210 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

At this point, note that the roots of p(x) in the splitting field, zi,z j are both solutions to theequation

x2− (z1 + z2)x+ z1z2 = 0,

which from the above has complex coefficients. By the quadratic formula the zi,z j are bothcomplex. Thus the original polynomial has a complex root. ■

With this lemma, it is easy to prove the fundamental theorem of algebra. The differ-ence between the lemma and this theorem is that in the theorem, the coefficients are onlyassumed to be complex. What this means is that if you have any polynomial with complexcoefficients it has a complex root and so it is not irreducible. Hence the field extension isthe same field. Another way to say this is that for every complex polynomial there exists afactorization into linear factors or in other words a splitting field for a complex polynomialis the field of complex numbers.

Theorem 9.3.3 Let p(x) ≡ anxn + an−1xn−1 + · · ·+ a1x+ a0 be any complex polynomial,n ≥ 1,an ̸= 0. Then it has a complex root. Furthermore, there exist complex numbersz1, · · · ,zn such that

p(x) = an

n

∏k=1

(x− zk)

Proof: First suppose an = 1. Consider the polynomial q(x)≡ p(x) p(x̄)(xn +an−1xn−1 + · · ·+a1x+a0

)·(

xn +an−1xn−1 + · · ·+a1x+a0)

This polynomial has real coefficients because the coefficient of xm is of the form

m

∑k=0

am−kak

and the sum involves adding terms of the form

aka j +aka j = aka j +aka j = aka j +aka j

so it is of the form of a complex number added to its conjugate. Hence q(x) has realcoefficients as claimed. Therefore, by by Lemma 9.3.2 it has a complex root z. Henceeither p(z) = 0 or p(z) = 0. Thus p(x) has a complex root.

Next suppose an ̸= 1. Then simply divide by it and get a polynomial in which an = 1.Denote this modified polynomial as q(x). Then by what was just shown and the Euclideanalgorithm, there exists z1 ∈ C such that

q(x) = (x− z1)q1 (x)

where q1 (x) has complex coefficients. Now do the same thing for q1 (x) to obtain

q(x) = (x− z1)(x− z2)q2 (x)

and continue this way. Thusp(x)an

=n

∏j=1

(x− z j) ■