Degree of Simple Algebraic Field Extension equals Degree of Algebraic Number

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 $\map F \theta$ is a finite extension of $F$ whose degree is:
 * $\index {\map F \theta} F = n$

Proof
Considered as a vector space over $F$, $\map F \theta$ is generated by the set $B$, where:
 * $B := \set {1, \theta, \theta^2, \ldots, \theta^{n - 1} }$

But $S$ is linearly independent over $F$, because otherwise:
 * $c_0 1 + c_1 \theta + c_2 \theta^2 + \dotsb + c_{n - 1} \theta^{n - 1} = 0$

with all the $c$s non-zero.

That would mean $\theta$ was the root of a polynomial whose degree was less than that of the minimal polynomial $\map m x$ of $\theta$, whose degree equals $n$.

Therefore $B$ is a basis of $\map F \theta$ over $F$.

Thus $\map F \theta$ is of dimension $\size B = n$.

Hence the $\index {\map F \theta} F = n$.