This is devoted to a mostly algebraic proof of the fundamental theorem of algebra. It depends on
the interesting results about symmetric polynomials which are presented above. I found it on the
Wikipedia article about the fundamental theorem of algebra. You google “fundamental theorem of
algebra” and go to the Wikipedia article. It gives several other proofs in addition to this one.
According to this article, the first completely correct proof of this major theorem is due
to Argand in 1806. Gauss and others did it earlier but their arguments had gaps in
them.
You can’t completely escape analysis when you prove this theorem. The necessary analysis is in
the following lemma.
Lemma F.2.1Suppose p
(x)
= x^{n} + a_{n−1}x^{n−1} +
⋅⋅⋅
+ a_{1}x + a_{0}where n is odd and thecoefficients are real. Then p
(x)
has a real root.
Proof:This follows from the intermediate value theorem from calculus.
Next is an algebraic consideration. First recall some notation.
m∏
ai ≡ a1a2 ⋅⋅⋅am
i=1
Recall a polynomial in
{z1,⋅⋅⋅,zn}
is symmetric only if it can be written as a sum of elementary
symmetric polynomials raised to various powers multiplied by constants.
The following is the main part of the theorem. In fact this is one version of the fundamental
theorem of algebra which people studied earlier in the 1700’s.
Lemma F.2.2Let p
(x)
= x^{n} + a_{n−1}x^{n−1} +
⋅⋅⋅
+ a_{1}x + a_{0}be a polynomial with realcoefficients. Then it has a complex root.
Proof:It is possible to write
n = 2km
where m is odd. If n is odd, k = 0. If n is even, keep dividing by 2 until you are left with an odd
number. If k = 0 so that n is odd, it follows from Lemma F.2.1 that p
(x)
has a real, hence
complex root. The proof will be by induction on k, the case k = 0 being done. Suppose then that
it works for n = 2^{l}m where m is odd and l ≤ k − 1 and let n = 2^{k}m where m is odd. Let
{z,⋅⋅⋅,z }
1 n
be the roots of the polynomial in a splitting field, the existence of this field being
given by the above proposition. Then
∏n ∑n k k
p (x) = (x − zj) = (− 1) pk(z1,⋅⋅⋅,zn)x (6.1)
j=1 k=0
(6.1)
where p_{k}
(z1,⋅⋅⋅,zn)
is the k^{th} elementary symmetric polynomial. Note this shows
an−k = pk(z1,⋅⋅⋅,zn) (− 1)k , a real number. (6.2)
(6.2)
There is another polynomial which has coefficients which are sums of real numbers times the
p_{k} raised to various powers and it is
∏
qt(x ) ≡ (x − (zi +zj +tzizj)),t ∈ ℝ
1≤i<j≤n
I need to verify this is really the case for q_{t}
(x)
. When you switch any two of the z_{i} in q_{t}
(x)
the
polynomial does not change. Thus the coefficients of q_{t}
(x)
must be symmetric polynomials
in the z_{i} with real coefficients. Hence by Proposition F.1.3 these coefficients are real
polynomials in terms of the elementary symmetric polynomials p_{k}. Thus by 6.2 the
coefficients of q_{t}
(x)
are real polynomials in terms of the a_{k} of the original polynomial.
Recall these were all real. It follows, and this is what was wanted, that q_{t}
(x)
has all real
coefficients.
Note that the degree of q_{t}
(x)
is
( )
n
2
because there are this number of ways to pick i < j
out of
{1,⋅⋅⋅,n}
. Now
( n ) n(n− 1) ( )
= --------= 2k−1m 2km − 1
2 2
k−1
= 2 (odd)
and so by induction, for each t ∈ ℝ,q_{t}
(x)
has a complex root.
There must exist s≠t such that for a single pair of indices i,j, with i < j,
(zi + zj + tzizj),(zi + zj + szizj)
are both complex. Here is why. Let A
(i,j)
denote those t ∈ ℝ such that
(z + z + tzz )
i j i j
is
complex. It was just shown that every t ∈ ℝ must be in some A
(i,j)
. There are infinitely many
t ∈ ℝ and so some A
(i,j)
contains two of them.
Now for that t,s,
zi + zj + tzizj = a
z+ z + sz z = b
i j i j
where t≠s and so by Cramer’s rule,
| |
|| a t ||
|| b s ||
zi + zj = ||----||∈ ℂ
|| 1 t ||
| 1 s |
and also
| |
|| 1 a ||
|| 1 b ||
zizj = |----|-∈ ℂ
|| 1 t||
|| 1 s||
At this point, note that z_{i},z_{j} are both solutions to the equation
x2 − (z1 + z2)x+ z1z2 = 0,
which from the above has complex coefficients. By the quadratic formula the z_{i},z_{j} are both
complex. Thus the original polynomial has a complex root. ■
With this lemma, it is easy to prove the fundamental theorem of algebra. The difference
between the lemma and this theorem is that in the theorem, the coefficients are only assumed to
be complex. What this means is that if you have any polynomial with complex coefficients it has a
complex root and so it is not irreducible. Hence the field extension is the same field. Another way
to say this is that for every complex polynomial there exists a factorization into linear
factors or in other words a splitting field for a complex polynomial is the field of complex
numbers.
Theorem F.2.3Let 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}suchthat
n
p (x) = an∏ (x− zk)
k=1
Proof:First suppose a_{n} = 1. Consider the polynomial q