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Let $K/L$ be a algebraic extension. Suppose that $\left|\cdot\right|$ is a absolute value in $K$ such that is trivially in $L$. Then is trivially in $K$.

Thanks for anny suggestion.

If is trivially in $L$ then is non-archimedian. Let $\alpha\in K$. Then $\alpha^n+a_{n-1}\alpha^{n-1}+\ldots+a_1\alpha+a_0=0$ with $a_i\in L$ ; this implies that for some $i\neq j$ $|a_i\alpha^i|=|a_j\alpha_j|\Rightarrow |\alpha^i|=|\alpha^j| \Rightarrow |\alpha|=1 $

This is correct?

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    $\begingroup$ Your tags could use updating I think.. $\endgroup$ – Cameron Williams Nov 30 '14 at 1:54
  • $\begingroup$ This problem is well outside my understanding. I was just suggesting that you update your tags to reflect the actual subject matter. $\endgroup$ – Cameron Williams Nov 30 '14 at 2:17
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I think your proof is correct. I would phrase it like this:

Since the trivial value is non-archimedean, its extension to $K$ is non-archimedean. Let $\alpha \neq 0$ be any element of $K$, and let $f(x) = \sum_{i=0}^n a_i x^i$ be its minimal polynomial over $L$. Then $f(\alpha)=0$, but because $|a_i| =1$ if $a_i \neq 0$, if $|\alpha| > 1$ we would have $|f(\alpha)| = |\alpha|^n \neq 0$, whereas $|\alpha|<1$ would imply $|f(\alpha)| = |a_0| \neq 0$. So necessarily $|\alpha| = 1$.

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