Hensel’s Lemma - Local Fields - Cambridge III notes

4 Hensel’s Lemma

Theorem 4.1 (Hensel’s Lemma version 1). Assuming that:

$(K, | ∙ |)$ is a complete discretely valued field
$f (X) \in O_{K} [X]$
assume $\exists a \in O_{K}$ such that $| f (a) | < | f^{'} (a) |^{2}$

Then there exists a unique

x \in O_{K}

such that

f (x) = 0

and

| x - a | < | f^{'} (a) |

Proof. Let $π \in O_{K}$ be a uniformiser and let $r = v (f^{'} (a))$ , with $v$ the normalised valuation ( $v (π) = 1$ ). We construct a sequence ${(x_{n})}_{n = 1}^{\infty}$ in $O_{K}$ such that:

(i) $f (x_{n}) \equiv 0 (m o d π^{n + 2 r})$
(ii) $x_{n + 1} \equiv x_{n} (m o d π^{n + r})$

Take $x_{1} = a$ : then $f (x_{1}) \equiv 0 (m o d π 1 + 2 r)$ .

Now we suppose we have constructed $x_{1}, \dots, x_{n}$ satisfying (i) and (ii). Define

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} .

Since $x_{n} \equiv x_{1} (m o d π^{r + 1})$ , we have

v (f^{'} (x_{n})) = v (f^{'} (x_{i})) = r,

and hence

\frac{f (x_{n})}{f^{'} (x_{n})} \equiv 0 (m o d π^{n + r})

by (i).

It follows that $x_{n + 1} \equiv x_{n} (m o d π^{n + r})$ , so (ii) holds. Note that letting $X, Y$ be indeterminates, we have

f (X + Y) = f_{0} (X) + f_{1} (X) Y + f_{2} (X) Y^{2} + \dots,

where $f_{i} (X) \in O_{K} [X]$ and $f_{0} (X) = f (X)$ , $f_{1} (X) = f^{'} (X)$ . Thus

f (x_{n + 1}) = f (x_{n}) + c f^{'} (x_{n}) + c^{2} f_{2} (x_{N}) + \underset{\in π^{n + 2 r + 1}}{\underset{⏟}{c^{2} f_{2} (x_{n}) + \dots}}

where $c = - \frac{f (x_{n})}{f^{'} (x_{n})}$ .

Since $c \equiv 0 (m o d π^{n + r})$ and $v (f_{i} (x_{n})) \geq 0$ we have

f (x_{n + 1}) \equiv f (x_{n}) + f^{'} (x_{n}) c \equiv 0 (m o d π^{n + 2 r + 1}),

so (i) holds.

Property (ii) implies that ${(x_{n})}_{n = 1}^{\infty}$ is Cauchy, so let $x \in O_{K}$ such that $x_{n} \to x$ . Then $f (x) = \lim_{n \to \infty} f (x_{n}) = 0$ by (i).

Moreover, (ii) impies that

\begin{array}{l} a = x_{1} & \equiv x_{n} (m o d π^{r + 1}) \forall n \\ ⟹ a & \equiv x (m o d π^{r + 1}) \\ ⟹ | x - a | & < | f^{'} (a) | \end{array}

This proves existence.

Uniqueness: suppose $x^{'}$ also satisfies $f^{'} (x) = 0$ , $| x^{'} - a | < | f^{'} (a) |$ . Set $δ = x^{'} - x \neq 0$ . Then

| x^{'} - a | < | f^{'} (a) | | x - a^{'} | < | f^{'} (a) |,

and the ultrametric inequality implies

| δ | = | x - x^{'} | < | f^{'} (a) | = | f^{'} (x) | .

But

0 = f (x^{'}) = f (x + δ) = \underset{= 0}{\underset{⏟}{f (x)}} + f^{'} (x) δ + \underset{| ∙ | \leq | δ |^{2}}{\underset{⏟}{\dots}} .

Hence $| f^{'} (x) δ | \leq | δ |^{2}$ , so $| f^{'} (x) | < | δ |$ , a contradiction. □

Corollary 4.2. Let $(K, | ∙ |)$ be a complete discretely valued field. Let $f (X) \in O_{K} [X]$ and $\bar{c} \in k : = O_{K} ∕ m$ a simple root of $\bar{f} (X) : = f (X) (m o d m) \in k [X]$ . Then there exists a unique $x \in O_{K}$ such that $f (x) = 0$ , $x \equiv \bar{c} (m o d m)$ .

