The Degree of an Extension

Every field extension E/F comes with a single number that measures exactly "how much bigger" the top field is than the base — and it is not some vague notion of size, it is a crisp integer (or \infty). The trick, as always in this module, is to forget that E is a field and remember only that it is a vector space over F. The degree of the extension is simply the dimension of that vector space.

The degree [E : F] is the dimension of E as a vector space over F.
The extension is finite if [E : F] < \infty and infinite otherwise.
For an algebraic \alpha, [F(\alpha) : F] = \deg m_\alpha, the degree of the minimal polynomial.

Read [E : F] aloud as "the degree of E over F." It counts the number of vectors in a basis of E over F — the number of independent "directions" you gain by climbing from the base field to the big one.

Reading the degree off a basis

Because degree is dimension, computing it means finding a basis and counting it. The core examples fall out at once:

[\mathbb{C} : \mathbb{R}] = 2, with basis \{1, i\} — every complex number is a \cdot 1 + b \cdot i.
[\mathbb{Q}(\sqrt 2) : \mathbb{Q}] = 2, with basis \{1, \sqrt 2\} — matching \deg(x^2 - 2) = 2.
[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3, with basis \{1, \sqrt[3]{2}, \sqrt[3]{4}\} — matching \deg(x^3 - 2) = 3.
[\mathbb{R} : \mathbb{Q}] = \infty — the reals are an infinite-dimensional rational vector space (they are even uncountable, so no finite basis could exist).

The link to the minimal polynomial is the workhorse. Adjoining a single algebraic \alpha whose minimal polynomial has degree n produces an extension of degree exactly n, because \{1, \alpha, \alpha^2, \dots, \alpha^{n-1}\} is a basis: every higher power of \alpha collapses back down using the minimal polynomial relation.

The Tower Law

The reason the degree is such a powerful bookkeeping device is one clean multiplicative rule. Stack three fields in a tower F \subseteq K \subseteq E, and the degrees multiply.

If F \subseteq K \subseteq E are fields, then [E : F] = [E : K] \cdot [K : F].
If \{\beta_i\} is a basis of E over K and \{\gamma_j\} a basis of K over F, then the products \{\beta_i \gamma_j\} form a basis of E over F.

The proof is the second bullet: you multiply the two bases together. There are [E : K] choices of \beta_i and [K : F] choices of \gamma_j, giving [E : K] \cdot [K : F] products, and one checks they are independent and spanning. The lattice below shows this in action for the tower \mathbb{Q} \subseteq \mathbb{Q}(\sqrt 2) \subseteq \mathbb{Q}(\sqrt 2, \sqrt 3): each edge carries its degree, and the degrees multiply as you climb.

Worked example: ℚ(√2, √3) over ℚ

Split the climb into two steps. First adjoin \sqrt 2: [\mathbb{Q}(\sqrt 2) : \mathbb{Q}] = 2. Now adjoin \sqrt 3 on top. Its minimal polynomial over the bigger base \mathbb{Q}(\sqrt 2) is still x^2 - 3 (one checks \sqrt 3 \notin \mathbb{Q}(\sqrt 2)), so [\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}(\sqrt 2)] = 2. The Tower Law finishes the job:

[\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}] = [\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}(\sqrt 2)] \cdot [\mathbb{Q}(\sqrt 2) : \mathbb{Q}] = 2 \cdot 2 = 4,

with basis \{1, \sqrt 2, \sqrt 3, \sqrt 6\} — precisely the products of \{1, \sqrt 2\} and \{1, \sqrt 3\}, exactly as the theorem promises.

A beautiful corollary drops out for free. Since [E : F] = [E : K] \cdot [K : F], the degree [K : F] of any intermediate field divides the total degree [E : F] — a Lagrange-flavoured divisibility fact. In our degree-4 tower the only intermediate degrees possible are 1, 2, 4; a field of degree 3 simply cannot fit inside.

The most seductive error is to add the two floor-heights instead of multiplying them. For the tower \mathbb{Q} \subseteq \mathbb{Q}(\sqrt 2) \subseteq \mathbb{Q}(\sqrt 2, \sqrt 3) both steps have degree 2, and 2 + 2 = 4 = 2 \cdot 2 — so the wrong method accidentally gives the right answer and reinforces the habit. Try it on a 3-then-2 tower and the mask slips: [\mathbb{Q}(\sqrt[6]{2}) : \mathbb{Q}] = 6 = 3 \cdot 2, not 3 + 2 = 5.

A second trap: the Tower Law needs a genuine tower, a chain of fields each contained in the next, F \subseteq K \subseteq E. If K is not a subfield sitting between F and E, there is no law to apply — you cannot multiply degrees of two extensions that do not stack.

The classical impossibility proofs are Tower-Law arithmetic in disguise. A length is constructible with straightedge and compass only if it lives in a field reached from \mathbb{Q} by a tower of degree-2 steps — so its degree over \mathbb{Q} must be a power of 2. Trisecting a 60^\circ angle would require building \cos 20^\circ, whose minimal polynomial has degree 3. But 3 does not divide any power of 2, so the number can never appear in such a tower. Two millennia of failed attempts, settled by a divisibility check.