Element of Simple Algebraic Field Extension of Degree n is Polynomial in Algebraic Number of Degree Less than n

Theorem
Let $F$ be a field.

Let $\theta \in \C$ be algebraic over $F$ of degree $n$.

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

Then any element of $\map F \theta$ can be written as $\map f \theta$, where $\map f x$ is a polynomial over $F$ of degree at most $n - 1$.

Proof
From Simple Algebraic Field Extension consists of Polynomials in Algebraic Number, an arbitrary element of $\map F \theta$ can be written as $\map f \theta$.

But:
 * $\map f x = \map m x \, \map q x + \map r x$

where:
 * $\map m x$ is minimal polynomial in $\theta$
 * $\map q x$ is a polynomial in $\map F \theta$
 * $\map r x$ is a polynomial in $\map F \theta$ such that either:
 * $\map \deg {\map r x} < \map \deg {\map m x}$
 * or:
 * $\map r x = 0$

Thus:
 * $\map f \theta = \map m \theta \, \map q \theta + \map r \theta$

and as $\map m \theta = 0$ we have:
 * $\map f \theta = \map r \theta$

So $\map f \theta$ can be expressed as $\map r \theta$ instead, which is of degree strictly less than that of $\map m \theta$.

Hence the result.