Proof of the tower law

Proof

1. Let $K \subseteq L \subseteq M$ be fields. By existence of bases, choose a basis $(u_\alpha)_{\alpha \in A}$ of $L$ over $K$ and a basis $(w_\beta)_{\beta \in B}$ of $M$ over $L$.
2. By the dimension theorem, the degrees are $[L : K] = \abs{A}$ and $[M : L] = \abs{B}$, as cardinalities. The product of cardinals is $\abs{A} \cdot \abs{B} = \abs{A \times B}$, with the Cartesian product.
3. We show that the family $(u_\alpha w_\beta)_{(\alpha, \beta) \in A \times B}$ is a basis of $M$ over $K$.
4. Spanning. Let $z \in M$. Since the $w_\beta$ span $M$ over $L$, $z = \sum_{\beta \in B_0} c_\beta w_\beta$ for a finite $B_0 \subseteq B$ and $c_\beta \in L$.
5. Since the $u_\alpha$ span $L$ over $K$, each $c_\beta$ is a $K$-combination of finitely many $u_\alpha$. Let $A_0 \subseteq A$ be the finite union of the index sets used. Adding zero coefficients, $c_\beta = \sum_{\alpha \in A_0} b_{\alpha \beta} u_\alpha$ with $b_{\alpha \beta} \in K$.
6. Then $z = \sum_{\beta \in B_0} \sum_{\alpha \in A_0} b_{\alpha \beta} \, u_\alpha w_\beta$, a linear combination over $K$ of the family of Step 3.
7. Linear independence. Let $A_0 \subseteq A$ and $B_0 \subseteq B$ be finite, let $b_{\alpha \beta} \in K$, and assume $\sum_{\alpha \in A_0} \sum_{\beta \in B_0} b_{\alpha \beta} u_\alpha w_\beta = 0$.
8. Regroup as $\sum_{\beta \in B_0} c_\beta w_\beta = 0$ with $c_\beta = \sum_{\alpha \in A_0} b_{\alpha \beta} u_\alpha \in L$.
9. The $w_\beta$ are linearly independent over $L$, so $c_\beta = 0$ for each $\beta \in B_0$.
10. The $u_\alpha$ are linearly independent over $K$, so $b_{\alpha \beta} = 0$ for all $\alpha \in A_0$ and $\beta \in B_0$.
11. Every finite subfamily of the family of Step 3 lies in one of the form of Step 7. So the family is linearly independent, and with Step 6 it is a basis.
12. Two members with different indices are different. Otherwise $1 \cdot v - 1 \cdot v = 0$ would be a nontrivial relation, against Step 11. So the basis has $\abs{A \times B}$ elements.
13. By the definition of dimension and Step 2, $[M : K] = \abs{A \times B} = \abs{B} \cdot \abs{A} = [M : L] \cdot [L : K]$.