# Simple Algebraic Field Extension consists of Polynomials in Algebraic Number

## Theorem

Let $F$ be a field.

Let $\theta \in \C$ be algebraic over $F$.

Let $\map F \theta$ be the simple field extension of $F$ by $\theta$.

Then $\map F \theta$ consists of polynomials that can be written in the form $\map f \theta$, where $\map f x$ is a polynomial over $F$.

## Proof

Let $H$ be the set of all numbers which can be written in the form $\map f \theta$.

We have that:

$H$ is closed under addition and multiplication.
$H$ contains $0$ and $1$
For every element of $H$, $H$ also contains its negative.

Let $\map f \theta \ne 0$.

Then $\theta$ is not a root of $\map f x$.

the minimal polynomial $\map m x$ in $\theta$ does not divide $\map f x$.

From Minimal Polynomial is Irreducible, the GCD of $\map m x$ and $\map f x$ is $1$.

Therefore:

$\exists \map s x, \map t x: \map s x \map m x + \map t x \map f x = 1$

Substituting for $\theta$:

$\map s \theta \, \map m \theta + \map t \theta \, \map f \theta = 1$

Because $\map m \theta = 0$ it follows that:

$\map t \theta \, \map f \theta = 1$

We have that $\map t \theta \in H$.

Thus $\map t \theta$ is the product inverse of $\map f x$ in $H$.

Thus $H$ is a field.

A field containing $F$ and $\theta$ must contain $1$ and all the powers of $\theta$ for positive integer index.

Hence such a field also contains all linear combinations of these, with coefficients in $F$.

So a field containing $F$ and $\theta$ contains all the elements of $H$:

$H \subseteq \map F \theta$

But by definition, $\map F \theta$ is the smallest field containing $F$ and $\theta$.

That is:

$\map F \theta \subseteq H$

Thus:

$\map F \theta = H$

and the result follows.

$\blacksquare$