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It was a exercise that one of our professors gave it to us and I don't know the solution of it:

Suppose that $v$ is a non-archimedean valuation on the field $F$ and $o(v)$ is the valuation ring of $v$. (It means $o(v)=\{x \in F : v(x)\leq1\}$.) If $x\in F$ is algebraic over $o(v)$, then prove that $x$ is in $o(v)$.

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    $\begingroup$ Dear @user : I suggest you consider the possibility of negative effects to the poster caused by editors swapping in their preferred notations. Edits like this have more potential for confusion than for help, and so are inadvisable. There are no conventions here dictating that posts must use "the most common notation google indicates". Regards $\endgroup$
    – rschwieb
    Jan 2, 2014 at 15:55
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    $\begingroup$ This question is odd. If $v(x) > 1$, then let $y = 1/x$. We have $y \in o(v)$, and $x$ is a root of $f(t) = yt + 1$ making it algebraic over $o(v)$, contradicting your claim. I assume you mean something stronger, like $x$ is also integral over $o(v)$. $\endgroup$
    – user14972
    Jan 3, 2014 at 1:19
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    $\begingroup$ @Hurkyl Good remark! (Nitpick: $yt+1$ is actually $yt-1$.) $\endgroup$
    – user89712
    Jan 3, 2014 at 18:16

1 Answer 1

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A non-archimedean valuation has the following property: $v(a+b)=\max(v(a),v(b))$ if $v(a)\neq v(b)$. If $x\in F$ is integral over $O_v$ then there exist $a_i\in O_v$ such that $\sum_{i=0}^{n-1}a_ix^i+x^n=0$. Suppose $v(x)>1$. Then $v(a_ix^i)<v(x^n)$ for all $i=0,1,\dots,n-1$ since $v(a_i)\le 1<v(x)$, and therefore $v(\sum_{i=0}^{n-1}a_ix^i+x^n)=v(x^n)>0$, a contradiction.

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  • $\begingroup$ I have some question: 1- in the Pierce we can see this definition: $v(a+b)‎\leq‎ a max(v(a),v(b))$, which a is a constant. with this definition we can't have any contradiction. 2- why we have $v(a_0)<v(a_1x)<...<v(a_nx^n)$, we know $v(a_nx^n)=v(a_n)v(x^n)$ and is it possible that $v(a_n)$ is very low such that $v(a_nx^n)<v(a_1x)$? $\endgroup$
    – mhf
    Jan 3, 2014 at 14:27

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