Proof of Hensel's lemma by using the deformation theory

Question

I am thinking about the simplest version of Hensel's lemma. Fix a prime $p$. Let $f(x)\in \mathbf{Z}[x]$ be a polynomial. Assume there exists $a_0\in \mathbf{F}_p$ such that $f(a_0)=0\mod p$, and $f'(a_0)\neq 0\mod p$. Then there exists a unique lift $a_n\in \mathbf{Z}/p^{n+1}\mathbf{Z}$ for every $n$. I know there is an elementary proof. However, I want to prove it by using standard deformation theory. It is simply a problem about extending the section $a_0$ order by order. Let $X$ be the scheme defined by $f(x)$ over $k=\mathbf{F}_p$. $I$ is the $k$-module $p\mathbf{Z}/p^2$. If I am correct, the obstruction class is in $Ext^1(a_0^*L_{X/k},I)$, and if it vanishes, the extension is classified by $Ext^0(a_0^*L_{X/k},I)$. Is there a proof of Hensel's lemma along this line?

I know this is like using a big machine to solve a simple problem. However, I really want to understand why $f'(a_0)\neq 0 \mod p$ implies that the obstruction class vanishes.

Answer 1

There is indeed a proof along these lines. Suppose one has a polynomial $f$ over $\mathbb{Z}_p$; I'll use $f$ to refer to its reductions mod $p^k$ as well. You have a diagram:

$$\text{Spec}((\mathbb{Z}/p^k)[t]/f(t))\to~\text{Spec}((\mathbb{Z}/p^{k+1})[t]/f(t))~~~~~~~~~~~~~$$ $$\downarrow\uparrow~~~~~~~~~~~~~~~~~~~~~~\downarrow$$ $$\text{Spec}(\mathbb{Z}/p^k)\to \text{Spec}(\mathbb{Z}/p^{k+1})$$

The vertical arrow on the left is a section $s$, e.g. a choice of root of $f$ in $\mathbb{Z}/p^k$. It is this section $s$ that we would like to show lifts uniquely. To do so, it suffices to show that the section $s: \text{Spec}(\mathbb{Z}/p^k)\to \text{Spec}((\mathbb{Z}/p^k)[t]/f(t))$ is etale, because this implies the associated $L_s=\Omega^1_s=0$; as you observe, $\Omega^1_s$ controls deformations and thus $\Omega^1_s=0$ implies deformations are unobstructed and unique.

In fact, we may see that $s$ is the inclusion of a connected component $X_k$ of $\text{Spec}((\mathbb{Z}/p^k)[t]/f(t))$ (which implies $\Omega^1_s=0$); to do so, it suffices to show that the map $\pi: X_k\to \mathbb{Z}/p^k$ is an isomorphism, as $s$ is a left inverse to this map. For $k=1$ this follows from the Chinese remainder theorem (using that $f'(a_0)\not=0$ implies $a_0$ is not a double root); in general, for $k>1$, this follows from $k=1$ case plus Nakayama.

Of course, this argument also shows that $\pi: X_{k+1}\to \text{Spec}(\mathbb{Z}/p^{k+1})$ is an isomorphism, whence its inverse is the required section. So deformation theory is strictly optional.

Answer 2

One can also show that any complete local ring $(R,\mathfrak{m})$ is Henselian using the infinitesimal lifting criterion for étale morphisms: Let $S$ be an étale $R$-algebra with a section $S \to R/\mathfrak{m}$. We want to show that there is a lift $S \to R = \varprojlim_n R/\mathfrak{m}^n$. But we can construct such a lift (a compatible family of lifts $S \to R/\mathfrak{m}^n$) inductively using the infinitesimal lifting criterion for étale morphisms $\mathrm{Hom}(S,R/\mathfrak{m}^n) = \mathrm{Hom}(S,R/\mathfrak{m}^{n+1})$.