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.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $73$