Hensel's lemma

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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 univariate polynomial has a simple root modulo a prime number p, then this root can be lifted to a unique root modulo any higher power of p. More generally, if a polynomial factors modulo p into two coprime polynomials, this factorization can be lifted to a factorization modulo any higher power of p (the case of roots corresponds to the case of degree 1 for one of the factors).

By passing to the "limit" (in fact this is an inverse limit) when the power of p tends to infinity, it follows that a root or a factorization modulo p can be lifted to a root or a factorization over the p-adic integers.

These results have been widely generalized, under the same name, to the case of polynomials over an arbitrary commutative ring, where p is replaced by an ideal, and "coprime polynomials" means "polynomials that generate an ideal containing 1".

Hensel's lemma is fundamental in p-adic analysis, a branch of analytic number theory.

The proof of Hensel's lemma is constructive, and leads to an efficient algorithm for Hensel lifting, which is fundamental for factoring polynomials, and gives the most efficient known algorithm for exact linear algebra over the rational numbers.

Modular reduction and lifting

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Hensel's original lemma concerns the relation between polynomial factorization over the integers and over the integers modulo a prime number p and its powers. It can be straightforwardly extended to the case where the integers are replaced by any commutative ring, and p is replaced by any maximal ideal (indeed, the maximal ideals of   have the form   where p is a prime number).

Making this precise requires a generalization of the usual modular arithmetic, and so it is useful to define accurately the terminology that is commonly used in this context.

Let R be a commutative ring, and I an ideal of R. Reduction modulo I refers to the replacement of every element of R by its image under the canonical map   For example, if   is a polynomial with coefficients in R, its reduction modulo I, denoted   is the polynomial in   obtained by replacing the coefficients of f by their image in   Two polynomials f and g in   are congruent modulo I, denoted   if they have the same coefficients modulo I, that is if   If   a factorization of h modulo I consists in two (or more) polynomials f, g in   such that  

The lifting process is the inverse of reduction. That is, given objects depending on elements of   the lifting process replaces these elements by elements of   (or of   for some k > 1) that maps to them in a way that keeps the properties of the objects.

For example, given a polynomial   and a factorization modulo I expressed as   lifting this factorization modulo   consists of finding polynomials   such that     and   Hensel's lemma asserts that such a lifting is always possible under mild conditions; see next section.

Statement

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Originally, Hensel's lemma was stated (and proved) for lifting a factorization modulo a prime number p of a polynomial over the integers to a factorization modulo any power of p and to a factorization over the p-adic integers. This can be generalized easily, with the same proof to the case where the integers are replaced by any commutative ring, the prime number is replaced by a maximal ideal, and the p-adic integers are replaced by the completion with respect to the maximal ideal. It is this generalization, which is also widely used, that is presented here.

Let   be a maximal ideal of a commutative ring R, and

 

be a polynomial in   with a leading coefficient   not in  

Since   is a maximal ideal, the quotient ring   is a field, and   is a principal ideal domain, and, in particular, a unique factorization domain, which means that every nonzero polynomial in   can be factorized in a unique way as the product of a nonzero element of   and irreducible polynomials that are monic (that is, their leading coefficients are 1).

Hensel's lemma asserts that every factorization of h modulo   into coprime polynomials can be lifted in a unique way into a factorization modulo   for every k.

More precisely, with the above hypotheses, if   where f and g are monic and coprime modulo   then, for every positive integer k there are monic polynomials   and   such that

 

and   and   are unique (with these properties) modulo  

Lifting simple roots

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An important special case is when   In this case the coprimality hypothesis means that r is a simple root of   This gives the following special case of Hensel's lemma, which is often called also Hensel's lemma.

With above hypotheses and notations, if r is a simple root of   then r can be lifted in a unique way to a simple root of   for every positive integer n. Explicitly, for every positive integer n, there is a unique   such that   and   is a simple root of  

Lifting to adic completion

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The fact that one can lift to   for every positive integer n suggests to "pass to the limit" when n tends to the infinity. This was one of the main motivations for introducing p-adic integers.

Given a maximal ideal   of a commutative ring R, the powers of   form a basis of open neighborhoods for a topology on R, which is called the  -adic topology. The completion of this topology can be identified with the completion of the local ring   and with the inverse limit   This completion is a complete local ring, generally denoted   When R is the ring of the integers, and   where p is a prime number, this completion is the ring of p-adic integers  

The definition of the completion as an inverse limit, and the above statement of Hensel's lemma imply that every factorization into pairwise coprime polynomials modulo   of a polynomial   can be uniquely lifted to a factorization of the image of h in   Similarly, every simple root of h modulo   can be lifted to a simple root of the image of h in  

Proof

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Hensel's lemma is generally proved incrementally by lifting a factorization over   to either a factorization over   (Linear lifting), or a factorization over   (Quadratic lifting).

