Degree of Field Extensions is Multiplicative

Theorem
Let $E, K$ and $F$ be fields.

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

Then $E / F$ is a finite finite field extensions, 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 that $\left[{E : K}\right] = m$ and $\left[{K : F}\right] = n$.

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:


 * $\gamma = \left\{{a_i b_j: 1 \le i \le m, 1 \le j \le n}\right\}$

is a basis of $E / F$.

As $\alpha$ is a basis of $E / K$, we have, for all $c \in E$:


 * $\displaystyle c = \sum_{i \mathop = 1}^m c_i a_i$, for some $c_i \in K$.

Define $\displaystyle b := \sum_{j \mathop = 1}^n b_i$ and $d_i := \dfrac{c_i} b$.

Note $b \ne 0$ since $\beta$ is linearly independent over $F$, and $d_i \in K$ since $b, c_i \in K$.

Now we have:


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

Thus $\gamma$ is seen to be a spanning set of $E / F$.

To show $\gamma$ is linearly independent, suppose that for some $c_{ij} \in F$:


 * $\displaystyle \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{ij} a_i b_j = 0$

Then we have (as fields are commutative rings):

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

Recalling the definition 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.

Also known as
This result is also known as the tower rule or tower law, from the definition of a tower of fields.