# 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

Assume is a field complete with respect to a normalised discrete valuation . Suppose, furthermore, that is the ring of integers of (i.e. all elements of with non-negative valuation), let be such that and let denote the residue field. Let be a polynomial with coefficients in . If the reduction has a simple root (i.e. there exists such that and ), then there exists a unique such that and the reduction in .[1]

### Alternative statement

Another way of stating this (in less generality) is: let be a polynomial with integer (or *p*-adic integer) coefficients, and let *m*,*k* be positive integers such that *m* ≤ *k*. 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

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

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

From we see that *s* − *r* = *tp ^{k}* 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

## Hensel lifting

Using the lemma, one can "lift" a root *r* of the polynomial *f* modulo *p ^{k}* to a new root

*s*modulo

*p*

^{k+1}such that

*r*≡

*s*mod

*p*(by taking

^{k}*m*=1; taking larger

*m*follows by induction). In fact, a root modulo

*p*

^{k+1}is also a root modulo

*p*, so the roots modulo

^{k}*p*

^{k+1}are precisely the liftings of roots modulo

*p*. The new root

^{k}*s*is congruent to

*r*modulo

*p*, so the new root also satisfies So the lifting can be repeated, and starting from a solution

*r*of we can derive a sequence of solutions

_{k}*r*

_{k+1},

*r*

_{k+2}, ... of the same congruence for successively higher powers of

*p*, provided for the initial root

*r*. This also shows that

_{k}*f*has the same number of roots mod

*p*as mod

^{k}*p*

^{k+1}, mod

*p*

^{k+2}, or any other higher power of

*p*, provided the roots of

*f*mod

*p*are all simple.

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

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*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. 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 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 *r _{k}* (roots mod

*p*) to

^{k}*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 *p ^{k}* here is not really a denominator since

*f*(

*r*) is divisible by

_{k}*p*):

^{k}Then set

This fraction may not be an integer, but it is a *p*-adic integer, and the sequence of numbers *r _{k}* 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

*r*is precisely Newton's method for finding roots to equations in the real numbers.

_{k}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

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 *r*_{1} = *r* we can recursively construct a sequence of integers such 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* in congruent to *r*_{1} 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 2^{k} to 2^{k+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 *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
^{3}and 25 = 1 + 2^{3}+ 2^{4}. - 7 = 1 + 2 + 2
^{2}and 23 = 1 + 2 + 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 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 *p*^{2} is a *p*-th power in (This is false for *p* = 2.)

## Generalizations

Suppose *A* is a commutative ring, complete with respect to an ideal and let *a* ∈ *A* is called an "approximate root" of *f*, if

If *f* has an approximate root then it has an exact root *b* ∈ *A* "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.**Suppose*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*A*to itself, and let denote its Jacobian matrix. Suppose^{n}**a**= (*a*_{1}, ...,*a*_{n}) ∈*A*is an approximate solution to^{n}**f**=**0**in the sense that

- Then there is some
**b**= (*b*_{1}, ...,*b*_{n}) ∈*A*satisfying^{n}**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.

## 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

## References

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

- Eisenbud, David (1995),
*Commutative algebra*, Graduate Texts in Mathematics,**150**, Berlin, New York: Springer-Verlag, doi:10.1007/978-1-4612-5350-1, ISBN 978-0-387-94269-8, MR 1322960 - Milne, J. G. (1980),
*Étale cohomology*, Princeton University Press, ISBN 978-0-691-08238-7