A field on top of a field
You already met field extensions in Volume I as a passing remark: ℚ sits inside ℝ, and ℝ sits inside ℂ. Now we make them the main characters. A field extension L/K is nothing more than a field L together with a subfield K. The slash is read “L over K”. The single best idea in the whole subject is this: because K is a field and L is a K-vector space (you can add elements of L and multiply them by scalars from K), an extension is automatically a vector space. So we can ask for its dimension.
That dimension is the degree of the extension, written [L:K]. When it is finite we call L/K a finite extension. The smallest nontrivial example: ℂ = ℝ(i) has basis {1, i} over ℝ, so [[degree-of-an-extension|[ℂ:ℝ] = 2]]. Likewise ℚ(√2) = {a + b√2 : a, b ∈ ℚ} has basis {1, √2}, giving degree 2. Degree counts how many “independent directions” L adds on top of K.
Adjoining one element
Given α in some bigger field, K(α) is the smallest field containing K and α — a simple extension. There are exactly two things α can be. If α satisfies no nonzero polynomial over K, it is transcendental and K(α) is a copy of the rational function field K(x), infinite-dimensional. If α satisfies some polynomial, it is algebraic, and then K(α) is finite over K. Almost everything in this track lives in the algebraic case.
An algebraic extension is one in which *every* element is algebraic over K. A clean fact you should internalize immediately: every finite extension is algebraic. The reason is short — if [L:K] = n, then for any α the n+1 powers 1, α, α², …, αⁿ cannot be linearly independent over K, so some K-linear combination of them vanishes, and that is a polynomial relation. The converse fails: an algebraic extension can be infinite, for example the field of all algebraic numbers over ℚ.
Degrees multiply
The workhorse theorem of the subject is the tower law: if K ⊆ M ⊆ L is a tower of fields, then [[tower-law|[L:K] = [L:M]·[M:K]]]. Degrees are multiplicative. The proof is pure linear algebra: take a basis {mᵢ} of M over K and a basis {lⱼ} of L over M; then the products {mᵢ lⱼ} form a basis of L over K. Count them: (number of mᵢ)·(number of lⱼ). That's it.
Compute [Q(2^(1/2), 3^(1/2)) : Q].
Tower: Q < Q(2^(1/2)) < Q(2^(1/2), 3^(1/2)) = L
Step 1: [Q(2^(1/2)) : Q] = 2 (basis {1, 2^(1/2)}; x^2 - 2 is the relation)
Step 2: [L : Q(2^(1/2))] = ?
Is 3^(1/2) already in Q(2^(1/2))? Suppose 3^(1/2) = a + b*2^(1/2), a,b in Q.
Square: 3 = a^2 + 2b^2 + 2ab*2^(1/2).
Since 2^(1/2) is irrational, need ab = 0.
b = 0 => 3 = a^2, no rational a.
a = 0 => 3 = 2b^2, no rational b.
Contradiction, so 3^(1/2) is NOT in Q(2^(1/2)), and x^2 - 3 stays irreducible there.
Hence [L : Q(2^(1/2))] = 2.
Tower law: [L : Q] = 2 * 2 = 4.
A Q-basis of L: {1, 2^(1/2), 3^(1/2), 6^(1/2)}.