The main ingredient of the proof is that coprime polynomials over a field satisfy Bézout's identity. That is, if f and g are coprime univariate polynomials over a field (here  ), there are polynomials a and b such that     and

 

Bézout's identity allows defining coprime polynomials and proving Hensel's lemma, even if the ideal   is not maximal. Therefore, in the following proofs, one starts from a commutative ring R, an ideal I, a polynomial   that has a leading coefficient that is invertible modulo I (that is its image in   is a unit in  ), and factorization of h modulo I or modulo a power of I, such that the factors satisfy a Bézout's identity modulo I. In these proofs,   means  

Linear lifting

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Let I be an ideal of a commutative ring R, and   be a univariate polynomial with coefficients in R that has a leading coefficient   that is invertible modulo I (that is, the image of   in   is a unit in  ).

Suppose that for some positive integer k there is a factorization

 

such that f and g are monic polynomials that are coprime modulo I, in the sense that there exist   such that   Then, there are polynomials   such that     and

 

Under these conditions,   and   are unique modulo  

Moreover,   and   satisfy the same Bézout's identity as f and g, that is,   This follows immediately from the preceding assertions, but is needed to apply iteratively the result with increasing values of k.

The proof that follows is written for computing   and   by using only polynomials with coefficients in   or   When   and   this allows manipulating only integers modulo p.

Proof: By hypothesis,   is invertible modulo I. This means that there exists   and   such that  

Let   of degree less than   such that

 

(One may choose   but other choices may lead to simpler computations. For example, if   and   it is possible and better to choose   where the coefficients of   are integers in the interval  )

As g is monic, the Euclidean division of   by g is defined, and provides q and c such that   and   Moreover, both q and c are in   Similarly, let   with   and  

One has   Indeed, one has

 

As   is monic, the degree modulo   of   can be less than   only if  

Thus, considering congruences modulo   one has

 

So, the existence assertion is verified with

 

Uniqueness

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Let R, I, h and   as a in the preceding section. Let

 

be a factorization into coprime polynomials (in the above sense), such   The application of linear lifting for   shows the existence of   and   such that     and

 

The polynomials   and   are uniquely defined modulo   This means that, if another pair   satisfies the same conditions, then one has

 

Proof: Since a congruence modulo   implies the same concruence modulo   one can proceed by induction and suppose that the uniqueness has been proved for n − 1, the case n = 0 being trivial. That is, one can suppose that

 

By hypothesis, has

 

and thus

 

By induction hypothesis, the second term of the latter sum belongs to   and the same is thus true for the first term. As   is invertible modulo I, there exist   and   such that   Thus

 

using the induction hypothesis again.

The coprimality modulo I implies the existence of   such that   Using the induction hypothesis once more, one gets

 

Thus one has a polynomial of degree less than   that is congruent modulo   to the product of the monic polynomial g and another polynomial w. This is possible only if   and implies   Similarly,   is also in   and this proves the uniqueness.

Quadratic lifting

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Linear lifting allows lifting a factorization modulo   to a factorization modulo   Quadratic lifting allows lifting directly to a factorization modulo   at the cost of lifting also the Bézout's identity and of computing modulo   instead of modulo I (if one uses the above description of linear lifting).

For lifting up to modulo   for large N one can use either method. If, say,   a factorization modulo   requires N − 1 steps of linear lifting or only k − 1 steps of quadratic lifting. However, in the latter case the size of the coefficients that have to be manipulated increase during the computation. This implies that the best lifting method depends on the context (value of N, nature of R, multiplication algorithm that is used, hardware specificities, etc.).[citation needed]

Quadratic lifting is based on the following property.

Suppose that for some positive integer k there is a factorization

 

such that f and g are monic polynomials that are coprime modulo I, in the sense that there exist   such that   Then, there are polynomials   such that     and

 

Moreover,   and   satisfy a Bézout's identity of the form

 

(This is required for allowing iterations of quadratic lifting.)

Proof: The first assertion is exactly that of linear lifting applied with k = 1 to the ideal   instead of  

Let   One has

 

where

 

Setting   and   one gets

 

which proves the second assertion.

Explicit example

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Let  

Modulo 2, Hensel's lemma cannot be applied since the reduction of   modulo 2 is simply[1]pg 15-16

 

with 6 factors   not being relatively prime to each other. By Eisenstein's criterion, however, one can conclude that the polynomial   is irreducible in  
Over  , on the other hand, one has

 

where   is the square root of 2 in  . As 4 is not a cube in   these two factors are irreducible over  . Hence the complete factorization of   in   and   is

 

where   is a square root of 2 in   that can be obtained by lifting the above factorization.
Finally, in   the polynomial splits into

 

with all factors relatively prime to each other, so that in   and   there are 6 factors   with the (non-rational) 727-adic integers

 

Using derivatives for lifting roots

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Let   be a polynomial with integer (or p-adic integer) coefficients, and let m, k be positive integers such that mk. If r is an integer such that

 

then, for every   there exists an integer s such that

 

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

 

where   is an integer satisfying

 

Note that   so that the condition   is met. As an aside, if  , then 0, 1, or several s may exist (see Hensel Lifting below).

