By Long Luo

知乎上有个问题是高考数学最后一题可以有多难？ ¹，而公认史上最难高考数学题就是 2008 年江西高考理科数学压轴题。2005 - 2014 年是江西自主命题，而江西卷也以题难计算量大著称，尤其是数学和理综。陶平生老师 是 2008 - 2011 年江西高考数学命题组长，参与了 2005 - 2011 年江西高考命题，他出江西数学卷时，是江西30多万学生被支配的恐惧！

作为一名来自十八线农村做题家，高考时赶上了江西自主命题，在考场上也体会到了数学和理综卷题目居然没做完没思路的恐惧！中学时没能感受到数学的乐趣，最近几年看了一些数学书之后，重新拾起了数学的乐趣，经常找些数学题来训练下我的思维，今天就来挑战一下 2010 年江西高考理科数学压轴题：

这道题其实是数论( \(\textit{Number theory}\) ) ² 背景，除了搞竞赛的同学，谁学过数论呢？这种构造性( \(\textit{Constructive proof}\) ) ³ 题目，没有接触过类似题目的话，根本不知道如何下手。要是我在考场上也会一脸懵圈，甚至在几年前我也是一筹莫展，不过现在倒是有勇气挑战这类题目了！网上关于这道题的参考答案太简略了，主要是如何找到答案的构造不清楚。这道题真是一道绝妙的题，我花了比较长时间来思考这道题，下面详细描述我的解题思路。

第一问

根据题意，正整数 \(a, b, c\) 且 \(b < c\) ，满足 \(a^2, b^2, c^2\) 为等差数列，即：

\[ b^2 - a^2 = c^2 - b^2 \Rightarrow 2b^2 = a^2 + c^2 \]

观察上式，我们可以得到 \(2\) 个推论：

1. \(a < b < c\) ；
2. \(a\) 和 \(c\) 要么都是偶数，要么都是奇数。

因为 \(a\) 为任意正整数，那么就从最简单的开始，不妨设 \(a = 1\) ，则有：

1. 令 \(c = 1\) , 得 \(b = 1\) ，与 \(a < b < c\) 矛盾；
2. 令 \(c = 3\) , 得 \(b = \sqrt {5}\) ，与 \(b\) 为正整数矛盾；
3. 令 \(c = 5\) , 得 \(b = \sqrt {13}\) ，与 \(b\) 为正整数矛盾；
4. 令 \(c = 7\) , 得 \(b = 5\) ，满足条件。

简单猜测实验得到一组要求的值： \(a = 1\) , \(b = 5\) , \(c = 7\) 。

再设 \(a = 2\) ，同理有：

1. 令 \(c = 4\) , 得 \(b = \sqrt {10}\) ，与 \(b\) 为正整数矛盾；
2. 令 \(c = 6\) , 得 \(b = 2 \sqrt {5}\) ，与 \(b\) 为正整数矛盾；
3. 令 \(c = 8\) , 得 \(b = \sqrt {34}\) ，与 \(b\) 为正整数矛盾；
4. 令 \(c = 10\) , 得 \(b = 2 \sqrt {13}\) ，与 \(b\) 为正整数矛盾；
5. 令 \(c = 12\) , 得 \(b = \sqrt {74}\) ，与 \(b\) 为正整数矛盾；
6. 令 \(c = 14\) , 得 \(b = 10\) ，满足条件。

通过实验又得到一组要求的值： \(a = 2\) , \(b = 10\) , \(c = 14\) 。

综合 \(a\) 为奇数和偶数的情况，猜想：

对于 \(\forall a = n \in \mathbb{N}^+\) ，令 \(b = 5a\) , \(c = 7a\) ，易得：

\[ 2 \cdot (5n)^2 = n^2 + (7n)^2 \]

故命题(1)得证。

Introduction

Let $G$ be a locally compact abelian group (for example, $\mathbb{R}$, $\mathbb{Z}$, $\mathbb{T}$, $\mathbb{Q}_p$). Then every irreducible unitary representation $\pi:G \to U(\mathcal{H}_\pi)$ is one dimensional, where $\mathcal{H}_\pi$ is a non-zero Hilbert space, in which case we take it as $\mathbb{C}$. It follows that $\pi(x)(z)=\xi(x)z$ for all $z \in \mathbb{C}$ where $\xi \in \operatorname{Hom}(G,\mathbb{T})$, viewing $\mathbb{T}$ as the unit circle in the complex plane. Such homomorphisms are called (unitary) characters, and we denote all characters of $G$ by $\widehat{G}$, calling it the Pontryagin dual group. It should ring a bell of the representation theory of finite groups. For convenience, instead of $\xi(x)$, we often write $\langle x,\xi \rangle$. We also write $\langle x,\xi\rangle\langle y,\xi \rangle=\langle x+y ,\xi\rangle$.

