Hensel's lemma (Understanding it's Proof)

  • Thread starter Thread starter sidm
  • Start date Start date
  • Tags Tags
    Proof
sidm
Messages
16
Reaction score
0
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 don't know of a way to talk about the convergence of polynomials other than coefficient-wise.

Any help would be appreciated.
 
Physics news on Phys.org
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?
 
sidm said:
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?
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?
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
Back
Top