# Proving that V+J=k[x]

1. May 6, 2005

### grimster

J=sum from i=1->n of <X_i^q - X_i>
V={polynomials f with deg in X_i < q)

k is a field with q elements. k[X] is the polynomial ring in "n" variables".

i am supposed to prove that V+J=k[x].

i was told that this could be done with induction for the number "k", by using the following notion.given a polynomial f E k[x]:

k0=the sum of L_j. j=1..."number of monomials".
L_j is the sum of of the "degrees" of the "indeterminates/variables" which are >q in monomial "j".

so L_j=sum of all s_i, where s_i > q. s_i are the exponents of the monomial which are >q.

so somehow i have to show that this is true for <k0 and then it is true for k0. or something like that?

does anyone know what i'm saying here? it might be a little confusing. just ask if there's anything you don't understand.

2. May 6, 2005

### grimster

$k0=\sum\limits_{j\in \left\{ 1,...,\text{number of monomials}\right\} }L_{j}$

3. May 6, 2005

### grimster

i was thinking that if given a polynomial f in k[x].

then if i reduce all monomials in f such that all exponents are < q. then the reduced f will be in V, not?