# Are all nonarchimedean valuations discrete?

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).

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see this: jstor.org/pss/2040901 – PEV Feb 3 '11 at 21:59
The algebraic closure of a discrete valuation ring is nondiscrete valuation ring. So, the p-adic complex numbers gives an example. – George Lowther Feb 3 '11 at 22:02
I mean, the extension of a (nontrivial) discrete valuation to the algebraic closure is a nondiscrete valuation. – George Lowther Feb 3 '11 at 22:21

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.

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I should have worded the question more carefully, but I meant "why do we ignore nondiscrete nonarchimedean valuations in class field theory?". Your example shows that the answer here is also "we don't", but at least there are no such things on finite extensions. Thanks! – Vitaly Lorman Feb 7 '11 at 23:58

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.

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Thanks for the link to the notes. I suspect I will find them very useful for a lot of different things. – Vitaly Lorman Feb 7 '11 at 23:53
@Vitaly Lorman: Out of curiosity, I want to know if you have studied the foundation of algebraic number theory, algebraic geometry, et cetera, i.e. commutative algebra to get started to the theory of class fields? – awllower Mar 6 '11 at 3:34
@awllower: I had a class on algebraic number theory (ideal class group, group of units, how primes behave in extensions, etc.), and I am currently taking a class on local class field theory. I studied commutative algebra from Atiyah-Macdonald in undergrad. I am now studying algebraic geometry from a few different places: Liu, Hartshorne, and Vakil's notes online. So far the algebraic geometry has not been essential to understanding class field theory--mostly Galois theory and algebraic number theory are helpful. – Vitaly Lorman Mar 7 '11 at 23:35
@Vitaly Lorman: thanks for your elaboration and I must say that algebraic geometry is closely related to class field theory though it isn't directly related to the theory. Like J. Neukirch once said, Number theory is geometry which may imply the content will be helpful in understanding the class fields. Also, I am not good in algebraic geometry, hence you may ignore my opinion. – awllower Mar 8 '11 at 11:22
@awllower: Sure, it has been nice to have some geometric intuition, and I am sure there is a great deal of overlap between CFT and algebraic geometry. I only meant that to begin studying (local) CFT, it seems to me that one does not need to know algebraic geometry. – Vitaly Lorman Mar 10 '11 at 17:40