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I'm reading a paper where the author appeals to Hensel's lemma, but it is not clear to me quite how it is meant to be applied (or, for that matter, which version!). My commutative algebra background is unfortunately not as good as I'd like, and I can't find references between plain old Hensel's lemma for $p$-adics and Henselian rings.

The setup is this. I have a split exact sequence of (unital, and otherwise nice) rings $$ A \stackrel{r}{\to} B \to 0 $$ where $N = ker(r)$ is a nil-ideal, and $s$ is a section of $r$. Let $\Sigma$ be a set of primes (in the end it will probably just be a single prime), $\mathbb{Z}_\Sigma = \{a/b\in\mathbb{Q} \mid p\not| b,\ \forall p\in\Sigma\}$ and $R_\Sigma = R\otimes_\mathbb{Z} \mathbb{Z}_\Sigma$ for $R$ a ring (sim. for an ideal). Then we still have $N_\Sigma$ a nil-ideal, $r_\Sigma$ is still onto. Let $U_\Sigma = r_\Sigma^{-1}(1) \subset A_\Sigma$. We also have that $a\in A_\Sigma$ is a unit if and only if $r_\Sigma(a)$ is a unit.

Now comes the bit I don't understand.

The author claims, using Hensel, that given a unit $n\in \mathbb{Z}_\Sigma^\times$ with $n\gt 0$ and some $a\in U_\Sigma$, there is a unique $b\in U_\Sigma$ such that $b^n = a$.

I can think of several relevant points, but I can't connect them up into a proof in my mind.

  1. As $N$ is a nil-ideal, so $A$ should be complete in the $N$-adic topology.

  2. Since $a\in U_\Sigma$, I'm lifting the solution $1$ to $x^n - 1=0$ in $B_\Sigma$ through the various projections $A_\Sigma/N_\Sigma^{k+1} \to A_\Sigma/N_\Sigma^k$.

  3. The rings $A$ and $B$ are quite nice ($B$ is isomorphic to $\mathbb{Z}^m$, for instance, but $A$ can be more complicated), so I can probably assume they're Noetherian...

Is the proof there under my nose? Or is a bit more subtle?

theHigherGeometer
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1 Answers1

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After inverting primes outside $\Sigma$, we can assume $A=A_\Sigma$. Hensel's lemma is being used as the following assertion: if $C$ is a finite étale $A$-algebra, then any section $C \to B$ lifts uniquely to a section $C \to A$. Apply this with $C = A[x]/(x^n - a)$, for any $n$ prime to $\Sigma$ and the section $C\to B$ corresponding to the identity element.

EDIT: To see why $A[x]/(x^n - a)$ is étale, one uses the differential criterion. The derivative of $x^n$ is $nx^{n-1}$. If $x$ is a solution to $x^n - a$, then it is n particular a unit, and therefore so is $nx^{n-1}$, which is then non-invertible everywhere.

In any case, the proof of the assertion that I gave is the same as that of the regular old Hensel's lemma via Newton's method, so it's really no different. But I prefer this formulation.

Keerthi Madapusi
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  • I didn't say $n$ was prime. Just that it was prime to $\Sigma$. I did mess up and invert primes in $\Sigma$ rather than primes not in $\Sigma$. The edit corrects this. – Keerthi Madapusi Jul 02 '15 at 20:58
  • Ah, yes, just saw it. Thanks. Do you have a reference? I'd hate to just repeat the uninformative mantra 'mumble mimble Hensel's lemma' when I use this. – theHigherGeometer Jul 02 '15 at 21:02
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    The most complete reference is probably Raynaud's 'Anneaux locaux henseliens', which is a pretty fantastic commutative algebra volume in general. The chapter that's most relevant to your question is the very last one on Henselian couples (most of the book deals with the local situation, but this section shows how the results generalize without the localness assumption). – Keerthi Madapusi Jul 02 '15 at 21:05
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    The relevant result is Proposition 1 on p. 121 of the book, which is the exact result I quoted (but phrased in terms of idempotent elements instead of splittings). – Keerthi Madapusi Jul 02 '15 at 21:08
  • Ah, excellent. I happen to have that book in a box somewhere, I was looking after it for someone else. – theHigherGeometer Jul 03 '15 at 00:09