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LaGrange's Theorem

Date: 03/12/2001 at 17:45:53
From: Joan
Subject: Number theory

Please explain LaGrange's Theorem on the number of roots of a 

Thank you.

Date: 03/13/2001 at 10:35:36
From: Doctor Rob
Subject: Re: Number theory

Thanks for writing to Ask Dr. Math, Joan.

This theorem states that a polynomial of degree n over a field has at 
most n roots in that field.

Usually this is proved by induction on n. Let f(x) be the polynomial.  
If n = 0, then f(x) is a nonzero constant, and so has no roots, so the 
statement is true for n = 0.  Now suppose that n > 0, and that the 
statement is true for all polynomials of degree n - 1. Then f(x) may 
have no roots in the field, in which case the theorem holds. If it has 
roots in the field, let r be one of them. Then divide f(x) by x - r 
with remainder:

   f(x) = (x-r)*g(x) + c,  c a constant.

Evaluate this at x = r:

   0 = f(r) = (r-r)*g(r) + c = c,

since r is a root of f(x). Since c = 0, that means that

   f(x) = (x-r)*g(x).

Suppose that s is any root of f(x) in the field. Then

   0 = (s-r)*g(s).

This implies, since we are working in a field, that either

   s - r = 0,  or  g(s) = 0

(or both). Thus either s = r, or s is a root of g(x), or both. Now 
the induction hypothesis applies to g(x) whose degree is n - 1.

I leave the rest of the proof to you.

By the way, this is false if you are not working over a field. The 
polynomial x^2 - x has four roots in the ring of integers modulo 6, 
namely 0, 1, 3, and 4.

- Doctor Rob, The Math Forum   
Associated Topics:
College Number Theory
High School Number Theory

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