Linear Algebra, Theory And Applications, 2012a

450 FIELDS AND FIELD EXTENSIONS
Theorem F.2.3 Let p (x) ≡ a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 be any complex polynomial,
n ≥ 1,a n ≠0. Then it has a complex root. Furthermore, there exist complex numbers
z 1 , ··· ,z n such that
p (x) =a n
n ∏
k=1
(x − z k )
Proof: First suppose a n =1. Consider the polynomial
q (x) ≡ p (x) p (x)
this is a polynomial and it has real coefficients. This is because it equals
(
x n + a n−1 x n−1 + ···+ a 1 x + a 0
)
·
(
x n + a n−1 x n−1 + ···+ a 1 x + a 0
)
The x j+k term of the above product is of the form
and
a k x k a j x j + a k x k a j x j = x k+j (a k a j + a k a j )
a k a j + a k a j = a k a j + a k a j
so it is of the form of a complex number added to its conjugate. Hence q (x) has real
coefficients as claimed. Therefore, by by Lemma F.2.2 it has a complex root z. Hence either
p (z) =0orp (z) =0. Thusp (x) has a complex root.
Next suppose a n ≠0. Then simply divide by it and get a polynomial in which a n =1.
Denote this modified polynomial as q (x). Then by what was just shown and the Euclidean
algorithm, there exists z 1 ∈ C such that
q (x) =(x − z 1 ) q 1 (x)
where q 1 (x) has complex coefficients. Now do the same thing for q 1 (x) to obtain
and continue this way. Thus
q (x) =(x − z 1 )(x − z 2 ) q 2 (x)
p (x)
a n
=
n∏
(x − z j )
j=1
Obviously this is a harder proof than the other proof of the fundamental theorem of
algebra presented earlier. However, this is a better proof. Consider the algebraic numbers
A consisting of the real numbers which are roots of some polynomial having rational
coefficients. By Theorem 8.3.32 they are a field. Now consider the field A + iA with the
usual conventions for complex arithmetic. You could repeat the above argument with small
changes and conclude that every polynomial having coefficients in A + iA has a root in
A + iA. Recall from Problem 41 on Page 223 that A is countable and so this is also the case
for A + iA. Thus this gives an algebraically complete field which is countable and so very
different than C. Of course there are other situations in which the above harder proof will
work and yield interesting results.