# Hensel's lemma

# Hensel's lemma

In mathematics, **Hensel's lemma**, also known as **Hensel's lifting lemma**, named after Kurt Hensel, is a result in modular arithmetic, stating that if a polynomial equation has a simple root modulo a prime number *p*, then this root corresponds to a unique root of the same equation modulo any higher power of *p*, which can be found by iteratively "lifting" the solution modulo successive powers of *p*. More generally it is used as a generic name for analogues for complete commutative rings (including *p*-adic fields in particular) of the Newton method for solving equations. Since *p*-adic analysis is in some ways simpler than real analysis, there are relatively neat criteria guaranteeing a root of a polynomial.

Statement

Many equivalent statements of Hensel's lemma exist. Arguably the most common statement is the following.

General statement

`Assumeis a field complete with respect to a normalised discretevaluation. Suppose, furthermore, thatis the ring of integers of(i.e. all elements ofwith non-negative valuation), letbe such thatand letdenote theresidue field. Letbe apolynomialwith coefficients in. If the reductionhas a simple root (i.e. there existssuch thatand), then there exists a uniquesuch thatand the reductionin.`

^{[1]}Alternative statement

`Another way of stating this (in less generality) is: letbe apolynomialwithinteger(or`

*p*-adic integer) coefficients, and let*m*,*k*be positive integers such that*m*≤*k*. If*r*is an integer such thatthen there exists an integer *s* such that

Furthermore, this *s* is unique modulo *p**k*+*m*, and can be computed explicitly as the integer such that

`whereis an integer satisfying`

`Note thatso that the conditionis met. As an aside, if, then 0, 1, or several`

*s*may exist (see Hensel Lifting below).Derivation

We use the Taylor expansion of *f* around *r* to write:

`Fromwe see that`

*s*−*r*= *tpk- for some integer
*t*. Let

`Forwe have:`

`The assumption thatis not divisible by`

*p*ensures thathas an inverse modwhich is necessarily unique. Hence a solution for*t*exists uniquely moduloand*s*exists uniquely moduloHensel lifting

`Using the lemma, one can "lift" a root`

*r*of the polynomial*f*modulo *pk- to a new root
*s*modulo*p*

*k*+1such that*r*≡*s*mod *pk- (by taking
*m*=1; taking larger*m*follows by induction). In fact, a root modulo*p*

*k*+1is also a root modulo *pk*, so the roots modulo*p**k*+1are precisely the liftings of roots modulo *pk*. The new root*s*is congruent to*r*modulo*p*, so the new root also satisfiesSo the lifting can be repeated, and starting from a solution *rk- of

*r**k*+1,*r**k*+2, ... of the same congruence for successively higher powers of*p*, providedfor the initial root *rk*. This also shows that*f*has the same number of roots mod *pk- as mod
*p*

*k*+1, mod*p**k*+2, or any other higher power of*p*, provided the roots of*f*mod *pk- are all simple.

What happens to this process if *r* is not a simple root mod *p*? Suppose

`ThenimpliesThat is,for all integers`

*t*. Therefore, we have two cases:If then there is no lifting of

*r*to a root of*f*(*x*) modulo*p**k*+1.If then every lifting of

*r*to modulus*p**k*+1 is a root of*f*(*x*) modulo*p**k*+1.

**Example.** To see both cases we examine two different polynomials with *p* = 2:

`and`

*r*= 1. ThenandWe havewhich means that no lifting of 1 to modulus 4 is a root of*f*(*x*) modulo 4.`and`

*r*= 1. ThenandHowever, sincewe can lift our solution to modulus 4 and both lifts (i.e. 1, 3) are solutions. The derivative is still 0 modulo 2, so*a priori*we don't know whether we can lift them to modulo 8, but in fact we can, since*g*(1) is 0 mod 8 and*g*(3) is 0 mod 8, giving solutions at 1, 3, 5, and 7 mod 8. Since of these only*g*(1) and*g*(7) are 0 mod 16 we can lift only 1 and 7 to modulo 16, giving 1, 7, 9, and 15 mod 16. Of these, only 7 and 9 give*g*(*x*) = 0 mod 32, so these can be raised giving 7, 9, 23, and 25 mod 32. It turns out that for every integer*k*≥ 3, there are four liftings of 1 mod 2 to a root of*g*(*x*) mod 2*k*.Hensel's lemma for *p*-adic numbers