Some easily accessible examples are:

• $\widehat{\mathbb{R}} \cong \mathbb{R}$, with $\langle x,\xi \rangle = e^{2\pi i \xi x}$.
• $\widehat{\mathbb{T}} \cong \mathbb{Z}$, with $\langle z, n \rangle = z^n$.
• $\widehat{\mathbb Z} \cong \mathbb{T}$, with $\langle n,z \rangle = z^n$.
• $\widehat{\mathbb{Z}/k\mathbb{Z}} \cong \mathbb{Z}/k\mathbb{Z}$, with $\langle m,n\rangle =e^{2\pi i m n / k}$.

The Dual of p-adic Field

But we want to show that

The proof is broken down into several steps. But it shall be clear that $\mathbb{Q}_p$ is a topological group with respect to addition.

Step 1 - Find the simplest character

Every $p$-adic number $x \in \mathbb{Q}_p$ can be written in the form

where $m \in \mathbb{Z}$, $x_j \in \{1,2,\dots,p-1\}$ for all $j$. We define

and claim that $\xi_1$ is a character. Notice that the right hand side is always well-defined, because all summands when $j \ge 0$ contributes nothing as $\exp(2\pi i x_jp^j)=1$. That is to say, the right hand side can be understood as a finite product: when $m \ge 0$, i.e. $x \in \mathbb{Z}_p$, the pairing $\langle x, \xi \rangle = 1$; when $m<0$ however, $\langle x,\xi_1 \rangle = \exp\left( 2\pi i \sum_{j=m}^{-1}x_jp^j\right)$. Therefore it is legitimate to write

From this it follows immediately that

The function $\xi_1$ is continuous because it is continuous on $\mathbb{Z}_p$, being constant. Therefore it is safe to say that $\xi_1$ is a character with kernel $\mathbb{Z}_p$.

A quick thought would be, generating all characters out of $\xi_1$, to get something like $\xi_p$, $\xi_{1+p+p^2+\dots}$. But that approach might lead us to a nightmare of subscripts. Instead, we try to discover as many characters as possible. For any $y \in \mathbb{Q}_p$, we define

In other words, $\xi_y$ is defined by $x \mapsto \langle xy,\xi_1\rangle$. Since the multiplication is continuous, we see immediately that $\xi_y$ is a character, not very more complicated than $\xi_1$. We will show that this is all we need. To do this, we need to characterise all characters. Notice that have the same image but their kernels differ. So we try to characterise the characters by characterising their kernels.

Step 2 - Study the kernels of characters

For the $\xi_y$ above, we notice that $\langle x,\xi_y\rangle=1$ if and only if $xy \in \ker\xi_1=\mathbb{Z}_p$, i.e. $|xy|_p \le 1$. Therefore

We expect that all characters are of the form $\xi_y$. Therefore their kernels shall be like $\ker\xi_y$ naturally. Notice that for fixed $y$, we have $|y|_p=p^m$ for some $m \in \mathbb{Z}$. As a result $\ker\xi_y = \overline{B}(0,p^{-m})$. For this reason we have the following (more obscure) argument

Lemma 1. If $\xi \in \widehat{\mathbb{Q}}_p$, there exists an integer $k$ such that $\overline{B}(0,p^{-k}) \subset \ker\xi$.

