Are all nonarchimedean valuations discrete?

Solution 1:

No, there is a ridiculous number (i.e., a big proper class) of nondiscrete non-Archimedean valuations. To see some of them, you need only consult a text or section of a text which treats general valuations, e.g. Chapter 17 of these notes.

In them I include a proof of the following fact: for any torsionfree commutative group $G$, there exists a total ordering $\leq$ on $G$ and a valuation ring $R$ with value group isomorphic to $(G,\leq)$.

I'm not sure what to make of the question "Why do we ignore them?" We don't. In some branches of mathematics -- like the theory of local fields -- discrete valuations are more important than non-discrete valuations, and in other branches of mathematics -- e.g. commutative algebra, certain parts of algebraic geometry -- one definitely needs to consider more general valuation rings.

Solution 2:

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.