Hensel's lemma and local extensions
Lift simple modular roots in a complete local ring and distinguish unramified extensions from Eisenstein ramification.
On this page
The question for today
What you’ll understand
- Perform a one-digit Hensel lift with a unit derivative.
- Distinguish the simple-root criterion from a stronger valuation criterion.
- Calculate ramification and residue degrees in two quadratic local extensions.
A modular root is a first approximation to a local-field root, not automatically a proof that an exact root exists. Hensel's lemma supplies a lifting theorem under derivative and completeness conditions. The simple-root version gives a unique root in a specified residue class, while stronger versions can handle some roots with nonunit derivatives. These are separate hypotheses, and failure of one test is not a theorem of nonexistence. We connect the lifting calculation with two common ways local extensions arise: enlarging the residue field or introducing a ramified uniformizer.
A unit derivative determines the next digit
For and , assume and . There is a unique root a in Z_p congruent to a₁ modulo p. To lift a root approximation modulo , write the next one as with t chosen modulo p. Taylor expansion modulo makes the correction equation linear in t, and the unit derivative makes its solution unique.
The resulting approximations are compatible and Cauchy, so completeness supplies their limit and continuity makes it an exact root. This proof does not claim that a polynomial has only one root globally: distinct simple residue roots can lift to different roots. Nor does it work by real bisection or by choosing the ordinary-real closest root. Its error control measures increasing p-adic divisibility, and its uniqueness is within the declared residue class.
Singular roots need a different theorem
A common stronger form assumes and for a₀ in Z_p. Then a root exists and is unique in the ball . This is narrower than claiming uniqueness in the entire residue class when the derivative is not a unit. Newton iteration proves the result using a strict improvement of the error valuation. The derivative and function valuations must use the same normalization.
If the simple-root hypothesis fails, roots may still exist, may branch, or may disappear at a higher congruence level. For example x²−1 has two 2-adic roots, plus and minus one, both congruent to one modulo two; a simple-root uniqueness claim at that residue would be false because the derivative is even. Different valuation-based versions of Hensel's lemma have different non-strict or ball-size conditions, so quote the actual variant being applied rather than replacing it by a slogan that every modular root lifts.
Residue extension and ramification are different
A monic polynomial in Z_p[x] whose reduction is irreducible and separable of degree f over F_p gives an unramified degree-f extension when a root is adjoined to Q_p. Its residue field enlarges to F_(p^f) while p remains a uniformizer. By contrast an Eisenstein polynomial is monic, all nonleading coefficients are divisible by p, and its constant term is not divisible by p². It is irreducible and generates a totally ramified extension; its root is a uniformizer. These are sufficient constructions, not a claim that every defining polynomial is already in one of these forms.
For x^e−p, Eisenstein gives degree e, ramification index e, and residue degree one, including cases where ramification is wild. With the extension's integer-valued uniformizer normalization, the root has valuation one and p has valuation e. Under the valuation extending v_p, the root instead has valuation 1/e. Comparing these two conventions explicitly prevents a correct local-extension construction from appearing to contradict the base-field normalization used in the root-lifting inequalities.
Work it through
Lift a square root of two at seven
For f(x)=x²−2 over Z7, start with the root a1=3 modulo seven. Lift it to a root modulo 49 and then modulo 343.
The initial residual is f(3)=7 and derivative f'(3)=6, a unit modulo seven. Write the next approximation as 3+7t. Dividing its residual congruence by seven gives 1+6t=0 modulo seven, so t=1 and a2=10.
Check f(10)=98=2 times 49. Write the next approximation as 10+49t. Modulo 343, divide by 49 to get 2+20t=0 modulo seven, equivalently 2+6t=0, giving t=2.
Thus a3=108, and 108²−2=11662=34 times 343. The compatible residues converge to the unique Z7 root congruent to three modulo seven; the root in the other initial residue class is not ruled out.
The lift is 3 modulo 7, 10 modulo 49, and 108 modulo 343. Exact integer residuals verify the claimed precision at each step.
The living notebook
Explore the model
Interactive lab · modular
Compare compatible root residues at two precisions
Pause and Reset, then set both controls to modulus m=3 and additive step a=1; after one Step the residue is 1, a root of x²−7 modulo three. Reset and set m=9 and a=4; after one Step predict residue 4, whose square minus seven is nine and which reduces to 1 modulo three. The instrument displays these selected residues by addition, not by solving a polynomial or executing Hensel iteration; verify the residual arithmetic yourself.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA valid lift must both solve the higher-precision congruence and reduce to the chosen earlier root.
Pause & reason
Check your understanding
Your turn
Try it yourself
Compare L1=Q3(i) with L2=Q3(sqrt(3)). Determine degree, ramification index, residue degree, and the extension-normalized valuation of three in each. Also give the valuation of sqrt(3) that extends v3.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Reduce x²+1 modulo three for the first extension and use Eisenstein on x²−3 for the second.
Compare with the solution
The polynomial x²+1 is irreducible and separable modulo three, so L1 has degree two, e=1 and f=2. Its residue field is F9 and its integer-normalized valuation gives v_L1(3)=1. The polynomial x²−3 is Eisenstein, so L2 has degree two, e=2 and f=1. Its root is a uniformizer with v_L2(sqrt(3))=1 and v_L2(3)=2. Dividing that valuation by e gives the extension of v3, which assigns sqrt(3) the value 1/2.
Keep the idea
Take it with you
- Unit derivatives make correction digits unique in a fixed residue class.
- Stronger Hensel variants have their own valuation and uniqueness-ball hypotheses.
- Unramified residue extensions and Eisenstein ramification affect different local invariants.
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