I've been reading some papers on Igusa zeta functions, and they seem to be implicitly using a "quantitative version" of Hensel's Lemma, which also asserts the number of lifts of a $\mathbb{Z}/p\mathbb{Z}$-point to a $\mathbb{Z}/p^k\mathbb{Z}$-point. I'm looking for something like the following:



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> Let $X$ be a smooth irreducible separated scheme of finite type of relative dimension $n$ over the ring of $p$-adic integers $\mathbb{Z}\_p$. Then for any $k>0$ do we have
>  $$\# X(\mathbb{Z}/p^k\mathbb{Z}) = p^{n(k-1)}\# X(\mathbb{Z}/p\mathbb{Z}) \quad ?$$
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I'm looking for either a proof or a reference where I can find a proof.



A quantitative version of Hensel's Lemma
Let me try a very explicit proof. Consider the problem of taking a solution modulo $p^k$ and lifting to solutions modulo $p^{k+1}$. For this, we may assume that $X \subset \mathbb{A}^m$ is smooth and affine of dimension $n$, given by polynomials $f\_1, \dotsc, f\_r \in \mathbb{Z}\_p[x\_1, \dotsc, x\_m]$. Let $\mathbf{a} \in \mathbb{Z}\_p^m$ satisfy $f\_1(\mathbf{a}) \equiv \dotsb \equiv f\_r(\mathbf{a}) \equiv 0 \pmod{p^k}$.


To find points modulo $p^{k+1}$ that lift $\mathbf{a}$, you put $\mathbf{x} = \mathbf{a} + p^k \mathbf{y}$ and look for vectors $\mathbf{y}$ modulo $p$ giving solutions to $f\_i(\mathbf{x}) \equiv 0 \pmod{p^{k+1}}$. Taylor expansion gives



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> $(f\_1(\mathbf{x}), \dotsc, f\_r(\mathbf{x})) = (f\_1(\mathbf{a}, \dotsc, f\_r(\mathbf{a})) + p^k \mathbf{J}(\mathbf{a}) \mathbf{y} + O(p^{k+1})$
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where $\mathbf{J}$ is the Jacobian matrix $(\partial(f\_i)/\partial x\_j)$. Dividing by $p^k$ and reducing modulo $p$ gives an inhomogeneous linear equation over $\mathbf{F}\_p$:



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> $\mathbf{J}(\mathbf{a}) \mathbf{y} \equiv -p^{-k} (f\_1(\mathbf{a}, \dotsc, f\_r(\mathbf{a})) \pmod{p}$ .
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That $X$ is smooth over $\mathbb{Z}\_p$ at $\mathbf{a}$ implies that $\mathbf{J}(\mathbf{a})$, when reduced modulo $p$, has rank $m-n$. The space of solutions $\mathbf{y}$ is non-empty, by Hensel's Lemma, so is a linear space of dimension $n$ over $\mathbb{F}\_p$. In this way you see that every solution modulo $p^k$ lifts to precisely $p^n$ solutions modulo $p^{k+1}$, and you get your formula by induction.


