Tag Archives: fundamental-theorem-of-algebra

Polynomials and Commutativity

Standard

In high school, I came to know about the statement of the fundamental theorem of algebra:

Every polynomial of degree n with integer coefficients have exactly n 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 n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots.

A more general statement about the number of roots of a polynomial in one variable is the Factor Theorem:

Let R be a commutative ring with identity and let p(x)\in R[x] be a polynomial with coefficients in R. The element a\in R is a root of p(x) if and only if (x-a) divides p(x).

A corollary of above theorem is that:

A polynomial f of degree n over a field F has at most n roots in F.

(In case you know undergraduate level algebra, recall that R[x] is a Principal Ideal Domain if and only if R 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 \mathbb{C} 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 X^2 + X + 1 has only 2 complex roots, namely \omega = \frac{-1+i\sqrt{3}}{2} and \omega^2 = \frac{-1-i\sqrt{3}}{2}. 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.)

\displaystyle{A=\begin{bmatrix} 0 & -1 \\1 & -1 \end{bmatrix}, B=\begin{bmatrix} \omega & 0 \\0 & \omega^2 \end{bmatrix}, C=\begin{bmatrix} \omega^2 & 0 \\0 & \omega \end{bmatrix}}

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.

Advertisements

Real vs Complex numbers

Standard

I want to talk about the algebraic and analytic differences between real and complex numbers. Firstly, let’s have a look at following beautiful explanation by Richard Feynman (from his QED lectures) about similarities between real and complex numbers:

img_20170304_1237442

From Chapter 2 of the book “QED – The Strange Theory of Light and Matter” © Richard P. Feynman, 1985.

Before reading this explanation, I used to believe that the need to establish “Fundamental theorem Algebra” (read this beautiful paper by Daniel J. Velleman to learn about proof of this theorem) was only way to motivate study of complex numbers.

The fundamental difference between real and complex numbers is

Real numbers form an ordered field, but complex numbers can’t form an ordered field. [Proof]

Where we define ordered field as follows:

Let \mathbf{F} be a field. Suppose that there is a set \mathcal{P} \subset \mathbf{F} which satisfies the following properties:

  • For each x \in \mathbf{F}, exactly one of the following statements holds: x \in \mathcal{P}, -x \in \mathcal{P}, x =0.
  • For x,y \in \mathcal{P}, xy \in \mathcal{P} and x+y \in \mathcal{P}.

If such a \mathcal{P} exists, then \mathbf{F} is an ordered field. Moreover, we define x \le y \Leftrightarrow y -x \in \mathcal{P} \vee x = y.

Note that, without retaining the vector space structure of complex numbers we CAN establish the order for complex numbers [Proof], but that is useless. I find this consequence pretty interesting, because though \mathbb{R} and \mathbb{C} are isomorphic as additive groups (and as vector spaces over \mathbb{Q}) but not isomorphic as rings (and hence not isomorphic as fields).

Now let’s have a look at the consequence of the difference between the two number systems due to the order structure.

Though both real and complex numbers form a complete field (a property of topological spaces), but only real numbers have least upper bound property.

Where we define least upper bound property as follows:

Let \mathcal{S} be a non-empty set of real numbers.

  • A real number x is called an upper bound for \mathcal{S} if x \geq s for all s\in \mathcal{S}.
  • A real number x is the least upper bound (or supremum) for \mathcal{S} if x is an upper bound for \mathcal{S} and x \leq y for every upper bound y of \mathcal{S} .

The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
This least upper bound property is referred to as Dedekind completeness. Therefore, though both \mathbb{R} and \mathbb{C} are complete as a metric space [proof] but only \mathbb{R} is Dedekind complete.

In an arbitrary ordered field one has the notion of Dedekind completeness — every nonempty bounded above subset has a least upper bound — and also the notion of sequential completeness — every Cauchy sequence converges. The main theorem relating these two notions of completeness is as follows [source]:

For an ordered field \mathbf{F}, the following are equivalent:
(i) \mathbf{F} is Dedekind complete.
(ii) \mathbf{F} is sequentially complete and Archimedean.

Where we defined an Archimedean field as an ordered field such that for each element there exists a finite expression 1+1+\ldots+1 whose value is greater than that element, that is, there are no infinite elements.

As remarked earlier, \mathbb{C} is not an ordered field and hence can’t be Archimedean. Therefore, \mathbb{C}  can’t have least-upper-bound property, though it’s complete in topological sense. So, the consequence of all this is:

We can’t use complex numbers for counting.

But still, complex numbers are very important part of modern arithmetic (number-theory), because they enable us to view properties of numbers from a geometric point of view [source].