Derivation

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

 

From   we see that sr = tpk for some integer t. Let

 

For   we have:

 

The assumption that   is not divisible by p ensures that   has an inverse mod   which is necessarily unique. Hence a solution for t exists uniquely modulo   and s exists uniquely modulo  

Observations

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Criterion for irreducible polynomials

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Using the above hypotheses, if we consider an irreducible polynomial

 

such that  , then

 

In particular, for  , we find in  

 

but  , hence the polynomial cannot be irreducible. Whereas in   we have both values agreeing, meaning the polynomial could be irreducible. In order to determine irreducibility, the Newton polygon must be employed.[2]: 144 

Frobenius

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Note that given an   the Frobenius endomorphism   gives a nonzero polynomial   that has zero derivative

 

hence the pth roots of   do not exist in  . For  , this implies that   cannot contain the root of unity  .

Roots of unity

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Although the pth roots of unity are not contained in  , there are solutions of  . Note that

 

is never zero, so if there exists a solution, it necessarily lifts to  . Because the Frobenius gives   all of the non-zero elements   are solutions. In fact, these are the only roots of unity contained in  .[3]

Hensel lifting

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Using the lemma, one can "lift" a root r of the polynomial f modulo pk to a new root s modulo pk+1 such that rs mod pk (by taking m = 1; taking larger m follows by induction). In fact, a root modulo pk+1 is also a root modulo pk, so the roots modulo pk+1 are precisely the liftings of roots modulo pk. The new root s is congruent to r modulo p, so the new root also satisfies   So 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, provided that   for the initial root rk. This also shows that f has the same number of roots mod pk as mod pk+1, mod pk+2, or any other higher power of p, provided that the roots of f mod pk are all simple.

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

 

Then   implies   That 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 pk+1.
  • If   then every lifting of r to modulus pk+1 is a root of f(x) modulo pk+1.

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

  and r = 1. Then   and   We have   which means that no lifting of 1 to modulus 4 is a root of f(x) modulo 4.

  and r = 1. Then   and   However, since   we 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

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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 rk+1 (roots mod pk+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 rk+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 that   We just need to make sure the number   is 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 and   The 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 condition   needs 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 and   say that   and  

Examples

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Suppose that p is an odd prime and a is a non-zero quadratic residue modulo p. Then Hensel's lemma implies that a has a square root in the ring of p-adic integers   Indeed, let   If 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 integers   such that:

 

This sequence converges to some p-adic integer b which satisfies b2 = a. In fact, b is the unique square root of a in   congruent to r1 modulo p. Conversely, if a is a perfect square in   and it is not divisible by p then it is a nonzero quadratic residue mod p. Note that the quadratic reciprocity law allows 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 find   as follows:

 

Based on which the expression

 

turns into:

 

which implies   Now:

 

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

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 in   which has initial 7-adic expansion

 

If we started with the initial choice   then Hensel's lemma would produce a square root of 2 in   which 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, let   and   Then   and   so

 

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 of   from modulus 2k to 2k+1, the lifts starting with the root 1 mod 2 are as follows:

1 mod 2 → 1, 3 mod 4
1 mod 4 → 1, 5 mod 8 and 3 mod 4 → 3, 7 mod 8
1 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 16
9 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 + 23 and 25 = 1 + 23 + 24.
7 = 1 + 2 + 22 and 23 = 1 + 2 + 22 + 24.

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 in   Let   and take initial approximation a = 1. The basic Hensel's lemma cannot be used to find roots of f(x) since   for every r. To apply the general version of Hensel's lemma we want   which means   That 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 p2 is a p-th power in   (This is false for p = 2.)

Generalizations

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Suppose A is a commutative ring, complete with respect to an ideal   and let   aA 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, if   is not a zero-divisor then b is unique.

This result can be generalized to several variables as follows:

Theorem. Let A be a commutative ring that is complete with respect to ideal   Let   be a system of n polynomials in n variables over A. View   as a mapping from An to itself, and let   denote its Jacobian 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, if   for all i and   is a unit in A then there is a solution to f(b) = 0 with   for all i.

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

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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  while still retaining the Henselian property and remaining in the same category[clarification needed].

See also

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References

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  1. ^ Gras, Georges (2003). Class field theory : from theory to practice. Berlin. ISBN 978-3-662-11323-3. OCLC 883382066.{{cite book}}: CS1 maint: location missing publisher (link)
  2. ^ Neukirch, Jürgen (1999). Algebraic Number Theory. Berlin, Heidelberg: Springer Berlin Heidelberg. ISBN 978-3-662-03983-0. OCLC 851391469.
  3. ^ Conrad, Keith. "Hensel's Lemma" (PDF). p. 4.