Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A set of notes I'm reading says the following: let $K$ be complete with respect to discrete valuation $v$, with finite residue field $k$ and valuation ring $\mathcal{O}$. Then by Hensel's lemma, for each $\alpha \in k$, there is a unique $a \in \mathcal{O}$ such that $a^q = a,\,a \equiv \alpha \, (\operatorname{mod} \pi)$ where $\pi$ is a uniformiser. $a$ is the Teichmuller representative for $\alpha \in k$; we write $a = [\alpha]$. Then $[\cdot]: k \to \mathcal{O}$ is multiplicative.

As a corollary, it claims that for the sets of units we have $k^* \hookleftarrow \mathcal{O}^*$. What I don't see is, how is this a corollary of the multiplicativity of $[\cdot]$? I don't see how it follows from that at all. It's probably very obvious but I'm not sure why - many thanks for the help!

share|improve this question
Are you sure you've transcribed that correctly? The claim of an injection $\mathcal{O}^\times \mapsto k^\times$ is very wrong. Consider $\mathcal{O} = \mathbb{Z}_2$, the 2-adic integers. Not only do the roots of unity $-1$ and $1$ have the same image in $\mathbb{F}_2$, but so does every number of the form $1 + 2x$ (for any 2-adic integer $x$), and all of those are units too. –  Hurkyl Mar 11 '12 at 3:24
The lifting map $k^\times \to \mathcal{O}^\times$ is monic, though, and the proof of this fact is contained in the proof you cited. Could this be what it meant? I've sometimes seen people use "corollary" when they mean it follows from the proof, rather than from the theorem. –  Hurkyl Mar 11 '12 at 3:28
I've certainly transcribed the notes correctly, but they're someone else's lecture notes from a few years ago and they have plenty of typos elsewhere so I wouldn't be surprised if they were wrong here, that would certainly explain my confusion! The exact phrasing is as follows: Lemma: $[\cdot]: k^* \to \mathcal{O}^*$ is multiplicative. Proof: (2 line proof) Corollary: $k^* \hookleftarrow \mathcal{O}^*$. E.g., $\mu_{p-1} \subset \mathbb{Q}_p$. ($\mathbb{Q}_p$ refers to the p-adic numbers here.) Does the example offer any clues to what might have been meant? –  Spyam Mar 11 '12 at 3:40
@AndréNicolas, while your comment is quite true, it is also quite off-topic! –  Mariano Suárez-Alvarez Mar 11 '12 at 4:01
I agree with Mariano and think this comment should be deleted. –  Pete L. Clark Mar 11 '12 at 4:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.