Timeline for Field complete with respect to inequivalent absolute values
Current License: CC BY-SA 4.0
16 events
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| Apr 24 at 16:06 | history | edited | KConrad | CC BY-SA 4.0 |
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| Nov 7, 2022 at 23:40 | comment | added | KConrad | @TorstenSchoeneberg aha. What a strange way to describe the units! | |
| Nov 7, 2022 at 22:35 | comment | added | Torsten Schoeneberg | $\mathcal{O}_K^\times = \{ x \in K^\times: \exists y \in K^\times \text{ such that for all } r \in \mathbb Z, \\ 1+x^ry \text{ is an } n\text{-th power for infinitely many } n \ge 1\}$. | |
| Nov 7, 2022 at 19:36 | comment | added | KConrad | @TorstenSchoeneberg what is your proposed purely algebraic description of the elements with valuation 0 in a complete discretely valued field (with no restriction on the residue field)? In any case, the theorem of F. K. Schmidt in Fehm's answer shows a field can't be complete with respect to two inequivalent discrete valuations, because if it were then it'd satisfy Hensel's lemma for both of them and that would make the field algebraically closed (not just separably closed -- see my comment to that answer). And an algebraically closed field doesn't admit any nontrivial discrete valuation. | |
| Nov 7, 2022 at 19:28 | comment | added | KConrad | @TorstenSchoeneberg the field $K = \mathbf Q((t))$ has its discrete $t$-adc valuation, for which $\mathcal O_K = \mathbf Q[[t]]$ with infinite residue field $\mathbf Q$, but the description I gave of $\mathcal O_K^\times$ doesn't work for this example: $\mathbf Q^\times \subset \mathcal O_K^\times$, but a nonzero rational number $r$ other than $\pm 1$ is not an $n$-th power in $K$ for infinitely many $n \geq 1$, since if $r = f(t)^n$ then $f(t) \in \mathcal O_K$ and that lets us say $r = f(0)^n$, so $r$ would be an $n$th power in $\mathbf Q^\times$ for infinitely many $n$, which is impossible. | |
| Nov 7, 2022 at 18:58 | comment | added | Torsten Schoeneberg | I guess what I should have said is that one could generalize the last argument, via the linked method by reuns, to: For any field complete with respect to some discrete valuation, that discrete valuation is uniquely determined up to equivalence. I.e. the answer to OP is negative not just if we restrict to valuations which make the field locally compact, but already if we restrict to discrete valuations. I wonder if one can characterize those fields which have a unique nontrivial discrete valuation, in particular if there are more beyond complete (maybe Henselian) ones? | |
| Nov 6, 2022 at 20:00 | comment | added | KConrad | @TorstenSchoeneberg yes, those conditions are a consequence of the assumed local compactness. I have made an edit to the first paragraph of the proof to make discreteness of the value group and finiteness of the residue field more explicit at the start. | |
| Nov 6, 2022 at 19:57 | history | edited | KConrad | CC BY-SA 4.0 |
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| Nov 6, 2022 at 19:49 | history | edited | KConrad | CC BY-SA 4.0 |
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| Nov 6, 2022 at 19:32 | comment | added | Torsten Schoeneberg | Nitpicking: For the final argument you assume all possible absolute values (which we aim to show to be equivalent) make the field Henselian and have a finite residue field. Otherwise, via Zorn and restricting different $q$-adic values from their "common" superfield $\mathbb C$, there are many other, inequivalent absolute values on $\mathbb Q_p$. Just none of them makes it locally compact (as you say). In fact, none of these restrictions can be discrete on $\mathbb Q_p$. There are more fields which have a unique nontrivial discrete value, cf. math.stackexchange.com/a/4076344/96384. | |
| Nov 5, 2022 at 3:44 | history | edited | KConrad | CC BY-SA 4.0 |
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| Nov 5, 2022 at 1:23 | history | edited | KConrad | CC BY-SA 4.0 |
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| Nov 5, 2022 at 0:04 | comment | added | Yuri Bilu | Sure, you are right. | |
| Nov 5, 2022 at 0:03 | comment | added | KConrad | I added the constraint of the absolute value being nontrivial to your question, in order to avoid really silly examples like any complete nontrivially valued field, equipped with either its given nontrivial absolute value or the trivial absolute value. | |
| Nov 5, 2022 at 0:01 | comment | added | Yuri Bilu | Thanks! I indeed overlooked a simple answer. | |
| Nov 4, 2022 at 23:56 | history | answered | KConrad | CC BY-SA 4.0 |