# Tube of a mod p point on a smooth Z_(p)-scheme

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.

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.
• @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. – Will Sawin Mar 22 '16 at 22:54