The Tower Law

The Tower Law is the workhorse behind essentially every degree computation in algebraic number theory: the degree of a number field $\mathbb{Q}(\alpha_1,\dots,\alpha_n)$ over $\mathbb{Q}$ is computed by adjoining generators one at a time and multiplying the resulting simple-extension degrees, exactly the strategy foreshadowed by Finitely Generated Algebraic Extensions. In classical geometry, the impossibility of trisecting an arbitrary angle or doubling the cube with straightedge and compass rests on showing that constructible numbers lie in towers of quadratic extensions of $\mathbb{Q}$, so any constructible number has degree over $\mathbb{Q}$ equal to a power of 2 by repeated application of the Tower Law — this is the technical heart of the theory of Constructible Numbers. In Galois theory proper, the Tower Law underlies the correspondence between subgroups of a Galois group and intermediate fields, since subgroup index and field degree must match via this multiplicative law. Beyond pure algebra, the same multiplicativity principle for module rank appears in the theory of vector bundles and in coding theory, where degrees of field extensions $\mathbb{F}_{q^n}/\mathbb{F}_q$ multiply exactly as predicted, governing the structure of finite fields used throughout cryptography (e.g. AES's field $\mathbb{F}_{2^8}$ sits atop towers of smaller extensions) and in the design of Reed–Solomon and BCH error-correcting codes.

:::quiz[If $[K:F] = 12$ and $E$ is a field with $F \subseteq E \subseteq K$, which of the following is NOT a possible value of $[E:F]$?]

• 1
• 3
• 5 ✓
• 6

:::quiz[Given $F \subseteq E \subseteq K$ with $[E:F]=4$ and $[K:E]=7$, what is $[K:F]$?]

• 11
• 28 ✓
• 3
• 74

:::quiz[Suppose $[K:F]=\infty$. What can you conclude about $[E:F]$ and $[K:E]$ for an intermediate field $E$?]

• Both must be infinite
• At least one of them must be infinite ✓
• Both must be finite
• Nothing can be concluded

:::quiz[Why does the counterexample block replace a field $E$ with a ring like $\mathbb{Q}[x]/(x^2)$ to illustrate the needed hypothesis?]

• To show the Tower Law fails only when degrees are irrational
• To show that "basis," "dimension," and their rigidity properties depend on the base being a field, not merely a ring ✓
• To show rings never have finite-dimensional vector space structure
• To show the Tower Law requires characteristic zero

:::faq[Does the existence of an intermediate field of degree $d$ follow whenever $d$ divides $[K:F]$?] No — this is a common and important misconception. The Tower Law is a statement about degrees of intermediate fields that do exist; it gives a necessary divisibility condition, not a sufficient existence guarantee. There can be extensions $K/F$ where $[K:F]=n$ but no intermediate field has degree equal to some divisor $d$ of $n$. Determining which divisors actually occur as degrees of intermediate fields generally requires much finer information, such as the subgroup structure of a Galois group when $K/F$ is Galois, via the Galois correspondence.

Ask the tutor: How do I compute $[\mathbb{Q}(\alpha):\mathbb{Q}]$ when $\alpha$'s minimal polynomial over an intermediate field is different from its minimal polynomial over $\mathbb{Q}$?

Think of degree as a measure of "how much information" it takes to build one field on top of another, counted in units of dimension. Adjoining $\sqrt{2}$ to $\mathbb{Q}$ doubles the dimension, because $1$ and $\sqrt{2}$ form a basis. If you then adjoin $\sqrt{3}$ to $\mathbb{Q}(\sqrt{2})$, you again double the dimension over that new base field, because $\sqrt{3}$ satisfies a degree-2 polynomial over $\mathbb{Q}(\sqrt{2})$ (one can check $\sqrt 3 \notin \mathbb Q(\sqrt2)$). The Tower Law says the total effect on $\mathbb{Q}$ is not additive — it is not $2+2=4$ by coincidence of arithmetic — but genuinely multiplicative: $2 \times 2 = 4$, matching the fact that $\{1, \sqrt2, \sqrt3, \sqrt6\}$ is an $\mathbb{Q}$-basis for $\mathbb{Q}(\sqrt2,\sqrt3)$. The picture to hold in your mind is a grid. If $\{e_1, \dots, e_m\}$ is a basis of $E$ over $F$, and $\{k_1, \dots, k_n\}$ is a basis of $K$ over $E$, then every element of $K$ can be written uniquely as an $E$-linear combination of the $k_j$'s, and each $E$-coefficient can in turn be written uniquely as an $F$-linear combination of the $e_i$'s. Substituting one expansion into the other produces a grid of $mn$ products $e_i k_j$, and this grid is exactly a basis for $K$ over $F$. Degree multiplies because bases multiply — the proof below is really nothing more than making this substitution completely rigorous.

