Degrees of field extensions multiply along a chain of fields, the same way dimensions would if you built a basis in two stages. This simple rule settles questions that puzzled mathematicians for two thousand years.
Why it works
Tower law. If $F \subseteq K \subseteq L$ are fields with finite degrees, then $$[L : F] = [L : K]\,[K : F].$$ Proof. Let $u_{1}, \dots, u_{m}$ be a basis of $L$ over $K$ and $v_{1}, \dots, v_{n}$ a basis of $K$ over $F$. Every $x \in L$ is $\sum_{i} c_{i}u_{i}$ with $c_{i} \in K$, and each $c_{i} = \sum_{j} a_{ij}v_{j}$ with $a_{ij} \in F$; so the $mn$ products $u_{i}v_{j}$ span $L$ over $F$. If $\sum a_{ij}u_{i}v_{j} = 0$, group as $\sum_{i}(\sum_{j} a_{ij}v_{j})u_{i} = 0$: independence of the $u_{i}$ over $K$ makes each inner sum $0$, and independence of the $v_{j}$ over $F$ makes every $a_{ij} = 0$. So the $u_{i}v_{j}$ are a basis. $\square$
Consequence. If $\alpha \in L$, then $F \subseteq F(\alpha) \subseteq L$, so the degree of $\alpha$ divides $[L : F]$. For example, $\sqrt{2} \notin \mathbb{Q}(\sqrt[3]{2})$, because $2$ does not divide $3$.
Worked example: Two square roots
Find $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]$.
- $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$.
- Is $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$? If $\sqrt{3} = a + b\sqrt{2}$ with rational $a, b$, squaring gives $3 = a^{2} + 2b^{2} + 2ab\sqrt{2}$, so $ab = 0$. If $b = 0$, $\sqrt{3} = a$ is rational; if $a = 0$, $\sqrt{3/2} = b$ is rational. Both are impossible.
- So $x^{2} - 3$ stays irreducible over $\mathbb{Q}(\sqrt{2})$, and $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$.
Answer: $2 \cdot 2 = 4$.
Worked example: Coprime degrees
Find $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3}) : \mathbb{Q}]$.
- $\sqrt[3]{2}$ has degree $3$ and $\sqrt{3}$ has degree $2$, and both divide the total degree, so $6$ divides it.
- Adjoining $\sqrt{3}$ to $\mathbb{Q}(\sqrt[3]{2})$ at most doubles the degree, so the total is at most $3 \cdot 2 = 6$.
Answer: $6$.
Worked example: Doubling the cube
Show that $\sqrt[3]{2}$ cannot be constructed with straightedge and compass.
- Each construction step solves at most a quadratic equation, so a constructible number lies in a tower $\mathbb{Q} = K_{0} \subset K_{1} \subset \cdots \subset K_{r}$ with each step of degree $2$; the top has degree $2^{r}$.
- $\sqrt[3]{2}$ has degree $3$ (its minimal polynomial is $x^{3} - 2$), and $3$ does not divide $2^{r}$.
Answer: Not constructible: the ancient problem of doubling the cube is impossible.
Common mistake
Adding degrees instead of multiplying, or forgetting to check whether the new element is already in the field. $[\mathbb{Q}(\sqrt{2}, \sqrt{8}) : \mathbb{Q}] = 2$, not $4$: $\sqrt{8} = 2\sqrt{2}$ is already in $\mathbb{Q}(\sqrt{2})$, so the second step has degree $1$.
Key idea
$[L : F] = [L : K][K : F]$: bases multiply. The degree of any element divides the degree of the field containing it; $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4$, and degree-3 numbers like $\sqrt[3]{2}$ are not constructible.