# Hensel's lemma (Understanding it's Proof)

1. Jul 20, 2010

### sidm

We're are looking at a field, K, complete with respect to a (normalized nonarchimedean) valuation, ||, and let A be it's discrete valuation ring (all elements of K with absolute value less than or equal to 1) with maximal ideal m=(p), it's residue field k=A/m...now Hensel's lemma can be stated as follows: Let f be monic in A[x] and f' its image in k[x], if f'=g'h' in k[x] with g',h' monic and relatively prime in k[x] then there are g,h in A[x] both monic such that f=gh and g and h are relatively prime with g and h congruent to g' and h' respectively modulo m.

First off let me see if i got this straight: Say we're looking at Zp (completion of integers w.r.t. p) then to factor a monic polynomial in Zp[x] we only need to factor it over Zp/(p)[x] (which is isomorphic to F_p[x]!) ?

Now the proof is long winded but I will sketch it, it involves showing first that if the g' and h' are relatively prime over the residue field then g and h are in A[x] with u and v in A[x] with degrees less than g and h respectively s.t. gv+hu=1. Then we prove that if such a pair exists then it's unique. Now the last part is what i'm having trouble with, existence:

They construct the polynomials inductively at each step producing a g_n,h_n such that g_n is congruent to g_0 modulo m and similarly for h_n & h_o also that f=h_ng_n modulo m^{n+1}. The base case is obviously what we've been given as a hypothesis so then they proceed to build the next polynomials, they say they need a u and v with degu<deg(g_0), degv(v)<deg(h_0) such that f=(h_n+p^{n+1}v)(g_n+p^{n+1}u) modulo p^{n+2}. Note here p is the generator for m.

This is where i'm confused and also where the proof ends, why do the degrees of the polynomials of u and v have to be controlled?

So say we have all our h_n's and g_n's then we would take their limit (i.e the limits of the coefficients) which are well defined because the sequence defined by say the coefficient of x^i in the h_n's are cauchy:

say (m>n)
|h_n-h_m| is no bigger than |p^{n+1}| and the absolute value consists precisely of the sum of the differences of coefficients. But how do we actually know that each of these differences is small? It's clear that the total sum of these differences goes to zero but i dont know of a way to talk about the convergence of polynomials other than coefficient-wise.

Any help would be appreciated.

2. Jul 20, 2010

### sidm

well i have an answer tomy last question: h_n-h_m=p^{n+1}(something) thus the difference between coefficients of say x_i is a multiple of p^{n+1} is thus small. Thus h and g exist and clearly their product is equal to f:

f-gh has arbitrarily small absolute value (triangle inequality with h_ng_n for large enough n) so must be zero.

Still i'm not absolutely certain as to why the degrees of u and v above have to be controlled: maybe this is to maintain the monicity and degree of each iterant of the polynomial?

3. Jul 20, 2010

### Hurkyl

Staff Emeritus
That sounds reasonable.

You know that g and h are monic, and you know their degrees. Modulo pk, gn couldn't possibly be congruent to g if gn wasn't also monic and of the same degree.

Sure, you could find a different Cauchy sequence that converges to g, but why?