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Ex. $2.3.3$ in Algebraic Number Theory by Neukirch is the following:

Let $k$ be a field and $K = k(t)$ the function field in one variable. Show that the valuations $v_{\mathfrak p} $ associated to the prime ideals $\mathfrak p = (p(t))$ of $k[t]$, together with the degree valuation $v_\infty$, are the only valuations of $K$, up to equivalence. What are the residue class fields?

By valuation, I mean the exponential valuation. For instance the 2-valuation on $\Bbb Z$ is $$v(2^sn) = s , (2,n) = 1$$

Take $k = \Bbb Q$ and a non archimedean valuation $v$ on it and extend it to $k[x]$ by $$f(t) = a_0+a_1t+ \dots +a_nt^n \in K, v(f) = \min\{v(a_0),\dots, v(a_n)\}$$ and to $k(x)$ by $$v(f/g) = v(f)-v(g).$$ This is proven to be a valuation earlier on in the chapter. Which of the valuations mentioned in the problem is this equivalent to?

user26857
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Asvin
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  • Could you please point out precisely where it is proved that this $v$ is a valuation? In particular, consider a primitive polynomial $f \in \Bbb{Z}[t]$, i.e. a polynomial with $\gcd(a_0,\dotsc,a_n) = 1$. Then for any choice of non-archimedan $v$ on $\Bbb{Q}$ we have $v(f) = 1$ and $v(p^s f) = s$, where $p$ is the prime corresponding to $v$. Then clearly $v$ cannot be the degree valuation, but it cannot be a $v_{\mathfrak{p}}$ either because (for $s \neq 0$) $p^s f$ is irreducible if and only if $f$ is irreducible. – A.P. Jun 29 '15 at 20:39
  • It is proved after proposition $3.6$ but is also mentioned here. The book does not give a complete proof but only outlines it. I haven't worked through the details yet. – Asvin Jun 29 '15 at 20:52
  • @A.P. I have worked through the details now and they seem to hold up. It seems like the exercise is wrong? I might be missing something though... – Asvin Jun 29 '15 at 21:09
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    I think in the highlighted part you described the $k$-valuations on $k(t)$. The last one doesn't seem to be one of them because it's not trivial on $k$. – user26857 Jun 29 '15 at 21:12
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    The question does not mention that it is a $k$-valuation, in fact I don't think the text has defined what a $k$-valuation is yet. That might be it though, thanks @user26857 – Asvin Jun 29 '15 at 21:23

1 Answers1

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As discussed in the comments, the highlighted part seems to be based on the extra assumption that the valuation is trivial on $k$, $v(k^\times) = \{0\}$. Under this assumption, the result is proved e.g. here (where $k$ is assumed to be finite at first, but what is actually used in the proof is that the value is trivial on $k$).

Without that assumption, the highlighted part is not true. Generalising your example, if you have a nonarchimedean value on $k$, you can assign to $t$ any value $r\in \mathbb{R}$ you like, and set $$v(\sum_{i=0}^n a_i t^i) = \min_{i} (v(a_i) +ir).$$

These valuations and their extensions to power series and Laurent series that converge on certain annuli are of high importance in $p$-adic functional analysis, number theory and representation theory. (Your example is $r=0$; written with multiplicative values, you are looking at polynomials with "$t$ evaluated on the unit circle".)

Torsten Schoeneberg
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