:::svg[Tower of fields with bases multiplying: $[K:F] = [K:E][E:F]$]

Example 1: An asymmetric biquadratic tower. Let $F = \mathbb{Q}$, $E = \mathbb{Q}(\sqrt[3]{2})$, and $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ where $\omega = e^{2\pi i/3} = \frac{-1+\sqrt{-3}}{2}$ is a primitive cube root of unity. First compute $[E:F]$. The element $\sqrt[3]{2}$ is a root of $x^3 - 2 \in \mathbb{Q}[x]$. This polynomial is irreducible over $\mathbb{Q}$ by Eisenstein's criterion at the prime $2$ (the coefficients are $1, 0, 0, -2$; $2 \mid 0, 2\mid -2$, $2^2 = 4 \nmid -2$). By the theory of Simple Extensions and F(alpha), $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = \deg(x^3-2) = 3$, with basis $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$. Next compute $[K:E]$. Over $E = \mathbb{Q}(\sqrt[3]2) \subset \mathbb{R}$, the element $\omega$ satisfies $x^2+x+1=0$ (since $\omega^3=1$, $\omega \ne 1$, and $x^3-1=(x-1)(x^2+x+1)$). This quadratic is irreducible over $E$ because $E$ is a subfield of $\mathbb{R}$ while the roots of $x^2+x+1$, namely $\omega$ and $\omega^2$, are non-real complex numbers; a real field cannot contain a root of a polynomial whose roots are all non-real, so $x^2+x+1$ has no root in $E$, and being degree 2 with no root in $E$ forces irreducibility over $E$. Hence $[K:E] = [E(\omega):E] = 2$, with basis $\{1, \omega\}$. By the Tower Law,

and moreover an explicit $\mathbb{Q}$-basis of $K$ is given by the theorem's construction:

This field $K$ is in fact the splitting field of $x^3-2$ over $\mathbb{Q}$, and the computation $[K:\mathbb{Q}]=6$ (rather than $3+2=5$ or some other guess) is exactly what the multiplicativity forces — a fact used repeatedly in Galois theory to identify $\mathrm{Gal}(K/\mathbb{Q})$ as a group of order 6.

Example 2: Using the Tower Law to rule out an intermediate degree. Let $F = \mathbb{Q}$ and $K = \mathbb{Q}(\zeta)$ where $\zeta = e^{2\pi i /7}$ is a primitive 7th root of unity. It is a standard fact (provable via irreducibility of the 7th cyclotomic polynomial $\Phi_7(x) = x^6+x^5+x^4+x^3+x^2+x+1$, itself irreducible over $\mathbb{Q}$ by an Eisenstein-after-substitution argument at $x \mapsto x+1$) that $[K:\mathbb{Q}] = 6$. Suppose someone proposes an intermediate field $E$ with $\mathbb{Q} \subseteq E \subseteq K$ and claims $[E:\mathbb{Q}] = 4$. By the Tower Law,

But $[K:\mathbb{Q}] = 6$ and $[K:E]$ must be a positive integer (as $[K:\mathbb{Q}]=6<\infty$ forces $[K:E]<\infty$ by Step 5 of the proof, and dimensions are non-negative integers). So we would need $4 \mid 6$ with quotient a positive integer, i.e. $[K:E] = 6/4 = 1.5$, which is not an integer. This is a contradiction. Hence no such intermediate field of degree 4 exists inside $K$. This illustrates the most common use of the Tower Law in practice: it is not just a computational tool for multiplying known degrees, but a constraint, via the forced divisibility $[E:F] \mid [K:F]$, that rules out entire families of candidate intermediate fields without ever having to exhibit or rule them out one at a time. (In fact the intermediate fields of $\mathbb{Q}(\zeta)/\mathbb{Q}$ correspond, via the Galois correspondence, to subgroups of the cyclic group of order 6, which has subgroups only of orders $1,2,3,6$ — exactly the divisors of 6, consistent with what the Tower Law predicts.)