Proof. Since $\xi$ is continuous, $\langle 0,\xi\rangle=1$ on the circle, there exists $k$ such that $\overline{B}(0,p^{-k}) \subset \xi^{-1}\{z \in \mathbb{T}:|z-1| < 1\}$ (this is to say the right hand side is an open set). But $\overline{B}(0,p^{-k})$ is a group (as $|\cdot|_p$ is non-Archimedean), therefore it maps into a subgroup of $\mathbb{T}$, which can only be $\{1\}$. $\square$

We cannot say the kernel of $\xi$ is exactly of the form $\overline{B}(0,p^{-k})$ yet, but we have a way to formalise them now. If $\overline{B}(0,p^{-k}) \subset \ker\xi$ for all $k$, then $\xi=1$ is the unit in $\widehat{\mathbb{Q}}_p$. Otherwise, for each $\xi$, there is a smallest $k_0$ such that $\overline{B}(0,p^{-k_0})\subset \ker\xi$ but $\overline{B}(0,p^{-k}) \not \subset \ker\xi$ whenever $k<k_0$. In another way around, we have $\langle p^{k_0-1},\xi\rangle \ne1$ but $\langle p^k,\xi\rangle=1$ whenever $k \ge k_0$. As one may guess, such $k_0$ subjects to the “size” of $\xi$. For convenience we study the case when $k_0=0$ first.

Lemma 2 (“Fourier series”). Suppose for given $\xi \in \widehat{\mathbb{Q}}_p$, $\langle 1,\xi \rangle = 1$ but $\langle p^{-1},\xi \rangle \ne 1$. There is a sequence $(c_j)$ taking values in $\{0,1,\dots,p-1\}$ such that $\langle p^{-k},\xi \rangle=\exp\left(2\pi i\sum_1^k c_{k-j}p^{-j}\right)$ for all $k=1,2,\dots$. In particular, $c_0 \ne 0$.

Proof. Put $\omega_k=\langle p^{-k},\xi\rangle$. Then $\omega_0=1$ but $\omega_k \ne 1$ for all $k \ge 1$. Since

each $\omega_{k+1}$ is a $p$-th root of $\omega_{k}$, and in particular $\omega_1$ is a $p$-th root of unity. There exists $c_0 \in \{1,\dots,p-1\}$ such that

and the overall formula for $\omega_k$ follows from induction. $\square$

One would guess that for the corresponding $k_0$, the “size” of $\xi$ should be $p^{k_0}$. This looks realistic, but will be tedious. Right now we still only study the case when $k_0=0$.

Lemma 3. Notation being in lemma 2, there exists $y \in \mathbb{Q}_p$ with $|y|_p=1$ such that $\xi = \xi_y$.

Proof. From lemma 2 we obtain a series $y=\sum_{j=0}^{\infty}c_jp^j$ with $c_0 \ne 0$. Then in particular $|y|_p=1$. By expanding the term, we see

It follows that $\langle x,\xi \rangle = \langle x,\xi_y \rangle$ for all $x \in \mathbb{Q}_p$. $\square$

Now we are ready to conclude our observation of the dual group.

Step 3 - Realise the dual group

Theorem. The map $\Lambda:y \mapsto \xi_y$ is an isomorphism of topological groups. Hence $\mathbb{Q}_p \cong \widehat{\mathbb{Q}}_p$.

Proof. First of all we study the algebraic isomorphism. First of all if $\xi_y=1$, then

Hence the map $\Lambda$ is injective. To show that $\Lambda$ is surjective, fix $\xi \in \widehat{\mathbb{Q}}_p$. By the comment below lemma 1, there is a smallest integer $k$ such that $\langle p^k,\xi \rangle = 1$. Then one considers the character $\eta$ defined by

It satisfies the condition in lemma 3, therefore there exists $z \in \mathbb{Q}_p$ such that $\eta=\xi_z$, and it follows that $\xi=\xi_{p^{-k}z}$.

Next we show that $\Lambda$ is a homeomorphism. Observe the following sets

ranging over $\ell \ge 1$ and $k \in \mathbb{Z}$. These sets constitute a local base at $1$ for $\widehat{\mathbb{Q}}_p$. We need to show that it corresponds to a local base of $\mathbb{Q}_p$ under the map $\Lambda$:

The image of the set $\{x:|x|_p \le p^k\}$ under $\xi_1$ is $\{1\}$ if $k \le 0$ and is the group of $p^k$-th roots of unity if $k>0$, and hence is contained in $\{z:|z-1|<\ell^{-1}\}$ if and only if $k \le 0$. It follows that $\xi_y \in N(\ell,k)$ if and only if $|y|_p \le p^{-k}$, i.e., $y \in \overline{B}(0,p^{-k})$. We are done. $\square$

Heights

Definition. For a polynomial with coefficients in a number field $K$

the height of $f$ is defined to be

where

is the Gauss norm for any place $v$.

