At the beginning of the section of a book I am reading it says $K$ is a local field complete with respect to a discrete valuation. Local fields can be archimedean or non-archimedean. Does the condition that it has a discrete valuation imply we are only dealing with non-archimedean local fields? Thank you very much.
2026-03-26 22:54:11.1774565651
Given a local field complete with respect to a discrete valuation, does that imply it is non-Archimedean?
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To turn comments into an answer:
An absolute value on a field $K$ is a map $|\cdot|: K \rightarrow \mathbb{R}_{\ge 0}$ such that $$|x| = 0 \Leftrightarrow x = 0,$$ $$|xy| = |x||y|,$$ $$|x+y| \le |x|+|y|.$$ An absolute value defines a metric on $K$ by $d(x,y) := |x-y|$, which in turn defines a topology on $K$.
The value is called nonarchimedean if $$|x+y| \le \max(|x|, |y|)$$ which is equivalent to its induced metric being an ultrametric. On the equivalence of alternative conditions/definitions for this, see here.
The trivial absolute value is $|x| = 1$ for all $x \in K^*$. It is nonarchimedean, but boring, and we exclude it.
A value is called discrete if its value group $|K^*|$ is discrete in $\mathbb{R}$. (Here is a possible source of confusion: The trivial value induces what is often called the discrete metric and discrete topology. Now the trivial value is also a discrete value, but what we are interested in from now on, on the other hand, are non-trivial discrete values, which do not induce the discrete topology.)
After all this preamble, a local field is a field with a non-trivial absolute value such that the induced topology is locally compact.
One can show that a local field whose value is archimedean is either $\mathbb{R}$ or $\mathbb{C}$. The value in theses cases, however, is non-discrete (the value group is all of $\mathbb{R}_{\ge 0}$.)
If, on the other hand, the value is discrete, then one can show that the field is either
-- and presumably, these are the fields your source talks about.
Note that in both cases the residue field is finite. In the first case, $K$ has characteristic $0$, (sometimes called "the mixed characteristic case"), whereas in the second case, $K$ and its residue field share the same positive characteristic. Further, the absolute value is (equivalent to) the $p$-adic value in the first case, and in both cases the value group is of the form $r^\mathbb{Z}$ where $r$ is some positive real number $\neq 1$. The most common choices for $r$ are the cardinality of the residue field $q$, or its characteristic $p$. As one learns early, this is of no real importance, and for these discrete values one most often switches from the multiplicative absolute value $|\cdot|$ to the additive valuation $val := -\log_r | \cdot |$, whose value group is just the additive group $\mathbb{Z}$.
Finally, I want to add that some (in particular French) sources use a broader definition of local fields, where the residue field is allowed to be any perfect field (not just finite). Then there are more cases, although they resemble the ones we have here. But I assume your book works with the stricter definition above.