Math 254A. Hensel’s Lemma

Proposition. Let K = (K, | · |) be a complete non-archimedean valued field, let A be its valuation ring {a ∈ K | |a| ≤ 1}, and let f(x) ∈ A[x] . Assume that α_{0} ∈ A satisfies

|f(α_{0})| < |f′(α_{0})|^{2}  (1)

(where f′ is the derivative taken formally). Then the sequence defined by

α_{i+1} = α_{i} − \frac{f(α_{i})}{f′(α_{i})} , i ∈ ℕ

converges to a root α of f satisfying

|α − α_{0}| ≤ \frac{|f(α_{0})|}{|f′(α_{0})|} < 1 .  (2)

This root is the only root of f satisfying (2); more generally it is the only root of f satisfying

|α − α_{0}| < |f′(α_{0})| .  (3)

Proof. First we claim that if |α − α_{0}| < |f′(α_{0})| then |f′(α)| = |f′(α_{0})|. To see this, we first note that since α_{0} ∈ A and f′(x) ∈ A[x] , f′(α_{0}) ∈ A and therefore also α ∈ A. By Taylor’s formula (for polynomials) there exists β ∈ A such that

f′(α) = f′(α_{0}) + β(α − α_{0}) .

Thus

|f′(α) − f′(α_{0})| ≤ |α − α_{0}| < |f′(α_{0})|

and therefore |f′(α)| = |f′(α_{0})| by the non-archimedean property of the valuation. In particular, by (1), this holds for all α satisfying (2).

Now let c = |f(α_{0})|/|f′(α_{0})|^{2} < 1 . By induction we will show that, for all i ≥ 0 , (i). |α_{i} − α_{0}| ≤ |f(α_{0})|/|f′(α_{0})| < 1 , (ii). |f′(α_{i})| = |f′(α_{0})|, and (iii). |f(α_{i})| ≤ c^{2^{i}}|f′(α_{0})|^{2} . The base case i = 0 holds trivially. For the inductive step, assume that (i)–(iii) hold for some value of i. First, by (ii) and (iii) for i, we have

|α_{i+1} − α_{i}| = \frac{|f(α_{i})|}{|f′(α_{i})|} ≤ \frac{c^{2^{i}}|f′(α_{0})|^{2}}{|f′(α_{0})|} = c^{2^{i}}|f′(α_{0})| .  (4)

Now we show (i) for i + 1 . By (4), the inequality c < 1 , and the definition of c,

|α_{i+1} − α_{i}| ≤ c^{2^{i}}|f′(α_{0})| ≤ c|f′(α_{0})| = \frac{|f(α_{0})|}{|f′(α_{0})|} .

Combining this with (i) for i then gives (i) for i + 1 .

To show (ii), we have

|α_{i+1} − α_{0}| ≤ \frac{|f(α_{0})|}{|f′(α_{0})|} < |f′(α_{0})| .

by (i) and (1). Therefore the claim applies, which gives (ii) for i + 1 .

Finally, we show (iii). By Taylor’s formula, there exists β ∈ A such that

\begin{aligned}f(α_{i+1}) & = f(α_{i}) + f′(α_{i})(α_{i+1} − α_{i}) + β(α_{i+1} − α_{i})^{2} \\ & = f(α_{i}) + f′(α_{i}) (−\frac{f(α_{i})}{f′(α_{i})}) + β(α_{i+1} − α_{i})^{2} \end{aligned}

= β(α_{i+1} − α_{i})^{2}.

Taking absolute values and applying (4) gives

|f(α_{i+1})| ≤ |α_{i+1} − α_{i}|^{2} ≤ (c^{2^{i}}|f′(α_{0})|)^{2} = c^{2^{i+1}}|f′(α_{0})|^{2} .

This proves (iii) for i + 1 .

The sequence (α_{i}) therefore is a Cauchy sequence by (4). By continuity and (iii), its limit α is a root of f .

Finally, we prove the uniqueness statement. Suppose α and α′ are distinct roots of f satisfying (3). We then have |α − α′| < |f′(α_{0})|. But by Taylor’s formula,

f(α′) = f(α) + f′(α)(α′ − α) + β(α′ − α)^{2}

for some β ∈ A. Since f(α) = f(α′) = 0 and α ≠ α′, this gives

f′(α) = −β(α′ − α);
|f′(α)| ≤ |α′ − α| < |f′(α_{0})| .

This is a contradiction since the claim at the beginning of the proof implies that |f′(α)| = |f′(α_{0})|. □