Here, $M_K$ refers to the canonical set of non-equivalent places on $K$. See first four pages of this document for a reference.

As one can expect, this can tell us about some complexity of a polynomial, just like how the height of an algebraic number tells us its complexity. Let us compute some examples.

Computing Heights

Let us consider the simplest one

first. Since $|x^2-1|_v=1$ for all places $v$, the height of $f$ is a sum of $0$, which is still $0$.

Next, we take care of a polynomial that involves prime numbers

We see $|g(x)|_\infty=2$, $|g(x)|_2=2^{-(-2)}=4$, $|g(x)|_3=3^{-(-1)}=3$, and the Gauss norm is $1$ for all other primes. Therefore

Put $u(x,y)=\sqrt{2}x^2 + 3\sqrt{2}xy+5y^2+7 \in \mathbb{Q}(\sqrt{2})[x,y]$, we can compute its height carefully. Notice that $|\sqrt{2}|_v=\sqrt{|2|_v}$ for all places $v$ and we therefore have

Height and Products

If $f \in K[s_1,\dots,s_n]$ and $g \in K[t_1,\dots,t_m]$ are two polynomials in different variables, then as a polynomial in $K[s_1,\dots,s_n;t_1,\dots,t_m]$, $fg$ has height $h(f)+h(g)$. This is immediately realised once we notice that the height of a polynomial is equal to the height of the vector of coefficients in appropriate projective space. The identity $h(fg)=h(f)+h(g)$ follows from the Segre embedding.

But if variables coincide, things get different. For example, $h(x+1)=0$ but $h((x+1)^2)=2$. This is because we do not have $|fg|_\infty=|f|_\infty|g|_\infty$. Nevertheless, for non-Archimedean places, things are easier.

Gauss’s lemma. If $v$ is not Archimedean, then $|fg|_v=|f|_v|g|_v$.

Proof. First of all, it suffices to prove it for univariable cases. If $f$ and $g$ have multiple variables $x_1,\dots,x_n$, let $d$ be an integer greater than the degree of $fg$. Then the Kronecker substitution

reduces our study into $K[t]$. This is because, with such a $d$, this substitution gives a univariable polynomial with the same set of coefficients.

Therefore we only need to show that $|f(t)g(t)|_v=|f(t)|_v|g(t)|_v$. Without loss of generality we assume that $|f(t)|_v=|g(t)|_v=1$. Write $f(t)=\sum a_k t^k$ and $g(t)=\sum b_k t^k$, we have $f(t)g(t)=\sum c_jt^j$ where $c_j=\sum_{j=k+l}a_kb_l$.

We suppose that $|fg|_v<1$, i.e., $|c_j|_v<1$ for all $j$, and see what contradiction we will get. If $|a_j|=1$ for all $j$, then $|c_j|_v<1$ implies that $|b_k|_v<1$ for all $k$ and therefore $|g|_v<1$, a contradiction. Therefore we may assume that, without loss of generality, $|a_0|_v<1$ but $|a_1|_v=1$. Then, since

we have $|a_1b_{j-1}|_v=|b_{j-1}|_v<1$ for all $j \ge 1$. It follows that $|g(t)|_v<1$, still a contradiction. $\square$

So much for non-Archimedean case. For Archimedean case things are more complicated so we do not have enough space to cover that. Nevertheless, we have

Gelfond’s lemma. Let $f_1,\dots,f_m$ be complex polynomials in $n$ variables an set $f=f_1\cdots f_n$, then

where $d$ is the sum of the partial degrees of $f$, and $\ell_\infty(f)=\max_j|a_j|=|f|_\infty$.

Combining Gelfond’s lemma and Gauss’s lemma, we obtain

Mahler Measure

Is not actually given by Mahler initially. It was named after Mahler because he successfully extended it to multivariable cases in an elegant way. We will cover the original motivation anyway.

