Fundamental Theorem of Algebra: Preliminaries

Cyclotomic integers allow Fermat's Last Theorem to be factored in the following way:

xⁿ + yⁿ = (x + y)(x + αy)(x + α²y)*...*(x + α^n-1y)

In the above factorization, n is an odd prime and α is a root of unity where αⁿ = 1 and αⁱ ≠ 1 for all 1 ≤ i ≤ n-1.

To establish this refactorization, I will be using the Fundamental Theorem of Algebra.

The main idea behind the Fundamental Theorem of Algebra is that for any given polynomial of a single variable of order n, there are at least n zeros to this equation in the complex domain.

This is an easy proof to state but difficult to prove. Rene Descartes and Jean le Rond d'Alembert knew about this result but it was not until Carl Friedrich Gauss that the fundamental theorem was rigorously proved. d'Alembert thought that the existence of a minimum point for a complex equation was obvious and needed no proof. I do not find this point so obvious so I begin with its proof.

The details in today's proof are taken from B. N. Delone's article on Algebra from Mathematics: Its Content, Methods, and Meaning by A. D. Aleksandrov, A. N. Kolmogorov, and M. A. Lavrent'ev and translated by S. H. Gould.

Theorem 1: Bolzano-Weierstrass Theorem

If a rectangle contains an infinite sequence of points (z₁, z₂, ..., z_n, ... in its interior, then there exists a point z₀ such that in any arbitrarily small neighborhood of z₀, there are infinitely many points of the sequence z₁, z₂, ..., z_n, ...

Proof:

(1) Let R₁ be a rectanlge that contains an infinite sequence of points.

(2) We divide it up into 4 equal parts using two lines parallel to each of its sides.

(3) At least one of these four parts contains infinitely many points, let us label it R₂

(4) We now divide up R₂ into 4 equal parts using two lines parallel to each of its sides.

(5) One of these four parts contains infinitely many points and we label it R₃

(6) In this way, we are able to generate a sequence of nested rectangles which we can label R₁, R₂, ..., R_n

(7) We can think of each side of this rectangle representing two nested intervals that exist on the x and y axises.

(8) Using Lemma 1 here, we there exists a point on the x-axis and a point on the y-axis that each of these nested intervals have in common.

(9) But this means that there is a point within the nested rectangles such that any arbitrary small neighborhood contains an infinity of points.

QED

Theorem 2: There exists a minimum point for c₀xⁿ + c₁x^n-1 + ... + c_n.

Let f(x) = c₀xⁿ + c₁x^n-1 + ... + c_n.

There exists a value x₀ such that w₀ = c₀(x₀)ⁿ + c₁(x₀)^n-1 + ... + c_n where absolute(w₀) is the minimum value.

Proof:

(1) Let g = absolute(f(0))

(2) Let G be a number greater than g.

(3) Let R be a number such that if absolute(x) > R, then absolute(f(x)) is greater than G

(4) Now, if f(0)=0, then x₀=0 and w₀=0 so we can assume going forward that f(0) ≠ 0

(5) If f(0) is greater than 0 and all f(x) ≥ g, then x₀=0 and w₀=g so we can assume that there exists at least 1 point x' such that absolute(f(x')) is less than g.

(6) Based on a nonzero g, we can set up the follow sequence:

0, g/n, 2g/n, ..., ng/n = g

(7) We can find a value i, c_n such that c_n = (i/n)g and all values absolute(f(x)) ≥ c_n.

(8) From i, we can also find a value c_n' such that c_n' = [(i+1)/n]g and there exists at least one value x' such that absolute(f(x')) is less than c_n'

(9) We can find c_n,i,c_n' regardless of the value of n so we can let n increase to infinity.

(10) Now for all values of n, we can assume that absolute(x) ≤ R since if absolute(x) is greater than R, then absolute(f(x)) is greater than G and therefore greater than g. For purposes here, let's call these values x_n

(11) So, we only need to consider the points x_n that lie inside a rectangle of sides 2R and with its center at the origin.

(12) By Theorem 1 above, there exists a point z₀ such that every neighborhood of z₀ contains infinitely many points of the sequence z₁, z₂, ..., z_n. Let us call this point x₀

(13) For any point x, we have:

absolute(f(x)) is greater than c_n = c_n' - g/n which is greater than absolute(f(x_n)) - g/n = absolute(f(x₀)) + absolute(f(x_n)) - absolute(f(x₀)) - g/n.

(14) This inequality is true for all values of n so as n increases toward infinity, we see that difference absolute(f(x_n)) - absolute(f(x₀)) becomes arbitrarily small in absolute value with g/n.

(15) Consequently, all absolute(f(x_n)) ≥ absolute(f(x₀)) so x₀ is the minimum point.

QED

References

• A. D. Aleksandrov, A. N. Kolmogorov, M. A. Lavrent'ev, Mathematics: Its Content, Methods, and Meaning.