TITLE: Are all nonarchimedean valuations discrete?
QUESTION [16 upvotes]: I am studying valuation theory on the way to local class field theory, and the texts I have looked at immediately focus on discrete valuations in developing the theory of nonarchimedean valuations.  Why? Are there nondiscrete nonarchimedean valuations? If so, why do we ignore them? (it is true that if a field is locally compact with respect to a nonarchimedean valuation, then that valuation must be discrete, and local compactness is very important, but I wonder if there isn't more to be said here).

REPLY [14 votes]: As Pete says, many of us do not ignore non-discrete valuations. However, I can explain why a text on class field theory might. 
If $K$ is a finite extension of $\mathbb{Q}$, then all nonarchimedean valuations on $K$ are discrete. If your text expects to spend most of its time focused on such fields, that would explain its focus.
Proof: Any valuation on $K$ gives rise to a valuation on $\mathbb{Q}$. By the classification of valuations on $\mathbb{Q}$, it must be the $p$-adic valuation for some $p$. Normalize $v(p)$ to $1$. If you read your textbooks description of extending valuations from $\mathbb{Q}$ to $K$, you should see that the image lands in $(1/e) \mathbb{Z}$, where $e$ is the ramification degree, and is bounded by $[K:\mathbb{Q}]$. QED
For an example of a non-discrete valuation of interest in number theory, let $K$ be the extension of $\mathbb{Q}$ obtained by adjoining every $p^k$ root of unity, for every $k$. If $\zeta_{p^k}$ is a $p^k$-th root of $1$, then $v_p(\zeta_{p^k} -1 ) = 1/((p-1)p^{(k-1)})$. In particular, the extension of $v_p$ to $K$ is not discrete.