Original Version and Lehmer’s Conjecture

Say we want to find prime numbers large enough. Pierce came up with an idea. Consider $p(x) \in \mathbb{Z}[x]$, which is factored into

Consider $\Delta_n=\prod_i(\alpha^n_i-1)$. Then by some Galois theory, this is indeed an integer. So perhaps we may find some interesting integers in the factors of $\Delta_n$. Also, we expect it to grow slowly. Lehmer studied $\frac{\Delta_{n+1}}{\Delta_n}$ and observed that

So it makes sense to compare all roots of $p(x)$ with $1$. He therefore suggested the following function related to $p(x)$:

This number appears if we consider $\lim_{n \to \infty}\Delta_{n+1}/\Delta_n$.

He also asked the following question, which is now understood as Lehmer conjecture, although in his paper he addressed it as a problem instead of a conjecture:

Is there a constant $c$ such that, $M(p)>1 \implies M(p)>c$?

It remains open but we can mention some key bounds.

• Lehmer himself found that

and actually this is the finest result that has ever been discovered. It was because of this discovery that he gave his problem.

This polynomial has also led to the discovery of a large prime number $\sqrt{\Delta_{379}}=1, 794, 327, 140, 357$, although by studying $x^3-x-1$, we have found a bigger prime number $\Delta_{127}=3, 233, 514, 251, 032, 733$.

• Breusch (and later Smyth) discovered that if $p$ is monic, irreducible and nonreciprocal, i.e. it does not satisfy $p(x)=\pm x^{\deg p}f(1/x)$, then

• E. Dobrowlolski found that, t if $p(x)$ is monic, irreducible and noncyclotomic, and
  has degree $d$ then

for some $c>0$.

The General Version and Jensen’s Formula

Definition. For $f \in \mathbb{C}[x_1,\dots,x_n]$, the Mahler measure is defined to be

where $d\mu_i=\frac{1}{2\pi}d\theta_i$, i.e., $d\mu_1\dots d\mu_n$ corresponds to the (completion of) Harr measure on $\mathbb{T}^n$ with total measure $1$.

We see through Jensen’s formula that when $n=1$ this coincides with what we have defined before. Observe first that $M(fg)=M(f)M(g)$. Consider $f(t)=a\prod_{i=1}^{d}(t-\alpha_i)$, then

On the other hand, as an exercise in complex analysis, one can show that

Combining them, we see

Taking the logarithm we also obtain Jensen’s formula

We first give a reasonable and useful estimation of $M(f)$, which will be used to prove the Northcott’s theorem.

Definition. For $f(t)=a_dt^d+\dots+a_0$, the $\ell_p$-norm of $f$ is naturally defined to be

For $p=\infty$, we have $\ell_\infty(f)=\max_j|a_j|$.

Lemma 1. Notation being above, $M(f) \le \ell_1(f)$ and

Proof. To begin with, we observe those obvious ones. First of all,

Therefore

Next, by Jensen’s inequality

However, by Parseval’s formula, the last term equals

For the remaining inequality, we use Vieta’s formula

and therefore

for all $0 \le r \le d$. Replacing $|a_{d-r}|$ with $\ell_\infty(f)$, we have finished the proof. $\square$

Before proving Northcott’s theorem, we show the connection between Mahler measure and heights.

Proposition 1. Let $\alpha \in \overline{\mathbb{Q}}$ and let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Z}$. Then

and

Proof. Put $d=\deg(\alpha)$ and write

Choose a number field $K$ that contains $\alpha$ and is a Galois extension of $\mathbb{Q}$, with Galois group $G$. Then $(\sigma\alpha:\sigma \in G)$ contains every conjugate of $\alpha$ exactly $[K:\mathbb{Q}]/d$ times. Since $a_0,\dots,a_d$ are coprime, for any non-Archimedean absolute value $v \in M_K$, we must have $\max_i|a_i|_v=|f|_v=1$. Combining with Gauss’s lemma and Galois theory, we see

Now we are ready to compute the height of $\alpha$ to rediscover the Mahler’s measure. Notice that

We therefore obtain

The last term corresponds to what we have computed above about non-Archimedean absolute values so we break it down a little bit:

for some $u \mid \infty$, according to the product formula. On the other hand, for $v \mid \infty$,

All in all,

The second assertion follows immediately because

Northcott’s Theorem

The set of non-zero algebraic integers of height $0$ lies on the unit circle, and they are actually roots of unit, by Kronecker’s theorem. However keep in mind that algebraic integers on the unit circle are not necessarily roots of units. See this short paper.