*p*-adic numbers

In the *p*-adic numbers, where we can make sense of rational numbers modulo powers of *p* as long as the denominator is not a multiple of *p*, the recursion from *rk* (roots mod *pk*) to *r**k*+1 (roots mod *p**k*+1) can be expressed in a much more intuitive way. Instead of choosing *t* to be an(y) integer which solves the congruence

let *t* be the rational number (the *pk* here is not really a denominator since *f*(*rk*) is divisible by *pk*):

Then set

This fraction may not be an integer, but it is a *p*-adic integer, and the sequence of numbers *rk* converges in the *p*-adic integers to a root of *f*(*x*) = 0. Moreover, the displayed recursive formula for the (new) number *r**k*+1 in terms of *rk* is precisely Newton's method for finding roots to equations in the real numbers.

`By working directly in the`

*p*-adics and using the*p*-adic absolute value, there is a version of Hensel's lemma which can be applied even if we start with a solution of*f*(*a*) ≡ 0 mod*p*such thatWe just need to make sure the numberis not exactly 0. This more general version is as follows: if there is an integer*a*which satisfies:`then there is a unique`

*p*-adic integer*b*such*f*(*b*) = 0 andThe construction of*b*amounts to showing that the recursion from Newton's method with initial value*a*converges in the*p*-adics and we let*b*be the limit. The uniqueness of*b*as a root fitting the conditionneeds additional work.`The statement of Hensel's lemma given above (taking) is a special case of this more general version, since the conditions that`

*f*(*a*) ≡ 0 mod*p*andsay thatandExamples

`Suppose that`

*p*is an odd prime and*a*is a non-zeroquadratic residuemodulo*p*. Then Hensel's lemma implies that*a*has a square root in the ring of*p*-adic integersIndeed, letIf*r*is a square root of*a*modulo*p*then:`where the second condition is dependent on the fact that`

*p*is odd. The basic version of Hensel's lemma tells us that starting from*r*1=*r*we can recursively construct a sequence of integerssuch that:`This sequence converges to some`

*p*-adic integer*b*which satisfies*b*2=*a*. In fact,*b*is the unique square root of*a*incongruent to*r*1modulo*p*. Conversely, if*a*is a perfect square inand it is not divisible by*p*then it is a nonzero quadratic residue mod*p*. Note that thequadratic reciprocity lawallows one to easily test whether*a*is a nonzero quadratic residue mod*p*, thus we get a practical way to determine which*p*-adic numbers (for*p*odd) have a*p*-adic square root, and it can be extended to cover the case*p*= 2 using the more general version of Hensel's lemma (an example with 2-adic square roots of 17 is given later).`To make the discussion above more explicit, let us find a "square root of 2" (the solution to) in the 7-adic integers. Modulo 7 one solution is 3 (we could also take 4), so we set. Hensel's lemma then allows us to findas follows:`

Based on which the expression

turns into:

`which impliesNow:`

`And sure enough,(If we had used the Newton method recursion directly in the 7-adics, thenand)`

`We can continue and find. Each time we carry out the calculation (that is, for each successive value of`

*k*), one more base 7 digit is added for the next higher power of 7. In the 7-adic integers this sequence converges, and the limit is a square root of 2 inwhich has initial 7-adic expansion`If we started with the initial choicethen Hensel's lemma would produce a square root of 2 inwhich is congruent to 4 (mod 7) instead of 3 (mod 7) and in fact this second square root would be the negative of the first square root (which is consistent with 4 = −3 mod 7).`

`As an example where the original version of Hensel's lemma is not valid but the more general one is, letandThenandso`

which implies there is a unique 2-adic integer *b* satisfying

i.e., *b* ≡ 1 mod 4. There are two square roots of 17 in the 2-adic integers, differing by a sign, and although they are congruent mod 2 they are not congruent mod 4. This is consistent with the general version of Hensel's lemma only giving us a unique 2-adic square root of 17 that is congruent to 1 mod 4 rather than mod 2. If we had started with the initial approximate root *a* = 3 then we could apply the more general Hensel's lemma again to find a unique 2-adic square root of 17 which is congruent to 3 mod 4. This is the other 2-adic square root of 17.

