Is every discrete absolute value non-archimedean?

I've been reading Milne's notes on absolute values and local fields (Chapter 7 of the book on algebraic number theory), and a possibly minor point is confusing me.

On page 126, Milne writes, "$K$ is a field complete with respect to a discrete absolute value $|\cdot|$..." and then he talks about things that only make sense for non-archimedean absolute values, like the ring of integers and its maximal ideal.

Is every discrete absolute value on a field $K$ also non-archimedean? This would explain a lot, but I also suspect that it is not true, because I can't find any statement of this fact anywhere.

EDIT: An absolute value is discrete if the value group $|K^\times|$ is a discrete subgroup of the multiplicative group $\mathbb{R}_{>0}$. An absolute value is non-archimedean if it satisfies the stronger triangle inequality $$|x+y| \le \max (|x|,|y|)$$ or equivalently if the set $\{|1|,|1+1|,|1+1+1|,\ldots\}$ is a bounded subset of $\mathbb{R}$.

• My questions are much like those of @mathmo, now deleted. I suppose that “discrete valuation” means that the value group is cyclic. My specific question is just what makes a valuation archimedean. I would suppose that in such a valuation, $\Bbb Z$ is unbounded, that is, that $\vert\Bbb Z\vert$ is an unbounded set of reals. But I’m not sure that this is your definition. – Lubin Apr 9 '18 at 1:58
• @Lubin I deleted my comment after checking out Milne's notes. By discrete, Milne literally means that the image of the absolute value is discrete. By non-archimedean, he means that the absolute value is ultrametric. – Mathmo123 Apr 9 '18 at 2:00
• @Mathmo123, thanks for the clarification. Pleading age, I’m much too lazy to go to Milne’s work. I await OP’s illumination. – Lubin Apr 9 '18 at 2:04
• If the image of $|\cdot|$ is discrete, then the set $|K^\times|\cap(0,1)\subset \mathbb R_{>0}$ has a maximal element, $\lambda$. Then $v(x) = \log_\lambda(x)$ is a valuation which corresponds to $|\cdot|$. – Mathmo123 Apr 9 '18 at 2:18
• @Mathmo123 I agree, how does this resolve the question? – Joshua Ruiter Apr 9 '18 at 2:24

You’re asking for a proposition that if $K$ has an absolute value $\vert\!\star\!\vert$ that’s archimedean, then the value group $\vert K^\times\vert\subset\Bbb R^{>0}$ is not cyclic.
In other words, you want a sequence of elements $z_n\in K$ with all $\vert z_n\vert<1$, but $\lim_n\vert z_n\vert=1$.
My proof below seems sloppy, and I welcome improvements. When I say “numbers” here, I mean non-negative integers, the set of all these being denoted $\Bbb N$, and the fact that $K$ is of characteristic zero implies that there is a unique map $\Bbb Z\to K$, so I’ll treat $\Bbb N$ as a subset of $K$.
Our hypothesis is that $\vert\Bbb N\vert$ is an unbounded set of reals. Among the numbers, I will call $m\in\Bbb N$ a “jump number” if $\vert m-1\vert<\vert m\vert$. Standardly, of course, every positive number is a jump number, but I don’t see an immediate reason to see that under our hypotheses. Clearly, though, there are infinitely many jump numbers, and their absolute values form an unbounded subset of $\Bbb R$. This is how we use the archimedean hypothesis.
For any jump number $m$, we get $\vert1-\frac1m\vert<1$. That’s the easy inequality. But there is a jump number $a_1$ with $1\le\vert a_1\vert$, and similarly there is for each $n\in\Bbb N$, there is a jump number $a_n$ with $n\le\vert a_n\vert$. Now, using the form of the triangle inequality that says $\vert x-y\vert\ge\vert x\vert-\vert y\vert$, we have \begin{align} n&\le\vert a_n\vert\\ \frac1n&\ge\frac1{\vert a_n\vert}\\ 1-\frac1n&\le1-\left\vert\frac1{a_n}\right\vert\le\left\vert1-\frac1{a_n}\right\vert<1\,, \end{align} so that we take $z_n=1-\frac1{a_n}$ to complete the proof.