_{(L_{\mathfrak{P}_j}, \hat{v}_{\mathfrak{P}_j})}. $$

( C.f. The equalities are true by Valuation ring of completion of a field ). If this isomorphism also fixes $\hat{\mathcal{O}}_{(K,v_{\mathfrak{p}})} = \mathcal{O}_{(K_{\mathfrak{p}},\hat{v}_\mathfrak{p})}$, then by taking $\operatorname{Frac}$, we have an isomorphism $\operatorname{Frac} (\hat{\sigma}) : L_{\mathfrak{P}_1} \to L_{\mathfrak{P}_j}$, which fixes $K_{\mathfrak{p}} = \operatorname{Frac}(\mathcal{O}_{(K_{\mathfrak{p}}, \hat{v}_{\mathfrak{p}})})$.