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.



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


