# Degree of Element of Finite Field Extension divides Degree of Extension

## Theorem

Let $F$ be a field whose zero is $0$ and whose unity is $1$.

Let $K / F$ be a finite field extension of degree $n$.

Let $\alpha \in K$ be algebraic over $F$.

Then the degree of $\alpha$ is a divisor of $n$.

## Proof

Let $\alpha \in K$.

The dimension of $K / F$ considered as a vector space is $n$.

Let $S$ be the set defined as:

- $S := \set {1, \alpha, \alpha^2, \ldots, \alpha_n}$

We have that:

- $\card S = n + 1$

From Cardinality of Linearly Independent Set is No Greater than Dimension:

- $S$ is linearly dependent on $F$.

Hence there are scalars $c_0, c_1, \ldots, c_n \in F$, not all zero, such that:

- $c_0 + c_1 \alpha + c_2 \alpha^2 + \dotsb + c_n \alpha_n = 0$

which is a polynomial of degree $n$.

So $\alpha$ is algebraic over $F$ with degree no greater than $n$.

Consider $\map F \alpha$, the simple algebraic extension of $F$ on $\alpha$.

We have that:

- $F \subseteq \map F \alpha \subseteq K$

Let $\alpha$ be algebraic over $F$ with degree $m$.

Then $\map F \alpha$ is a finite extension of $F$ with degree $\index {\map F \alpha} F = m$.

We also have that $K$ is a finite extension of $F$ with degree $n$.

So by Degree of Field Extensions is Multiplicative:

- $\index K F = \index K {\map F \alpha} \index {\map F \alpha} F$

But $\index K F = n$ and $\index {\map F \alpha} F = m$

so it follows that:

- $m \divides n$

The result follows.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $74 \ \text {(i)}$