most recent 30 from http://mathoverflow.net 2013-05-22T22:13:20Z http://mathoverflow.net/feeds/question/20987 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20987/trivial-valuation kwan 2010-04-11T06:14:36Z 2010-04-11T15:03:20Z

<p>I'm pretty sure trivial valuation over a field cannot be extended to a non-trivial one in a bigger field. Is there a simple way to show this without using the sledge hammer theorem on valuation extension over complete valued field?</p>

http://mathoverflow.net/questions/20987/trivial-valuation/20988#20988 Cam McLeman 2010-04-11T06:23:24Z 2010-04-11T06:23:24Z

<p>Sure it can. Consider the x-adic valuation on the field of Laurent series $k((x))$ over a field $k$, extending the trivial absolute value on $k$.</p>

http://mathoverflow.net/questions/20987/trivial-valuation/20989#20989 Pete L. Clark 2010-04-11T06:31:02Z 2010-04-11T06:42:11Z

<p>Maybe you meant for the extension $L/K$ to be algebraic, in which case it is true that any extension of the trivial valuation on $K$ to $L$ is trivial. This clearly reduces to the case of a finite extension, and then -- since a trivially valued field is complete -- this follows from the uniqueness of the extended valuation in a finite extension of a complete field. Maybe you view this as part of the sledgehammer, but it's not really the heavy part: see e.g. p. 16 of </p> <p><a href="http://math.uga.edu/~pete/8410Chapter2.pdf" rel="nofollow">http://math.uga.edu/~pete/8410Chapter2.pdf</a></p> <p>for the proof. (These notes then spend several more pages establishing the existence part of the result.) </p> <p><b>Addendum</b>: Conversely, if $L/K$ is transcendental, then there exists a nontrivial extension on $L$ which is trivial on $K$. Indeed, let $t$ be an element of $L$ which is transcendental over $K$, and extend the trivial valuation to $K(t)$ by taking $v_{\infty}(P/Q) = deg(Q) - deg(P)$. (The completion of $K$ with respect to $v_{\infty}$ is the Laurent series field $K((t))$, so this is really the same construction as in Cam's answer.) Then I prove* in the same set of notes that any non-Archimedean valuation can be extended to an arbitrary field extension, so $v$ extends all the way to $L$ and is certainly nontrivial there, being already nontrivial on $K(t)$. </p> <p>*: not for the first time, of course, though I had a hard time finding exactly this result in the texts I was using to prepare my course. (This <strong>does</strong> use the sledgehammer.)</p>

http://mathoverflow.net/questions/20987/trivial-valuation/21011#21011 ashpool 2010-04-11T15:03:20Z 2010-04-11T15:03:20Z

<p>Thank you both! Your answers were enormously helpful.</p>