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Let $K/\mathbb Q$ be a finite algebraic extension. Consider a prime $p$ and $v$ be a valuation of $K$ above the valuation $v_p$ of $\mathbb Q_p$. Denote by $K_v$ a completion for the valuation $v$ of $K$ and $\mathbb D_v$ be a completion of an algebraic closure of $K_v$. Is there an injection $\imath$ of $\mathbb C_p$ in $\mathbb D_v$? If yes, if one denotes by $v_1$ the unique valuation of $\mathbb D_v$ that extends $v$ and such that $v_1(p)=v(p)$ and still denotes $v_p$ the unique valuation of $\mathbb C_p$ that extends $v_p$ and such that $v_p(p)=1$, can one assert that there exists a real number $e$ such that for all $x\in\mathbb C_p$, $v_1(\imath(x))=ev_p(x)$?

Thanks in advance.

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  • $\begingroup$ An algebraic closure of $K_v$ is isomorphic to $\bar{\mathbb{Q}_p}$ compatibly with the valuations because $K_v$ is a finite extension of $\mathbb{Q}_p$, so $D_v$ is isomorphic to $\mathbb{C}_p$. $\endgroup$
    – SashaP
    Commented Jul 8, 2019 at 12:16
  • $\begingroup$ What do you mean by "compatibly with the valuations"? Do you mean that $v_1(f(x))=v_p(x)$ for all $x\in\mathbb C_p$ where $f$ is an isomorphism of $\mathbb C_p$ into $\mathbb D_v$? $\endgroup$
    – joaopa
    Commented Jul 8, 2019 at 12:30
  • $\begingroup$ Your question is weird because ($\Bbb{Q}_p$ being complete discrete valuation field) $v$ extends in a unique way to finite extensions of $\Bbb{Q}_p$ thus to $\overline{\Bbb{Q}}_p$ and there is a unique continuous (continuity of $a \mapsto 2^{-v(a)}$) extension to $\Bbb{C}_p$. So everything is about the normalization, the valuation you chose for $p$ (or $\pi_v$). $K_v$ is a finite extension of $\Bbb{Q}_p$, if $K/\Bbb{Q}$ is Galois then $K_v$ depends only on $K,p$, if it is not Galois it may depend on the prime of $O_K$ (the elements with valuation $> 0$) above $p$ you chose. $\endgroup$
    – reuns
    Commented Jul 8, 2019 at 13:41

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