When it comes to algebraic integers of small heights, things may get complicated, but Northcott’s theorem assures that we will be studying a finite set.

Northcott’s Theorem. Given an integer $N>0$ and a real number $H \ge1$, there are only a finite number of algebraic integers $\alpha$ satisfying $\deg(\alpha) \le N$ and $h(\alpha) \le \log H$.

Proof. Let $\alpha$ be a algebraic integer of degree $d<N$ and height $h(\alpha) \le \log H$. Suppose $f(t)=a_dt^d+\dots+a_0 \in \mathbb{Z}[t]$ is the minimal polynomial of $\alpha$. Then lemma 1 shows us that

On the other hand, by proposition 1,

we have actually

This gives rise to no more than $(2\lfloor (2H)^d \rfloor+1)^{d+1}$ distinct polynomials $f$, which produces at most $d(2\lfloor (2H)^d \rfloor+1)^{d+1}<\infty$ algebraic integers. Ranging through all $d \le N$ we get what we want. $\square$

We also have the Northcott property, where we do not care about degrees. A set $L$ of algebraic integers is said to satisfy Northcott property if, for every $T>0$, the set

is finite. Such a set $L$ is said to satisfy Bogomolov property if, there exists $T>0$ such that the set

is empty. As a matter of elementary topology, Northcott property implies Bogomolov property. It would be quite interesting if $L$ is a field. This paper can be quite interesting.

References / Further Reading

• Erico Bombieri, Walter Gubler, Heights in Diophantine Geometry.

• Michel Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, Transcendence Properties of the Exponential Function in Several Variables.

• Chris Smyth, THE MAHLER MEASURE OF ALGEBRAIC NUMBERS: A SURVEY.

Introduction

Let $F$ be a non-Archimedean local field, meaning that $F$ is complete under the metric induced by a non-Archimedean absolute value $|\cdot|$. Consider the ring of integers

and its unique prime (hence maximal) ideal

The residue field $k=\mathfrak{o}_F/\mathfrak{p}$ is finite because it is compact and discrete. For compactness notice that $\mathfrak{o}_F$ is compact, and the canonical projection $\mathfrak{o}_F \to k$ is open. For discreteness, notice that $\mathfrak{p}$ is open, connected and contains the unit.

Let $f \in \mathfrak{o}_F[x]$ be a polynomial. Hensel’s lemma states that, if $\overline{f} \in k[x]$, the reduction of $f$, has a simple root $a$ in $k$, then the root can be lifted to a root of $f$ in $\mathfrak{o}_F$ and hence $F$. This blog post is intended to offer a well-organised proof of this lemma.

To do this, we need to use Newton’s method of approximating roots of $f(x)=0$, something like

We know that $a_n \to \zeta$ where $f(\zeta)=0$ at a $A^{2^n}$ speed for some constant $A$, in calculus (do Walter Rudin’s exercise 5.25 of Principles of Mathematical Analysis if you are not familiar with it, I heartily recommend.). Now we will steal Newton’s method into number theory to find roots in a non-Archimedean field, which is violently different from $\mathbb{R}$, the playground of elementary calculus.

We will also use induction, in the form of which I would like to call “double induction”. Instead of claiming that $P(n)$ is true for all $n$, we claim that $P(n)$ and $Q(n)$ are true for all $n$. When proving $P(n+1)$, we may use $Q(n)$, and vice versa.

This method is inspired by this lecture note, where actually a “quadra induction” is used, and everything is proved altogether. Nevertheless, I would like to argue that, the quadra induction is too dense to expose the motivation and intuition of this proof. Therefore, we reduce the induction into two arguments and derive the rest with more reasonings.

Hensel’s Lemma

Hensel’s Lemma. Let $F$ be a non-Archimedean local field with ring of integers $\mathfrak{o}_F=\{\alpha \in F:|\alpha| \le 1\}$ and prime ideal $\mathfrak{p}=\{\alpha \in F:|\alpha|<1\}$. Let $f \in \mathfrak{o}_F[x]$ be a polynomial whose reduction $\overline{f} \in k[x]$ has a simple root $a \in k$, then $a$ can be lifted to $\alpha \equiv a \mod \mathfrak{p}$, such that $f(\alpha)=0$.

