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# 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 mk. If r is an integer such that

then 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 sr = *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 modulo

## Hensel 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 rs 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 pk+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
we can derive a sequence of solutions rk+1, rk+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 pk+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 2k.

## Hensel's lemma for 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 thatand

## Examples

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 r1= r we can recursively construct a sequence of integerssuch that:
This sequence converges to some p-adic integer b which satisfies b2= a. In fact, b is the unique square root of a incongruent to r1modulo 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 2kto 2k+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 x2 − 17 mod 2k, 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 + 23and 25 = 1 + 23
• 2
4.7 = 1 + 2 + 22and 23 = 1 + 2 + 22
• 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 p2is a p-th power in(This is false for p = 2.)

## Generalizations

Suppose A is acommutative ring, complete with respect to anidealand letaA is called an "approximate root" of f, if

If f has an approximate root then it has an exact root bA "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:

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 *An
• to itself, and let
denote itsJacobian matrix. Suppose a = (a1, ..., an) ∈ *An
• is an approximate solution to f = 0 in the sense that
Then there is some b = (b1, ..., bn) ∈ *An
• satisfying f(b) = 0, i.e.,
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.

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 Ah containing A such that Ah is Henselian with respect to mAh. This Ah is called the Henselization of A. If A is noetherian, Ah will also be noetherian, and Ah is manifestly algebraic as it is constructed as a limit of étale neighbourhoods. This means that Ah is usually much smaller than the completion Â while still retaining the Henselian property and remaining in the same category.

• Hasse–Minkowski theorem

• Newton polygon

## References

[1]
Citation Linkopenlibrary.orgSerge Lang, Algebraic Number Theory, Addison-Wesley Publishing Company, 1970, p. 43
Sep 21, 2019, 6:05 PM
[2]
Sep 21, 2019, 6:05 PM
[3]
Sep 21, 2019, 6:05 PM
[4]