**1.0 Introduction**

This is some well-established mathematics. I do not know of any use in economics.

**2.0 The Field of Real Numbers**

I start by taking real numbers **R**, with binary operations for addition
and multiplication, as known.

**3.0 A Ring of Polynomials**

Consider polynomials with coefficients taken from the reals as a formal object:

R[X] = {a_{n}X^{n}+ ... +a_{1}X+a_{0}|n≥ 0,coefficients from the reals. }

The symbol *X* is known as an indeterminate. One does not consider it here as a member of some
set. Polynomial addition is defined to yield another polynomial, with addition of coefficients
in the reals. Polynomial multiplication is also defined as usual.

**3.1 Evaluation Homomorphisms**

Still, one would like to talk about evaluating polynomials. For every real number *r*, there
exists an evaluation homomorphism φ_{r}( ) that maps **R**[*X*]
into the reals. This homomorphism is defined by:

φ_{r}(a_{n}X^{n}+ ... +a_{1}X+a_{0}) =a_{n}r^{n}+ ... +a_{1}r+a_{0}

Addition and multiplication on the right-hand side above is performed in the reals. The map is a homomorphism because it preserves addition and multiplication:

φ_{r}(f(X) +g(X) ) = φ_{r}(g(X) ) + φ_{r}(g(X) )

φ_{r}(f(X)g(X) ) = φ_{r}(g(X) ) φ_{r}(g(X) )

In words, it does not matter, in evaluating the sum or product of polynomials if:

- You perform the operation in the polynomial ring first and then evaluate the sum, or
- You evaluate the polynomials and then sum or multiply in the reals.

A homomorphism that is one-to-one is an isomorphism. These evaluation homomorphisms are not isomorphisms since more than one polynomial may be evaluated to have the same value. An example follows:

φ_{2}(X^{2}) = φ_{2}(X+ 2) = 4

**3.2 A Polynomial Without a Multiplicative Inverse**

The constant polynomials 0 and 1 are the additive and multiplicative identities for polynomial addition and multiplication, respectively. These identities are distinct.

Not all polynomials have a multiplicative inverse. A simple example is
the polynomial *X*. Suppose *f*(*X*) were a polynomial
in **R**[*X*] that was the multiplicative inverse of *X*.
Then:

Xf(X) = 1

Consider the evaluation homomorphism for the additive identity in the reals.

1 = φ_{0}(Xf(X)) = φ_{0}(X) φ_{0}(f(X)) = 0 φ_{0}(f(X)) = 0

So the non-existence of a multiplicative inverse for *X* is proven by a
proof by contradiction.

It has been demonstrated that **R**[*X*] cannot be a field, since
not every non-zero element has a multiplicative inverse. I believe it
is actually an integral domain. Just as the field of rational numbers can
be constructed as equivalence classes of ordered pairs of integers,
a field of rational polynomials with real coefficients can be
constructed. I do not pursue this construction here.

**4.0 Polynomial Addition and Multiplication Modulo**

*p*(*X*)
One can define the quotient *q*(*X*) and remainder *r*(*X*)
for any polynomials *f*(*X*) and *g*(*X*) in **R**[*X*]:

f(X) =q(X)g(X) +r(X)

where *r*(*X*) is of degree less than *g*(*X*).
Since the reals are a field, the quotient and remainder are unique.

The above theorem allows one to define polynomial addition and
multiplication modulo *p*(*X*). In particular,
consider:

p(X) =X^{2}+ 1

*p*(*X*) is irreducible. There do not
exist non-constant polynomials *f*(*X*)
and *g*(*X*) in **R**[*X* such that:

p(X) =f(X)g(X)

I now define the set **C** of polynomials

C= {r(X) | there exists af(X) inR[X]such thatr(X) =f(X) modp(X)}

All polynomials in **C** are at most of degree one.