By simple root we mean $\overline{f}(a)=0$ but $\overline{f}’(a) \ne 0$. Before we prove this lemma, we see some examples.

Examples and Applications

Square Root of 2 in 7-adic Numbers

Put $F=\mathbb{Q}_7$. Then $\mathfrak{o}_F=\mathbb{Z}_7$, $\mathfrak{p}=7\mathbb{Z}_7$ and $k=\mathbb{F}_7$. We show that square roots of $2$ are in $F$. Note $\overline{f}(x)=x^2-2=(x-3)(x+3) \in k[x]$, we therefore two simple roots of $\overline{f}$, namely $3$ and $-3$. Lifting to $\mathfrak{o}_F$, we have two roots $\alpha_1 \equiv 3 \mod 7\mathbb{Z}_7$ and $\alpha_2 \equiv -3 \mod 7\mathbb{Z}_7$, of $f$. For $\alpha_1$, we have

Hence we can put $\alpha=\sqrt{2}=3+7+2\cdot 7^2+6\cdot 7^3\cdots\in\mathbb{Z}_7 \subset \mathbb{Q}_7$. Likewise $\alpha_2$ can be understood as $-\sqrt{2}$. This expansion is totally different from our understanding in $\mathbb{Q}$ or $\mathbb{R}$.

Roots of Unity

Since $k$ is a finite field, we see $k^\times$ is a cyclic group of order $q-1$ where $q=p^n=|k|$ for some prime $p$. It follows that $x^{q-1}=1$ for all $x \in k^\times$. Therefore $f(x)=x^{q-1}-1$ has $q-1$ distinct roots in $k$. By Hensel’s lemma, $F$ contains all $(q-1)$st roots of unity. It does not matter whether $F$ is isomorphic to $\mathbb{Q}_p$ or $\mathbb{F}_q((t))$.

Proof of Hensel’s Lemma (with Explanation)

Pick any $a_0 \in \mathfrak{o}_F$ that is a lift of $a\mod\mathfrak{p}$. Define

then we claim that $a_n$ converges to the root we are looking for.

Step 1 - Establishing A Sequence by Newton’s Method

First of all, we need to show that $a_n \in \mathfrak{o}_F$, i.e., $|a_n| \le 1$ for all $n$. It suffices to show that $|f(a_{n-1})/f’(a_{n-1})| \le 1$. We firstly observe the case when $n=1$.

Since $\overline{f}(a)=0$ but $\overline{f}’(a) \ne 0$, we have $f(a_0) \in \mathfrak{p}$ but $f’(a_0)\not\in\mathfrak{p}$. As a result, $|f(a_0)|<1$ but $|f’(a_0)|=1$. As a result, $|f(a_0)/f’(a_0)|<1$, which implies that $f(a_0)/f’(a_0) \in \mathfrak{o}_F$ and therefore $a_1 \in \mathfrak{o}_F$.

By Taylor’s theorem.

for some $g_n \in \mathfrak{o}_F[x]$. When $n=1$, we see $g_1(a_1) \in \mathfrak{o}_F$ and as a result $|g_1(a_1)| \le 1$. Therefore

Since $a_1 \in \mathfrak{o}_F$, we also see that $f(a_1) \in \mathfrak{o}_F$ hence its absolute value is not greater than $1$. As a result $|f(a_1)/f’(a_1)| \le 1$, which implies that $a_2 \in \mathfrak{o}_F$.

This inspires us to claim the following two statements:

(a) $|f(a_n)| < 1$ for all $n \ge 0$.

(b) $|f’(a_n)|=|f’(a_0)|=1$ for all $n \ge 0$.

We have verified (a) and (b) for $n=0$ and $n=1$. Now assume that (a) and (b) are true for $n-1$, then, for $n$, we will verify as follows.

First of all, by (a) and (b) for $n-1$, we see $a_n \in \mathfrak{o}_F$.

