Proving V+J=k[x] using induction for k with polynomial ring in n variables

  • Context: Graduate 
  • Thread starter Thread starter grimster
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving that the sum of the vector space V and the ideal J equals the polynomial ring k[x] for a field k with q elements. Here, J is defined as the sum of polynomials of the form X_i^q - X_i, and V consists of polynomials with degrees in X_i less than q. The proof involves using induction on the number k, specifically analyzing the sum of degrees of monomials exceeding q. The goal is to demonstrate that if the property holds for k0, it also holds for k0 itself.

PREREQUISITES
  • Understanding of polynomial rings, specifically k[X] in n variables.
  • Familiarity with vector spaces and ideals in algebra.
  • Knowledge of induction principles in mathematical proofs.
  • Concept of degrees of polynomials and monomials.
NEXT STEPS
  • Study the properties of polynomial rings in multiple variables.
  • Learn about vector spaces and their relationship with ideals in algebra.
  • Explore induction techniques in algebraic proofs.
  • Investigate the implications of reducing polynomials based on their degrees.
USEFUL FOR

Mathematicians, algebraists, and students studying abstract algebra, particularly those interested in polynomial rings and vector space theory.

grimster
Messages
39
Reaction score
0
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.
 
Physics news on Phys.org
[itex]$ k0=\sum\limits_{j\in \left\{ 1,...,\text{number of monomials}\right\} }L_{j}$[/itex]
 
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?
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K