9
$\begingroup$

Let $R$ be a smooth, integral, finite-type $\mathbb{Z}_{(p)}$-algebra of relative dimension $n$ and $\overline{f} \colon R \to \mathbb{F}_p$. Then Hensel's lemma tells us that this lifts to a map $R \to \mathbb{Z}_p$. My understanding is that the space of lifts looks like an affine space, but I would like to understand this more explicitly.

I'm in particular hoping that it's always possible to choose an injective lift. In geometric terms, this is asking for a $\mathbb{Z}_p$-point such that the associated $\mathbb{Q}_p$-point maps to the generic point of $X:=\mathrm{Spec}(R)$.

I'm hoping this has something to do with the tangent space - like we want a tangent vector whose coordinates in $\mathbb{Z}_p$ are algebraically independent over $\mathbb{Q}$. But I don't understand this deformation space.

I'm listed this as "reference-request" because it might be fairly standard from deformation theory, but I couldn't find a reference and don't know where to look.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

Take $x_1,\dots,x_n$ in $R$ that lie in the kernel of $f$ and generate the tangent space a the point $f$. By Hensel's lemma, the map $(\frac{x_1}{p},\dots,\frac{x_n}{p})$ from the space of $\mathbb Z_p$-points of $R$ to $\mathbb Z_p^n$ is a bijection.

In fact the inverse function can be seen to be analytic. So each nonzero element of $R$ may be seen as a power series in $x_1,\dots,x_n$, hence a convergent power series in $\frac{x_1}{p},\dots,\frac{x_n}{p}$. By the integrality assumption none of these power series are identically zero. There are countably many power series.

Hence there is a point that is not in the vanishing locus of any of them by the standard cardinality / induction on the dimension argument. View each power series as a power series in $x_n$ with coefficients in $x_1, \dots,x_{n-1}$. There are countably many of these, so by induction there must be a tuple of values of $x_1,\dots,x_{n-1}$ where each power series has some nonvanishing coefficient of a power of $x_n$. Hence each has finitely many zeros as a function of $x_n$, so some point is a zero of none.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Thanks, your first paragraph is precisely what I needed. Do you have a reference for this version of Hensel's lemma? I have not seen such a thing mentioned. $\endgroup$
    – David Corwin
    Mar 22, 2016 at 21:05
  • $\begingroup$ I'm also wondering if there's an easy way to see how this fits into the formalism of deformation theory. $\endgroup$
    – David Corwin
    Mar 22, 2016 at 22:16
  • $\begingroup$ @DavidCorwin 1. The equations $x_i - p y_i$ for $y_1,\dots,y_n \in \mathbb Z_p$ in $\operatorname{Spec} R$ define a smooth scheme of dimension $0$, hence with a unique lift of the $\mathbb F_p$-point by the usual Hensel's lemma 2. In deformation theory, you want to take the completion of $R$ at the kernel of $f$. There is an obvious map from $\mathbb Z_p[[x_1,\dots,x_n]]$ to this completion. Because $x_1,\dots,x_n$ generate the cotangent space you can see that this is surjective, and by smoothness the kernel must vanish, so it is an isomorphism. That isomorphism tells you everything you want. $\endgroup$
    – Will Sawin
    Mar 22, 2016 at 22:54
  • $\begingroup$ Sorry this question is so basic, but what is the precise relationship between completions and deformations of points? $\endgroup$
    – David Corwin
    Mar 24, 2016 at 1:50
  • $\begingroup$ @DavidCorwin What do you mean by deformations of points? I meant the deformation ring of a point on a scheme. That's just the unique complete local ring that represents the scheme as a functor from art in local rings, which is of course the completion of the local ring. $\endgroup$
    – Will Sawin
    Mar 24, 2016 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.