Suppose you want to solve $x^2 - 2 = 0$. The rational numbers $\mathbb{Q}$ contain no solution: there is no fraction whose square is $2$. You could declare this a failure and move on, but the more productive response is to ask a different question — what is the *smallest* collection of numbers that contains $\mathbb{Q}$ and also contains a solution? The answer is $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$, a field in its own right, built by adjoining the missing root. This act of enlargement — taking a field and constructing a larger one that contains it — is the subject of field extension theory.

The same construction arises throughout algebra. The complex numbers $\mathbb{C}$ are an extension of $\mathbb{R}$, formed precisely because $x^2 + 1$ has no real root. The field $\mathbb{F}_{p^n}$ with $p^n$ elements is an extension of the prime field $\mathbb{F}_p$. Number fields — the central objects of algebraic number theory — are finite extensions of $\mathbb{Q}$. In every case, the motivation is the same: the base field is too small to contain all roots of certain polynomials, and we enlarge it to accommodate them.

What makes the theory deep is that extensions carry rich structure. An extension $K/k$ is simultaneously a field and a vector space over $k$. Its dimension as a vector space — called the degree — gives a precise measure of how much $K$ exceeds $k$. Elements of $K$ can be algebraic (roots of polynomials over $k$) or transcendental. The algebraic elements are organised by their minimal polynomials. And when the extension is Galois, its symmetries — the automorphisms of $K$ fixing $k$ pointwise — form a group whose subgroup lattice mirrors the lattice of intermediate fields. This last phenomenon, the Galois correspondence, is the culmination of the entire theory.

[example: Irrationality of the Square Root of Two] We verify that $\sqrt{2} \notin \mathbb{Q}$, confirming that $\mathbb{Q}$ genuinely lacks a square root of $2$. Suppose for contradiction that $\sqrt{2} = p/q$ with $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p, q) = 1$. Then $2q^2 = p^2$, so $2 \mid p^2$, and since $2$ is prime, $2 \mid p$. Write $p = 2m$; then $2q^2 = 4m^2$, giving $q^2 = 2m^2$, so $2 \mid q$. This contradicts $\gcd(p, q) = 1$. Therefore $\sqrt{2}$ is irrational and $\mathbb{Q}(\sqrt{2})$ is a strictly larger field than $\mathbb{Q}$. [/example]

## Definition

The central object is the field extension itself. Two fields are involved: a smaller one $k$ that sits inside a larger one $K$. What matters structurally is not just the containment, but how $K$ behaves over $k$ — and the right language for this is the language of vector spaces.

[definition: Field Extension] A **field extension** $K/k$ consists of two fields $K$ and $k$ together with a field homomorphism $\iota: k \to K$. When $\iota$ is the inclusion map (i.e., $k \subset K$ as sets with the same operations), we say $k$ is a **subfield** of $K$ and write $K/k$ directly. The field $K$ is called the **extension field** (or **top field**) and $k$ is called the **base field** (or **ground field**). [/definition]

Note that the slash in $K/k$ does not denote a quotient — it is purely notational and should be read "$K$ over $k$."

Because any field homomorphism is injective, the map $\iota$ embeds $k$ as a subfield of $K$. We may therefore always assume $k \subset K$ without loss of generality (identifying $k$ with its image under $\iota$). When $K$ is a field extension of $k$, the field $K$ carries the structure of a vector space over $k$: the underlying abelian group of $K$ is the vector space, with scalar multiplication $k \times K \to K$ given by the field multiplication restricted to $k \times K$.

The dimension of this vector space is the most fundamental numerical invariant of the extension.

Before defining it, consider why dimension matters. In $\mathbb{Q}(\sqrt{2})$, every element has the form $a + b\sqrt{2}$ with $a, b \in \mathbb{Q}$. The set $\{1, \sqrt{2}\}$ spans $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ and is linearly independent over $\mathbb{Q}$ (since $\sqrt{2} \notin \mathbb{Q}$). So the extension has dimension $2$. In contrast, $\mathbb{C}/\mathbb{R}$ has dimension $2$ as well (basis $\{1, i\}$), while $\mathbb{R}/\mathbb{Q}$ has infinite dimension — indeed uncountably infinite. The degree tells us precisely how many "new directions" the extension adds.

[definition: Degree of a Field Extension] Let $K/k$ be a field extension. The **degree** of $K$ over $k$, denoted $[K : k]$, is the dimension of $K$ as a vector space over $k$: \begin{align*} [K : k] := \dim_k K. \end{align*} If $[K : k]$ is finite, the extension is called **finite**; otherwise it is called **infinite**. [/definition]

[remark: Degree Is Not the Same as the Number of Elements] The degree $[K : k]$ measures the vector space dimension, not the cardinality of $K$. A degree-$2$ extension of $\mathbb{Q}$ is an infinite set, but a degree-$2$ extension of $\mathbb{F}_p$ has exactly $p^2$ elements. The degree captures how $K$ sits over $k$ as a $k$-module, not the size of $K$ itself. [/remark]

The degree is multiplicative in towers, a fact that turns out to be one of the most useful tools in the theory. If $k \subset F \subset K$ are fields, then $K$ sits over $k$ in two steps: first $F/k$, then $K/F$. How do their degrees relate?

[quotetheorem:1248]

The Tower Law is indispensable. It tells us that the degree of a composite extension is the product of the degrees of the intermediate steps. As an immediate consequence: if $[K : k] = p$ is prime, then the only intermediate fields between $k$ and $K$ are $k$ and $K$ themselves, since any intermediate field $F$ would satisfy $p = [K : k] = [K : F][F : k]$, forcing one factor to equal $1$.

## Algebraic and Transcendental Elements

Having defined field extensions in the abstract, the next task is to understand the elements of the extension field. An element $\alpha \in K$ can relate to $k$ in two fundamentally different ways: it may satisfy a polynomial equation with coefficients in $k$, or it may not. This dichotomy — algebraic versus transcendental — drives most of the structure theory.

Why does this distinction matter? The algebraic elements are tame: they are controlled by polynomials, their minimal polynomials carry precise information, and extensions generated by algebraic elements are always finite. The transcendental elements, by contrast, behave like free variables — they generate extensions isomorphic to rational function fields, which are infinite and lack the rich polynomial structure.

[definition: Algebraic Element] Let $K/k$ be a field extension and let $\alpha \in K$. The element $\alpha$ is **algebraic** over $k$ if there exists a nonzero polynomial $f \in k[x]$ such that $f(\alpha) = 0$. If no such polynomial exists, $\alpha$ is called **transcendental** over $k$. [/definition]