I am reading over the theorem of extensions of valuation as you can find it for example in Neukirch's book about number theory:
$ $Let $(K,v)$ be a (archimedian or non-archimedian) valued field, $f \in K[X]$ an irreducible polynomial and $L=K(\alpha)$ a simple field extension, where $f(\alpha)=0$. Assume that over $K_v[X]$, where $K_v$ denotes the completion of $K$ w.r.t. $v$, $f$ decomposes into $$f(X)=\prod_{i=1}^r f_i(X)^{m_i}$$ with $r$ distinct irreducible factors $f_i \in K_v[X]$. Then these factors $f_1, \ldots, f_r$ uniquely correspond to the extensions $w_1,\ldots, w_r$ of $v$ on $L$.Before that we have already seen the following:
$ $Any extension $w$ of the valuation $v$ on $L$ is of the form $w=\overline{v} \circ \tau$, where $\overline{K}_v$ denotes the algebraic closure of $K_v$, $\tau: L \hookrightarrow \overline{K}_v$ is a $K$-embedding of $L$ and $\overline{v}$ denotes the unique extension of $v$ on $\overline{K}_v$. If $w_1=\overline{v} \circ \tau_1$, $w_2=\overline{v}\circ \tau_2$ are two extension, then they agree if and only if $\tau_1$ and $\tau_2$ are conjugate, i.e. if there is some $\sigma \in \mathrm{Gal}(\overline{K}_v/K_v)$ with $\sigma(\tau_1) =\tau_2$.Now my original problem occurred when trying to compute an explicit example.
$ $Let $K=\mathbb{Q}$ equipped with the usual archimedean valuation $v=\lvert \cdot \rvert_{\infty}$ and let $f=X^3-5$ and $L=\mathbb{Q}(\sqrt[3]{5})$. Over $K_v=\mathbb{R}$ we have a decomposition of $f$ into $$f(X)=(X-\sqrt[3]{5})(X^2+\sqrt[3]{5}X+\sqrt[3]{25})$$ with two irreducible factors $f_1, f_2$ over $\mathbb{R}$. Thus, following the theorem above, we should have two extensions of $v$ on $L$. Moreover we know that these extensions must be of the form $w=\overline{v}\circ \tau$, where $\overline{v}$ is the extension of $v$ on $\overline{\mathbb{R}}=\mathbb{C}$ and $\tau$ are the $\mathbb{Q}$-embeddings of $L$. These are given by $$\tau_i: L \hookrightarrow \mathbb{C}, \qquad \sqrt[3]{5} \mapsto \begin{cases} \ \sqrt[3]{5}, & \textrm{ für }i=1, \\ \ \zeta \sqrt[3]{5}, & \textrm{ für }i=2, \\ \ \zeta^2 \sqrt[3]{5} & \textrm{ für }i=3. \end{cases}$$ with some primitive third root of unity $\zeta$. Further $\overline{v}$ is the usual complex valuation.Now the problem is:
$ $ $$w_i=\overline{v} \circ \tau_i$$ for $i=1,2,3$ agree, since on $\mathbb{Q}$ they are the identity and we have $$w_i(\sqrt[3]{5}) = (\overline{v} \circ \tau_i)(\sqrt[3]{5}) = \overline{v}(\zeta^{i-1} \sqrt[3]{5}) = \sqrt[3]{5}.$$So all in all for me there seems to be only one extension of $v$ and honestly I cannot think of a second one extending $v$. Where is my mistake?
Thank you!
Write $\newcommand{\al}{\alpha}\al=\sqrt[3]5$. A typical element of $L$ is $a+b\al+c\al^2$ where $a$, $b$, $c\in\Bbb Q$. The two Archimedean valuations (up to equivalence) on $L$ are $$v_1(a+b\al+c\al^2)=|a+b\sqrt[3]5+c\sqrt[3]{25}|$$ and $$\newcommand{\om}{\zeta} v_2(a+b\al+c\al^2)=|a+\om b\sqrt[3]5+\om^2 c\sqrt[3]{25}| =\sqrt{(a-b\sqrt[3]5/2-c\sqrt[3]{25}/2)^2+3(b\sqrt[3]5-c\sqrt[3]{25})^2/4}.$$ ($\om=\exp(2\pi i/3)$.) These are as inequivalent as they look.