Proof. Apply Theorem 4.1 to a lift $c \in O_{K}$ of $\bar{c}$ . Then $| f (c) | < 1 = | f^{'} (c) |^{2}$ since $\bar{c}$ is a simple root. □

Example. $f (X) = X^{2} - 2$ has a simple root modulo 7. Thus $\sqrt{2} \in ℤ_{7} \subseteq ℚ_{7}$ .

Corollary 4.3.

{ℚ_{p}}^{\times} ∕ {({ℚ_{p}}^{\times})}^{2} ≅ {\begin{matrix} {(ℤ ∕ 2 ℤ)}^{2} & if p > 2 \\ {(ℤ ∕ 2 ℤ)}^{3} & if p = 2 \end{matrix}

Proof. Case $p > 2$ : Let $b \in {ℤ_{p}}^{\times}$ . Applying to $f (X) = X^{2} - b$ , we find that $b \in {({ℤ_{p}}^{\times})}^{2}$ if and only if $b \in {(𝔽_{p}^{\times})}^{2}$ . Thus ${ℤ_{p}}^{\times} ∕ {({ℤ_{p}}^{\times})}^{2} ≅ 𝔽_{p}^{\times} ∕ {(𝔽_{p}^{\times})}^{2} ≅ ℤ ∕ 2 ℤ$ ( $𝔽_{p}^{\times} ≅ ℤ ∕ (p - 1) ℤ$ ).

We have an isomorphism

{ℤ_{p}}^{\times} \times (ℤ, +) ≅ {ℚ_{p}}^{\times}

given by $(u, n) \mapsto u p^{n}$ . Thus

{ℚ_{p}}^{\times} ∕ {({ℚ_{p}}^{\times})}^{2} ≅ {(ℤ ∕ 2 ℤ)}^{2} .

Case $p = 2$ : Let $b \in ℤ_{2}^{\times}$ . Consider $f (X) = X^{2} - b$ . Note $f^{'} (X) = 2 X \equiv 0 (m o d 2)$ . Let $b \equiv 1 (m o d 8)$ . Then

| f (1) | = 2^{- 3} < 2^{- 2} = | f^{'} (1) |^{2} .

Hensel’s Lemma version 1 gives

b \in {(ℤ_{2}^{\times})}^{2} ⟺ b \equiv 1 (m o d 8) .

Then

ℤ_{2}^{\times} ∕ {(ℤ_{2}^{\times})}^{2} ≅ {(ℤ ∕ 8 ℤ)}^{\times} \equiv {(ℤ ∕ 2 ℤ)}^{j} .

Again using $ℚ_{2}^{\times} \equiv ℤ_{2}^{\times} \times ℤ$ , we find that $ℚ_{2}^{\times} ≅ {(ℤ ∕ 2 ℤ)}^{3}$ . □

Remark. Proof uses the iteration

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})},

which is the non-archimedean analogue of the unewton Raphson method.

Theorem 4.4 (Hensel’s Lemma version 2). Assuming that:

$(K, | ∙ |)$ is a complete discretely valued field
$f (X) \in O_{K} [X]$
$\bar{f} (X) : = f (X) (m o d m) \in k [X]$ factorises as $\bar{f} (X) = \bar{g} (X) \bar{h} (X)$ in $k [X]$
$\bar{g} (X)$ and $\bar{h} (X)$ coprime.

Then there is a factorisation

f (X) = g (X) h (X)

in $O_{K} [X]$ , with $\bar{g} (X) \equiv g (X) (m o d m)$ , $\bar{h} (X) \equiv h (X) (m o d m)$ and $\deg \bar{g} = \deg g$ .

Proof. Example Sheet 1. □

Corollary 4.5. Let $(K, | ∙ |)$ be a complete discretely valued field. Let

f (X) = a_{n} X^{n} + \dots + a_{n} \in K [X]

with $a_{0}, a_{n} \neq 0$ . If $f (X)$ is irreducible, then $| a_{i} | \leq \max (| a_{0} |, | a_{n} |)$ for all $i$ .

Proof. Upon scaling, we may assume $f (X) \in O_{K} [X]$ with $\max_{i} (| a_{i} |) = 1$ . Thus we need to show that $\max (| a_{0} |, | a_{n} |) = 1$ . If not, let $r$ minimal such that $| a_{r} | = 1$ , then $0 < r < n$ . Thus we have

\bar{f} (X) = X^{r} (a_{r} + \dots + a_{n} X^{n - r}) (m o d m) .

Then Theorem 4.4 implies $f (X) = g (X) h (X)$ with $0 < \deg < n$ . □