Thursday, October 01, 2009

Nonzero Polynomials with Distinct Parameters

The following is taken from Harold M. Edwards in his book Galois Theory.


Let K be a field.

Let x1, x2, x3, ... be an infinite sequence of distinct elements of K

Let f(A,B,C,...) be a nonzero polynomial in n variables A,B,C,... with coefficients in K


It is possible to select values A=xj, B = xk, C = xm for the variables A,B,C from the sequence x1, x2, x3, ... so that F( xj, xk, xm, ...) ≠ 0


(1) Assume that f(x) is a nonzero polynomial of one variable with degree m.

(2) Using the Fundamental Theorem of Algebra (see Theorem, here), we know that f(x) has at most m distinct roots.

(3) If we list off m+1 distinct elements of K from the infinite sequence, it is clear that at least one (let us say xr) will not be a root.

(4) So that f(xr) ≠ 0

(5) Assume that this is true up to n-1 variables for F(A,B,C...,Y) so that we know that F(xi, xj, ..., xy) ≠ 0

(6) Let G be a function of n variables so that we have G(A,B,C,...Z)

(7) Let H be a function on the first n-1 variables so that we have H(A,B,C,...Y) = G(A,B,C,...,Y,1)

(8) By assumption, we can find xi, xj, ... xy such that:

H(xi, xj, ..., xy) ≠ 0

(9) But then G(xi, xj, ..., xy, 1) ≠ 0.




rasraster said...

Hi Larry,

This is a really nice and concise proof.

Edwards indicates the general direction the proof should go (induction, starting with A). But since the function in question involves polynomials with mixed terms (e.g., k *(A^2)*(B^3)*(C^4)), I really wasn't sure how to get from (n-1) to n. Your proof is nice and clean and simple and doesn't require imagining how to solve the mixed terms.


rasraster said...


Thinking about this further, there seems to be an interesting corollary to this theorem:

If one takes the argument further from the n_th parameter backward to the first, it seems that it should be true that:

G(x_r,1,1,1,...,1) <> 0, where x_r is an element of K that is not a root of the polynomial in A that one derives by setting all the other variables to 1.