In high school, I came to know about the statement of the fundamental theorem of algebra:
Every polynomial of degree
with integer coefficients have exactly
complex roots (with appropriate multiplicity).
In high school, a polynomial = a polynomial in one variable. Then last year I learned 3 different proofs of the following statement of the fundamental theorem of algebra [involving, topology, complex analysis and Galois theory]:
Every non-zero, single-variable, degree
polynomial with complex coefficients has, counted with multiplicity, exactly
complex roots.
A more general statement about the number of roots of a polynomial in one variable is the Factor Theorem:
Let
be a commutative ring with identity and let
be a polynomial with coefficients in
. The element
is a root of
if and only if
divides
.
A corollary of above theorem is that:
A polynomial
of degree
over a field
has at most
roots in
.
(In case you know undergraduate level algebra, recall that is a Principal Ideal Domain if and only if
is a field.)
The key fact that many times go unnoticed regarding the number of roots of a given polynomial (in one variable) is that the coefficients/solutions belong to a commutative ring (and is a field hence a commutative ring). The key step in the proof of all above theorems is the fact that the division algorithm holds only in some special commutative rings (like fields). I would like to illustrate my point with the following fact:
The equation has only 2 complex roots, namely
and
. But if we want solutions over 2×2 matrices (non-commutative set) then we have at least 3 solutions (consider 1 as 2×2 identity matrix and 0 as the 2×2 zero matrix.)
if we allow complex entries. This phenominona can also be illusttrated using a non-commutative number system, like quaternions. For more details refer to this Math.SE discussion.