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:

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 $$\# X(\mathbb{Z}/p^k\mathbb{Z}) = p^{n(k-1)}\# X(\mathbb{Z}/p\mathbb{Z}) \quad ?$$

I'm looking for either a proof or a reference where I can find a proof.

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    $\begingroup$ Searching, as I'm sure you did. shows that Hensel's-lemma questions are quite popular here. I wonder if Wanderer's answer, on the implicit-function theorem, to PeteL.Clark's old question has any relevance? $\endgroup$
    – LSpice
    Commented Nov 27, 2016 at 12:27
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    $\begingroup$ Check out N. Bourbaki, Commutative Algebra, III, S4.5, Corollary 3. I think the uniqueness in the corollary should be sufficient to prove that $\#X(\mathbb Z/p^k\mathbb Z)/\# X(\mathbb Z/p^{k-1}\mathbb Z)=p^n$, for all $k>1$, from which the statement follows (at least in the case where $X$ is affine). $\endgroup$
    – kneidell
    Commented Nov 27, 2016 at 12:28
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    $\begingroup$ It's smooth, so work with the completed local rings and then it's easy, no? $\endgroup$
    – znt
    Commented Nov 27, 2016 at 14:46
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    $\begingroup$ Smooth over Z_p means the completed local ring at a Z/pZ-point in the special fibre is Z_p[[X_1,X_2,...,X_n]] with n the rel dim of the morphism. Now count how many maps from that ring to Z/p^kZ. Any map from Spec(Z/p^k) to the scheme whose image is this point induces a map on the local rings and because Z/p^k is Artinian it will factor through the completion. Isn't this enough? Did I miss something? $\endgroup$
    – znt
    Commented Nov 27, 2016 at 16:08
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    $\begingroup$ Roughly speaking, since $X$ is smooth, the subset of $X({\mathbb Q}_p)$ consisting of points reducing mod $p$ to some specified point in $X({\mathbb F}_p)$ is analytically isomorphic to the corresponding set for affine $n$-space, for which the statement is pretty clear. (This is just an attempt at giving a geometric version of znt's previous comment.) $\endgroup$ Commented Nov 27, 2016 at 18:45

1 Answer 1


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

$(f_1(\mathbf{x}), \dotsc, f_r(\mathbf{x})) = (f_1(\mathbf{a}, \dotsc, f_r(\mathbf{a})) + p^k \mathbf{J}(\mathbf{a}) \mathbf{y} + O(p^{k+1})$

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 $\mathbf{F}_p$:

$\mathbf{J}(\mathbf{a}) \mathbf{y} \equiv -p^{-k} (f_1(\mathbf{a}, \dotsc, f_r(\mathbf{a})) \pmod{p}$ .

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.

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    $\begingroup$ I guess you can make this a bit neater by using that X is a local complete intersection, so you can take $r=m-n$. Then the Jacobian is surjective, and you don't need to appeal to Hensel's Lemma to show existence of a solution. $\endgroup$ Commented Nov 28, 2016 at 13:58

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