2. A Proof Based on the Polynomial Coefficient Function
A formulation of the fundamental theorem of algebra is the following.
Theorem 1. For any polynomial of degree , with complex coefficients, there are , such that .
This formulation is equivalent to claim the surjectivity of the
Polynomial coefficient function :
Namely, if
is surjective, then we can conclude that the complex coefficients of any nth-degree polynomial are the image of the
Polynomial coefficient function applied to
n complex numbers (the polynomial roots). In symbols, if
è surjective and
there exists a vector
such that
, that is:
In a sense, the function
inverts the equation-solving process in that of equation-making. Of course,
is easy to compute from the roots, while equation-solving is, in general, very difficult to obtain (
is a
one-way encoding). Here we develop a proof of
surjectivity by combining basic notions of general topology in hyperspaces of any finite dimension over complex numbers and a recurrent equation holding for
.
Theorem 2. is surjective.
Proof. Let us fix a value for n, the arguments we develop apply to any value of n. For the sake of brevity, in the following, we drop the exponent n of the function that provides the coefficients of the polynomial .
It is easy to verify that coefficients
are respectively given by the functions:
Of course, functions
, for
, are continuous, therefore also
is continuous. However,
F is not injective because permutations of the same vector will provide the same
F image.
Let
and
. For
with
m a natural number greater than 1, we call
m-pair the hyperplane pairs of
:
The
m-pairs, for
, determine the portions of the space between them. The set of points between all the
pairs mutually orthogonal defines, for every
m, the
n-cubic
m-block centered on the origin
of
:
where ≤ is the partial order over complex numbers
if
e
(which extends to
applying the condition on all the components). The family
provides a covering of the hyperspace
.
The block has vertices, and the vertex is called the superior vertex, while is called the inferior vertex. The remaining vertices are for all possible choices of signs + or − in any component.
The image is a compact set of because is compact and F is continuous (which sends compacts into compacts). Moreover, in any closed and bounded set is compact (Heine-Borel’s theorem). The image of has vertices:
………
We know that a metric complete space such as
contains all the limits of Cauchy sequences in the space, and any closed subset of the space, is complete [
7,
8]. Therefore, for any closed and bounded set
X of
,
.
Let be the radius of , as the minimum norm (distance from the origin) reached by the points of the frontier of .
We will prove that, for , . This means that for increasing values of m the radius of grows illimitably so that any vector in is an image of F, whence the surjectivity of F follows.
In the following, we use rather than n (with no loss of generality because n is a generic value, and the theorem we are proving holds trivially for ).
Let
, we denote by
the vector of
extending
, such that:
and, for
:
Symmetrically, for
:
and:
The following equations easily derive from the definition (
1) of
F, for
, with
and
:
which explicitly gives all the following equations:
Then, from the definitions of
,
, the equations above are synthesized, for any
, and
, by the
Recurrent Coefficient Equation:
Now, let
, then we can suppose with no loss of generality (the order of components is not relevant in the determination of the
F-image) that
, with
where
and
, for
, that is,
E is a point of the edge between vertices
and
of
. According to the Recurrent Coefficient Equation (
4) the following inequalities hold, which give a lower bound of the radius
:
In conclusion, the radius of grows with m, therefore any of the blocks of (which is a covering of ) is included in the image of some block of . Then, F is surjective.