Proving Wilson's Theorem: Finite Field Product of Non-Zero Elements is -1

Click For Summary

Homework Help Overview

The problem involves demonstrating that the product of all non-zero elements in a finite field is -1, referencing Wilson's Theorem as a related concept.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the structure of the multiplicative group of non-zero elements and its cyclic nature. Some explore the implications of the field's characteristic on the group's order, while others consider the properties of inverses within the group.

Discussion Status

The discussion is active, with various approaches being explored. Some participants suggest using isomorphisms and Wilson's Theorem, while others propose direct product calculations. Questions about the existence of elements with specific properties are also raised.

Contextual Notes

Participants note the characteristics of the finite field and the implications of its structure on the problem, including the nature of inverses and the order of elements.

teleport
Messages
240
Reaction score
0

Homework Statement



Let F be a finite field. Show that the product of all non-zero elements of F is -1.



Homework Equations



An example of this is Wilson's Theorem.



The Attempt at a Solution



Let G be the multiplicative group of non-zero elements of F. Then G is cyclic. Let a be the generator of G. Here I get stuck. I thought that by just taking the product of each element represented by some power of the generator would be enough, but hey, it isn't. Not sure what to do now. Thnx for any help.
 
Physics news on Phys.org
Can we just say that if F has characteristic p, then |G| = p - 1. Since G is cyclic then it isomorphic to Z*_p (which is also cyclic), and then use Wilson's Theorem and the isomorphism to conclude the product is -1.
 
actually it's p^n for some positive integer n, but the same thing...
 
There's a more direct way. Take the product of all the nonzero members a_1*a_2*...*a_n. For each a_i, a_i^(-1) is also in the list. Only +1 and -1 are their own inverses.
 
but why isn't it possible for an element besides 1, -1, to have order 2 so that its inverse is itself?
 
If a^2=1 then a satisfies a^2-1=0. Factor to (a-1)(a+1)=0. Since F is a field, one of those factors must be zero.
 
awesome :) just what i needed. thnx
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K