**4.1**

**C**as a Two-Dimensional Vector Space
Each element of **C** can be expressed as a linear combination of the
elements of the basis {*X*, 1}:

C= { (a_{1},a_{0}) |a_{1}X+a_{0}is inR[X]}

**4.2**

**C**as a Field Extension of the Reals
Consider **C** with addition and multiplication defined modulo *p*(*X*).
I claim this is a field. Consider:

f(X) =a_{1}X+a_{0}

g(X) = (-a_{1}/(a_{0}^{2}+a_{1}^{2}))X+a_{0}/(a_{0}^{2}+a_{1}^{2})

Their product in **R**[*X*] is:

f(X)g(X) = ((-a_{1}^{2}/(a_{0}^{2}+a_{1}^{2}))X^{2}+a_{0}^{2}/(a_{0}^{2}+a_{1}^{2}))

The quotient and remainder are found from:

f(X)g(X) =p(X) (-a_{1}^{2}/(a_{0}^{2}+a_{1}^{2})) + 1

Or:

(f(X)g(X)) modp(X) = 1

So every non-zero element of **C** has a multiplicative inverse.

The set of constant polynomials in **C** is isomorphic to the reals.
Thus, **C** extends the reals in a precise sense.

**4.3 Evaluation Homomorphisms in**

**C**
For every *a*_{1} *X* + *a*_{0} in **C**, one
can define an evaluation homomorphism φ_{(a1 X + a0)}( )
that maps **R**[*X*] into **C**.
For every constant polynomial in **C**, this evaluation homomorphism yields the same
answer as the corresponding evaluation homomorphism in Section 3.1.

As an example of this evaluation homomorphism, consider:

φ_{X}(p(X)) = φ_{X}(X^{2}+ 1)

Or:

φ_{X}(p(X)) = φ_{X}(X^{2}) + φ_{X}(X^{1})

Or:

φ_{X}(p(X)) = (XX) modp(X) + 1

With addition and multiplication in **R**[*X*]:

X^{2}=p(X) - 1

Thus:

φ_{X}(p(X)) = -1 + 1 = 0

In other words, the polynomial *X* in the field extension **C** is the square
root of -1.

**5.0 Summary**

The above has extended the field of reals to the field of complex numbers. This field extension contains a zero for the equation:

p(X) =X^{2}+ 1 = 0

Furthermore:

- The imaginary numbers are polynomials of degree one and no constant
term, with addition and multiplication defined modulo
*p*(*X*). - The real numbers are isomorphic to constant polynomials,
with addition and multiplication defined modulo
*p*(*X*).

That is, the extension field **C** is the field of complex numbers.
The complex numbers are only defined up to isomorphism. But their
existence is constructed here, not postulated.

**6.0 Other Field Extensions**

One need not begin this exposition with polynomials with coefficients from the real numbers. Coefficients can be drawn from other fields.

For example, consider the set {0, 1, 2, ..., *p* - 1}, with addition
and multiplication defined modulo *p*, and *p* prime.
This set with these operations is a field. Let *p*(*X*) be,
as above, an irreducible polynomial in the ring of polynomials
with coefficients in the set. Suppose *p*(*X*) is
of degree *n*.
Then the field extension is the
Galois Field, GF(*p*^{n}).
The set of elements of GF(2^{n}) - {0},
with multiplication, is
a cyclic group.
GF(2^{n}) has application in the Advanced
Encryption System (AES) and in
Reed-Solomon error correction
codes. (The latter has something to do with how checkout scanners
work in your neighborhood supermarket.)

On the other hand, consider the field of rational numbers with addition and subtraction defined as usual. There are at most a countably infinite number of polynomials with rational coefficients. An irreducible polynomial leads to an extension field for use in constructing real numbers. But this construction leaves out an uncountably infinite number of real numbers, namely the transcendental real numbers. A real number is algebraic if it is the root of some polynomial with rational coefficients. The real numbers, including transcendentals, can be constructed, instead, as Dedekind cuts or as equivalence classes of Cauchy-convergent sequences of rational numbers. (Cauchy often comes across as a villain in accounts of Galois and Abel's short lives.)

Finally, consider polynomials with coefficients drawn from the field
of complex numbers. (Since, under the above construction, a complex
number is, in some sense, a first-degree polynomial with real coefficients,
this may be a somewhat confusing construction to think about.)
Suppose one defines polynomial addition and multiplication modulo
*p*(*X*), where *p*(*X*) is a first degree
polynomial in the ring **C**[*X*]. Then one obtains a
field "extension" isomorphic to the field of complex numbers.

To find a bigger field extension, one needs to find an
irreducible polynomial of at least degree two in **C**[*X*].
But no such polynomial exists. Proof: Abel was something else,
wasn't he?