---
tag: 81JV1D6DG6
kind: proof
name: Proof of the tower law
msc: [12F05]
proves: 0A7KJ68XHT
uses: [0GS74FGFEK, 0QA8CHH8MG, 00MS9VMQ9Z, 0VTQTPQGAV, 0EVHJFTG9Z, 0V8F3E48E6, 088RYKAWK9, 0WK0Y8RH1M, 05G2G0P224, 0C81YBVTWH, 0JTGRV4WT4]
sources:
  - url: https://en.wikipedia.org/wiki/Degree_of_a_field_extension
    locator: Proof of the multiplicativity formula in the finite case
  - url: https://www.jmilne.org/math/CourseNotes/FT.pdf
    locator: Proposition 1.20
---

## Proof

1. Let $K \subseteq L \subseteq M$ be [[fields|0QA8CHH8MG]]. By [[existence of bases|00MS9VMQ9Z]], choose a [[basis|0VTQTPQGAV]] $(u_\alpha)_{\alpha \in A}$ of $L$ over $K$ and a basis $(w_\beta)_{\beta \in B}$ of $M$ over $L$.
2. By the [[dimension theorem|0WK0Y8RH1M]], the [[degrees|0GS74FGFEK]] are $[L : K] = \abs{A}$ and $[M : L] = \abs{B}$, as [[cardinalities|0JTGRV4WT4]]. The product of cardinals is $\abs{A} \cdot \abs{B} = \abs{A \times B}$, with the [[Cartesian product|0C81YBVTWH]].
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|088RYKAWK9]] $M$ 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|0V8F3E48E6]] over $K$ of the family of Step 3.
7. [[Linear independence|0EVHJFTG9Z]]. 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|05G2G0P224]] and Step 2, $[M : K] = \abs{A \times B} = \abs{B} \cdot \abs{A} = [M : L] \cdot [L : K]$.

## Idea

Write an element of $M$ in the basis of $M$ over $L$, and then write each coefficient in the basis of $L$ over $K$. This shows that the products $u_\alpha w_\beta$ span $M$ over $K$. Reading the same computation backwards shows that they are independent.