Degree of Field Extensions is Multiplicative

Theorem
Let $E / K$ and $K / F$ be finite field extensions.

Then $E/F$ is a finite field extension, and:
 * $\left[{E : F}\right] = \left[{E : K}\right] \left[{K : F}\right]$

Proof
First, note that $E / F$ is a field extension as $F \subseteq K \subseteq E$.

Suppose $\left[{E : K}\right] = m$, $\left[{K : F}\right] = n$.

Let $\alpha = \left\{{a_1, \ldots,a_m}\right\}$ be a basis of $E / K$ and $\beta = \left\{ {b_1, \ldots, b_n} \right\}$ be a basis of $K / F$.

We wish to prove the set:
 * $\gamma = \left\{{a_i b_j: 1 \leq i \leq m, 1 \leq j \leq n}\right\}$

is a basis of $E/F$.

As $\alpha$ is a basis of $E/K$, then $\forall ~ c\in E$, we have:
 * $\displaystyle c = \sum_{i=1}^m c_i a_i$, for some $c_i \in K$.

Letting $\displaystyle b = \sum_{j=1}^n b_i$ and $\dfrac{c_i} b = d_i$ (note $c_i, b \in K \implies d_i \in K$ and $b \neq 0$ as $\beta$ is linearly independent over $F$), we have:


 * $\displaystyle c = \sum_{i=1}^m \frac{c_i} b \cdot b \cdot a_i = \sum_{i=1}^m \sum_{j=1}^n d_i a_i b_j$

Thus $\gamma$ is a spanning set of $E/K$.

To show $\gamma$ is linearly independent, we first observe that as $\beta \subseteq E$, we can express each $b_j$ as $\displaystyle b_j = \sum_{i=1}^m d_{ij} a_i$ for some $d_{ij} \in K$.

Hence we have:

Now, as $\beta$ is linearly independent, $\forall j: b_j \neq 0$, hence $\displaystyle \sum_{i=1}^m d_{ij} a_i \ne 0$.

Therefore, as all fields are integral domains, we have:

Hence $\gamma$ is a linearly independent spanning set; thus it is a basis.

Recalling the defintion of $\gamma$ as $\left\{{a_i b_j: 1 \le i \le m, 1 \le j \le n}\right\}$, we have:
 * $\left\vert {\gamma}\right\vert = m n = \left[{E : K}\right] \left[{K : F}\right]$

as desired.