`In terms of lifting the roots offrom modulus 2`

*k*to 2*k*+1, the lifts starting with the root 1 mod 2 are as follows:- 1 mod 2 --> 1, 3 mod 41 mod 4 --> 1, 5 mod 8 and 3 mod 4 ---> 3, 7 mod 81 mod 8 --> 1, 9 mod 16 and 7 mod 8 ---> 7, 15 mod 16, while 3 mod 8 and 5 mod 8 don't lift to roots mod 169 mod 16 --> 9, 25 mod 32 and 7 mod 16 --> 7, 23 mod 16, while 1 mod 16 and 15 mod 16 don't lift to roots mod 32.

For every *k* at least 3, there are *four* roots of *x*2 − 17 mod 2*k*, but if we look at their 2-adic expansions we can see that in pairs they are converging to just *two* 2-adic limits. For instance, the four roots mod 32 break up into two pairs of roots which each look the same mod 16:

- 9 = 1 + 2
- 2
- 2

^{3}and 25 = 1 + 2

^{3}

^{4}.7 = 1 + 2 + 2

^{2}and 23 = 1 + 2 + 2

^{2}

^{4}.

The 2-adic square roots of 17 have expansions

`Another example where we can use the more general version of Hensel's lemma but not the basic version is a proof that any 3-adic integer`

*c*≡ 1 mod 9 is a cube inLetand take initial approximation*a*= 1. The basic Hensel's lemma cannot be used to find roots of*f*(*x*) sincefor every*r*. To apply the general version of Hensel's lemma we wantwhich meansThat is, if*c*≡ 1 mod 27 then the general Hensel's lemma tells us*f*(*x*) has a 3-adic root, so*c*is a 3-adic cube. However, we wanted to have this result under the weaker condition that*c*≡ 1 mod 9. If*c*≡ 1 mod 9 then*c*≡ 1, 10, or 19 mod 27. We can apply the general Hensel's lemma three times depending on the value of*c*mod 27: if*c*≡ 1 mod 27 then use*a*= 1, if*c*≡ 10 mod 27 then use*a*= 4 (since 4 is a root of*f*(*x*) mod 27), and if*c*≡ 19 mod 27 then use*a*= 7. (It is not true that every*c*≡ 1 mod 3 is a 3-adic cube, e.g., 4 is not a 3-adic cube since it is not a cube mod 9.)`In a similar way, after some preliminary work, Hensel's lemma can be used to show that for any`

*odd*prime number*p*, any*p*-adic integer*c*congruent to 1 modulo*p*2is a*p*-th power in(This is false for*p*= 2.)Generalizations

`Suppose`

*A*is acommutative ring, complete with respect to anidealand let*a*∈*A*is called an "approximate root" of*f*, ifIf *f* has an approximate root then it has an exact root *b* ∈ *A* "close to" *a*; that is,

`Furthermore, ifis not a zero-divisor then`

*b*is unique.This result can be generalized to several variables as follows:

- to itself, and let
- is an approximate solution to
**f**=**0**in the sense that

**Theorem.**Suppose

*A*be a commutative ring that is complete with respect to idealLetbe a system of

*n*polynomials in

*n*variables over

*A*. Viewas a mapping from *A

^{n}

**a**= (

*a*

_{1}, ...,

*a*

_{n}) ∈ *A

^{n}

- Then there is some
- satisfying
**f**(**b**) =**0**, i.e.,

**b**= (

*b*

_{1}, ...,

*b*

_{n}) ∈ *A

^{n}

- Furthermore this solution is "close" to

**a**in the sense that

`As a special case, iffor all`

*i*andis a unit in*A*then there is a solution to**f**(**b**) =**0**withfor all*i*.`When`

*n*= 1,**a**=*a*is an element of*A*andThe hypotheses of this multivariable Hensel's lemma reduce to the ones which were stated in the one-variable Hensel's lemma.Related concepts

Completeness of a ring is not a necessary condition for the ring to have the Henselian property: Goro Azumaya in 1950 defined a commutative local ring satisfying the Henselian property for the maximal ideal **m** to be a **Henselian ring**.

Masayoshi Nagata proved in the 1950s that for any commutative local ring *A* with maximal ideal **m** there always exists a smallest ring *A*h containing *A* such that *A*h is Henselian with respect to **m***A*h. This *A*h is called the **Henselization** of *A*. If *A* is noetherian, *A*h will also be noetherian, and *A*h is manifestly algebraic as it is constructed as a limit of étale neighbourhoods. This means that *A*h is usually much smaller than the completion *Â* while still retaining the Henselian property and remaining in the same category.

See also

Hasse–Minkowski theorem

Newton polygon

## References

*Algebraic Number Theory*, Addison-Wesley Publishing Company, 1970, p. 43