Consider the Taylor’s expansion

where $h_n \in \mathfrak{o}_F[x]$. It follows that $|h_n(a_n)| \le 1$. Since $|f’(a_{n-1})|=1$, by (b) we actually have

To prove (b) for $n$, we consider the Taylor’s expansion

Notice that since $a_n \in \mathfrak{o}_F$, we have $f’’(a_{n-1}),g_n(a_n) \in \mathfrak{o}_F$. By (a) and (b) for $n-1$, we see

Hence

bearing in mind that for a non-Archimedean absolute value, $|x+y|=\max\{|x|,|y|\}$ iff $|x| \ne |y|$. Through this process we have also proved (b).

Step 2 - Validating the Convergence

We need to show that $\{a_n\}$ is a Cauchy sequence. To do this, it suffices to show that $|f(a_n)| \to 0$ sufficiently quick. Recall in the proof of (a) we have shown that $|f(a_n)| \le |f(a_{n-1})|^2$ for all $n$. By applying this relation inductively, we see $|f(a_n)| \le |f(a_0)|^{2^n}$. Since $|f(a_0)|<1$, it follows that $|f(a_n)| \to 0$ as $n \to \infty$.

For any $\varepsilon>0$, there exists $N>0$ such that $|f(a_n)| <\varepsilon$ for all $n \ge N$. As a result, for all $m>n>N$, we have

Therefore $\{a_n\}$ is Cauchy. Since $F$ is complete, $a_n$ converges to some $\alpha \in \mathfrak{o}_F \subset F$ such that $f(\alpha)=\lim_{n \to \infty}f(a_n)=0$.

Step 3 - Validating the Congruence

In local fields, congruence is determined by inequality. In fact, we only need to show that $|\alpha-a_0|<1$, which means that $\alpha-a_0 \in \mathfrak{p}$, and therefore $\alpha \equiv a \mod \mathfrak{p}$ as expected. To do this, we show by induction that $|a_n-a_0|<1$. For $n=1$ we see $|a_1-a_0|=|f_0|<1$.

Suppose $|a_{n-1}-a_0|<1$ then

Therefore $|\alpha-a_0|=\lim_{n \to \infty}|a_n-a_0|<1$, from which the result follows. $\square$

Stronger Version

In fact we have not explicitly used the fact that $a$ is a simple root. We only used the fact that $|f(a_0)|<1$ but $|f’(a_0)|=1$. Moreover, what really matters here is that $|f(a_n)|$ converges to $0$ quick enough. Therefore $1$ may be replaced by a smaller constant. For this reason we introduce a stronger version of Hensel’s lemma.

Hensel’s lemma, stronger version. Let $F$ be a non-Archimedean local field with ring of integers $\mathfrak{o}_F$. Suppose there exists $a \in \mathfrak{o}_F$ such that $|f(a)|<|f’(a)|^2$, then there exists some $b \in \mathfrak{o}_F$ such that $f(b)=0$ and $|b-a|<|f’(a)|$.

Instead of asserting $|f’(a_n)|=1$ for all $n$, we claim that $|f’(a_n)|=|f’(a_0)|$ (as it should be!). Instead of asserting $|f(a_n)|<1$, we claim that $|f(a_n)| \le \lambda^{2^n}|f’(a_0)|$ where $\lambda=|f(a_0)|/|f’(a_0)|^2$. The proof will be nearly the same.

For example, we can find a square root of $257$ in $\mathbb{Z}_2 \subset \mathbb{Q}_2$. The polynomial $f(x)=x^2-257$ is reduced to $\overline{f}(x)=x^2-1=(x-1)^2$ in $\mathbb{F}_2[x]$, where $1$ is not a simple root. Therefore we cannot apply the original version of Hensel’s lemma to this polynomial. Nevertheless, we see $f(1)=-256$ and $f’(1)=2$. Therefore $|f(1)|=\frac{1}{2^8}$ while $|f’(1)|=\frac{1}{2}$. We can apply Newton’s method here to find a square root of $257$ without worrying about repeated roots.

Ending Remarks

There are a lot of variants of Hensel’s lemma, for example you can do exercise 10.9 of Atiyah-MacDonald. In fact, we later even have Henselian ring and Henselisation of a ring.

There are some other proofs of Hensel’s lemma in this post, for example, since Newton’s method can also be understood as a contraction mapping, we can also prove it using properties of contraction mapping (see K. Conrad’s note).