Hensel’s Lemma

Local fields satisfy a collection of related theorems, any of which can be called Hensel’s lemma, after the person who discovered/invented the p-adic numbers. In this post, K is always a local field, and v is the absolute value on it. R is the ring of all a in K such that v(a) <= 1; it’s called the discrete valuation ring of K, and is in many ways analogous to the ring of integers of a number field. P is the ideal of all a in R with v(a) < 1, which is called the valuation ideal. Two polynomials over R are said to be equivalent modulo an ideal I iff their difference only has coefficients in I.

Hensel’s lemma concerns itself with polynomials over R. It states that if f = g1h1 mod P, g0 is monic (recall that this means g‘s leading coefficient is 1), and g1 and h1 are coprime, then we can find g and h such that f = gh, g = g1 and h = h1 mod P, and g is monic.

The basic idea of the proof comes from the definition of the discrete valuation ring R as having elements defined by their classes mod P, P^2, P^3… Although these elements come from the residue classes of elements of a ring of integers, in fact it’s clear that if R is the completion of a ring of integers S, then R/P = S/P, R/P^2 = S/P^2, etc. The completeness of R means that every compatible set of a residue class mod P, a class mod P^2, etc., yields an element of R.

With that in mind, let m be the degree of f and r be the degree of g0. The degree of h0 is then at most mr. Suppose that for any n, there are polynomials g(n) and h(n) such that,

1. g(n) = g(n-1) and h(n) = h(n-1) mod P^(n-1);

2. g(n) is monic of degree r;

3. h(n) has degree at most mr; and

4. f = g(n)h(n) mod P^n.

Then condition 1 ensures that the sequences g(n) and h(n) converge in K to g and h respectively, with gh = f. In fact, it’s enough to prove this by induction, i.e. to show that if we can define the sequences of polynomials for all integers less than n, then we can also define them for n.

Now, let p be any element of P that is not contained in P^2. Then p is prime and P = (p) in R. We assume that g(n) and h(n) exist and satisfy all four conditions. Then fg(n)h(n) is in (p^n), so a(n) = (fg(n)h(n))/p^n is in R[x]. g(n) and h(n) are coprime, since any common factor would have to be congruent mod P to a common factor of g1 and h1, which is impossible since g(n) is monic (in fact, it’s enough for g(n) to have a coefficient not in P). Therefore, the ideal they generate over R/P is the whole ring, since R/P[x] is a polynomial ring over a field. Therefore, it’s possible to express a(n) as g(n)b(n) + h(n)c(n) mod P for some polynomials b(n) and c(n).

Now, a(n) has degree at most m since g(n)h(n) and f both have degree at most m. Since R/P[x] is a polynomial ring over a field, it’s Euclidean with respect to the degree function; therefore, it’s possible to write c(n) as g(n)d(n) + e(n) mod P where e(n) is 0 or has degree less than r. Thus, we can assume that deg(c(n)) < r, or else we can change b to bdh and c to e. Then deg(h(n)c(n)) < m, so deg(g(n)b(n)) <= m, which means that deg(b(n)) <= mr.

Now, let g(n+1) = g(n) + c(n)p^n and h(n+1) = h(n) + b(n)p^n. Obviously, condition 1 is satisfies. Condition 2 is satisfied since deg(c(n)) < r. Condition 3 is satisfied since deg(b(n)) <= mr. Condition 4 is satisfied since g(n+1)h(n+1) = g(n)h(n) + (p^n)(g(n)b(n) + h(n)c(n)) + (p^2n)b(n)c(n) = g(n)h(n) + a(n)p^n = f mod P^(n+1).

Roughly, Hensel’s lemma says that polynomials that factor mod P factor over R. This is immensely useful; for example, it implies that every element in 1 + P has roots of any order coprime to the size of R/P. To see why, first note that if R/P is finite, then it’s necessarily an extension of Z/pZ for some prime integer p. The polynomial x^n – 1, where n isn’t divisible by p, has no repeated roots in any extension of Z/pZ, because any repeated root would have to divide both the polynomial and its derivative nx^(n-1), which is of course coprime to x^n – 1. The reason it doesn’t work if p divides n is that then n = 0 mod p, so the derivative is 0.

Anyway, if a = 1 mod P, then x^na = x^n – 1 = (x-1)(x^(n-1) + … + x + 1) mod P, x-1 and x^(n-1) + … + x + 1 are coprime, and x-1 is of course monic. Then by Hensel’s lemma, x^na = g(x)h(x), where g(x) = x – 1 mod P, i.e. g(x) = xb where b = 1 mod P. Obviously b is a root of the polynomial, then, so that b^n = a.

Incidentally, this gives a characterization of 1 + P as the set of all elements in K that have roots of all orders not divisible by p. The reason for that is that any element of K that isn’t a unit of R, i.e. has valuation 1, has valuation (1/p)^m for some nonzero integer m. Since the valuation is multiplicative, the element has roots only to orders dividing m. If the element is a unit of R that is in a + P with a != 1, then it has no root of order |R/P| – 1, i.e. the size of the multiplicative group of R/P, because in R/P, a^(|R/P| – 1) = 1 for all a.

This is where ramification screws things up, by the way. If K is the completion of a number field K1, then it’s helpful to know that every automorphism of K (i.e. isomorphism from K to itself) preserves the valuation. The automorphism will send 1 + P to itself, because 1 + P has the purely algebraic characterization given above.

If P = (p), where p is a prime integer, then there’s no problem, since p has the purely algebraic characterization as the number given by adding 1 to itself p times. In that case, the automorphism sends R to itself, since a in R is characterized as 1 + ap is in 1 + P, and then P^n is characterized as (p^n)R. Given some elementary analysis, this shows that an automorphism is continuous with respect to the valuation, so it’s enough to look at its action on K1, which is dense in K. In plain language, every automorphism of K is induced by an automorphism of K1, and we’re done; if K1 = Q, then it proves Q(p) has no non-trivial automorphisms for any p.

Unfortunately, if p ramifies, things get complicated. In a P-adic valuation of K1 where P^2 divides (p), p will not generate P in K, but only some power of P.

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