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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?

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2 Answers

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 $k$.

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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

http://math.uga.edu/~pete/8410Chapter2.pdf

for the proof. (These notes then spend several more pages establishing the existence part of the result.)

Addendum: 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 $K(t)$.

*: 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 does use the sledgehammer.)

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