Hensel's Lemma - Examples

Examples

Suppose that p is an odd prime number and a is a quadratic residue modulo p that is nonzero mod p. Then Hensel's lemma implies that a has a square root in the ring of p-adic integers Z_p. Indeed, let f(x)=x2-a. Its derivative is 2x, so if r is a square root of a mod p we have

and ,

where the second condition depends on p not being 2. The basic version of Hensel's lemma tells us that starting from r₁= r we can recursively construct a sequence of integers { r_k } such that

This sequence converges to some p-adic integer b and b2=a. In fact, b is the unique square root of a in Z_p congruent to r₁ modulo p. Conversely, if a is a perfect square in Z_p 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:

that is,

And sure enough, . (If we had used the Newton method recursion directly in the 7-adics, then r₂ = r₁ - f(r₁)/f'(r₁) = 3 - 7/6 = 11/6, and 11/6 ≡ 10 mod 72.)

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 Z₇ 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 Z₇ 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 f(x) = x2 - 17 and a = 1. Then f(a) = -16 and f'(a) = 2, so |f(a)|₂ < |f′(a)|₂2, which implies there is a unique 2-adic integer b satisfying b2 = 17 and |b- a|₂ < |f'(a)|₂ = 1/2, 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 roots of x2 - 17 from one modulus 2k to the next 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

1 + 23 + 25 + 26 + 27 + 29 + 210 + ..., 1 + 2 + 22 + 24 + 28 + 211...

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 Z₃. Let f(x) = x3 - c and take initial approximation a = 1. The basic Hensel's lemma can't be used to find roots of f(x) since f'(r) ≡ 0 mod 3 for every r. To apply the general version of Hensel's lemma we want |f(1)|₃ < |f'(1)|₃2, which means c ≡ 1 mod 27. 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 which is 1 mod p2 is a p-th power in Z_p. (This is false when p is 2.)