Ring theory ended with a recipe: take an irreducible polynomial, divide it out, and you get a field. This page turns that recipe around. Instead of asking "what field did I build?", we ask "what new field contains the old one, and how much bigger is it?" The answer to "how much bigger" turns out to be a dimension, and that one idea (fields are vector spaces over smaller fields) drives everything from here to Galois theory.
Start with the vocabulary. Whenever one field sits inside another, we call the big one an extension of the small one.
If $F \subseteq K$ are fields, then $K$ is called an extension of $F$, denoted $K/F$ or $\dfrac{K}{F}$.
$K/F$ looks exactly like a quotient. It is not a quotient. A field's only ideals are $0$ and the whole field, so there's nothing interesting to divide out anyway. Whoever chose this slash owes me an apology.
The small field gets a name too, because everything about the extension will be measured relative to it.
In an extension $K/F$, the field $F$ is called the base field (or ground field).
Some extensions you already know.
Most extensions we care about are built by throwing a few new elements into a field and closing up under $+, -, \times, \div$. The smallest field you can get that way has a name.
For an element $\alpha \in \mathbb{C}$, $\mathbb{Q}(\alpha)$ denotes the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q} \cup \{\alpha\}$, i.e., the intersection of all subfields of $\mathbb{C}$ containing $\mathbb{Q} \cup \{\alpha\}$. It is called the field obtained by adjoining $\alpha$ to $\mathbb{Q}$.
More generally, if $K/F$ is an extension and $\alpha_1, \dots, \alpha_n \in K$, then $F(\alpha_1, \dots, \alpha_n)$ is the smallest subfield of $K$ containing $F \cup \{\alpha_1, \dots, \alpha_n\}$.
"Smallest" makes sense because an intersection of subfields is again a subfield. Notice the round brackets: $F(\alpha)$ is a field, while $F[x]$ (square brackets) is a polynomial ring.
Describe $\mathbb{Q}(\sqrt{2})$.
Any field containing $\mathbb{Q}$ and $\sqrt2$ contains every $a + b\sqrt2$ with $a, b \in \mathbb{Q}$. That set is already a field: it's closed under $+$ and $\times$ (because $(\sqrt2)^2 = 2 \in \mathbb{Q}$), and inverses stay inside by rationalizing the denominator:
$$\frac{1}{a + b\sqrt2} = \frac{a - b\sqrt2}{a^2 - 2b^2} \qquad (a^2 - 2b^2 \neq 0 \text{ since } \sqrt2 \notin \mathbb{Q}).$$
So $\mathbb{Q}(\sqrt2) = \{a + b\sqrt2 : a, b \in \mathbb{Q}\}$.
Adjoining several elements is the same as adjoining them one at a time.
If $K/F$ is a field extension and $\alpha_1, \dots, \alpha_n \in K$, then
$$F(\alpha_1, \dots, \alpha_n) = \big(F(\alpha_1, \dots, \alpha_{n-1})\big)(\alpha_n).$$
Proof. Call the left side $L$ and the right side $R$. $R$ is a field containing $F(\alpha_1, \dots, \alpha_{n-1}) \supseteq F \cup \{\alpha_1, \dots, \alpha_{n-1}\}$ and also $\alpha_n$, so $L \subseteq R$ by minimality of $L$. Conversely, $L$ is a field containing $F \cup \{\alpha_1, \dots, \alpha_{n-1}\}$, so $L \supseteq F(\alpha_1, \dots, \alpha_{n-1})$; it also contains $\alpha_n$, so $L \supseteq R$ by minimality of $R$. $\blacksquare$
So $\mathbb{Q}(\sqrt2, \sqrt3) = \mathbb{Q}(\sqrt2)(\sqrt3) = \{x + y\sqrt3 : x, y \in \mathbb{Q}(\sqrt2)\}$, provided $\sqrt3 \notin \mathbb{Q}(\sqrt2)$ (we'll check that in a minute).
There's a second way to get extensions that doesn't need an ambient field like $\mathbb{C}$ at all. It's the recipe from ring theory.
On Prime & Maximal Ideals: in a PID, an irreducible element generates a maximal ideal, and a ring mod a maximal ideal is a field.
Let $F$ be a field and $f \in F[x]$ an irreducible polynomial. Why is $K = F[x]/(f)$ an extension of $F$?
$(f)$ is a maximal ideal in the PID $F[x]$, so $K$ is a field. The map $\phi\colon F \to K$, $\phi(a) = a + (f)$, is a ring homomorphism. Its kernel is an ideal of the field $F$ not containing $1$ (since $1 \notin (f)$), so $\ker(\phi) = \{0\}$. Hence $\phi(F) \cong F$ is an isomorphic copy of $F$ inside $K$, and we identify $a$ with $a + (f)$ to view $K$ as an extension of $F$.
Strictly speaking, $F$ is not a subset of $F[x]/(f)$: the elements of the quotient are cosets. "Extension" here means "contains an isomorphic copy of $F$." Everyone makes this identification silently, and from now on, so do we.
Here's the key move. If $K/F$ is an extension, you can add elements of $K$, and you can multiply an element of $K$ by an element of $F$. That's exactly a vector space: $K$ is an $F$-vector space under field addition and scalar multiplication $(c, k) \mapsto ck$ for $c \in F$, $k \in K$. Any two $F$-bases for $K$ have the same size, so the dimension makes sense.
Bases and dimension on Bases & Dimension were for real scalars, but the proofs only add, multiply, and divide scalars, which works in any field. (That's also how the Finite Fields page counted $p^n$ elements.)
The degree of the extension $K/F$, denoted $[K:F]$, is $[K:F] := \dim_F(K)$.
Find $[\mathbb{C}:\mathbb{R}]$, $[\mathbb{Q}(\sqrt2):\mathbb{Q}]$, $[\mathbb{F}_{p^n}:\mathbb{F}_p]$, and $[\mathbb{R}:\mathbb{Q}]$.
Degrees multiply up a tower of fields. This is the single most-used fact in field theory, so it's worth seeing why.
Suppose $L/K$ and $K/F$ are field extensions. Let $\mathcal{A}$ be a $K$-basis for $L$ and $\mathcal{B}$ be an $F$-basis for $K$. Then:
Proof. Spanning. Let $x \in L$. Since $\mathcal{A}$ spans $L$ over $K$, $x = \sum_{\alpha} k_\alpha \alpha$ for some $k_\alpha \in K$ (finitely many nonzero). Since $\mathcal{B}$ spans $K$ over $F$, each $k_\alpha = \sum_{\beta} c_{\alpha\beta}\, \beta$ with $c_{\alpha\beta} \in F$. Substituting, $x = \sum_{\alpha, \beta} c_{\alpha\beta}\, \alpha\beta$.
Independence. Suppose $\sum_{\alpha, \beta} c_{\alpha\beta}\, \alpha\beta = 0$ with $c_{\alpha\beta} \in F$. Group by $\alpha$: $\sum_\alpha \big(\sum_\beta c_{\alpha\beta}\beta\big)\alpha = 0$. The inner sums lie in $K$ and $\mathcal{A}$ is $K$-independent, so each $\sum_\beta c_{\alpha\beta}\beta = 0$. Then $\mathcal{B}$ is $F$-independent, so every $c_{\alpha\beta} = 0$. The products $\alpha\beta$ are also pairwise distinct (if $\alpha\beta = \alpha'\beta'$ with $(\alpha,\beta) \neq (\alpha', \beta')$, the same argument applied to $\alpha\beta - \alpha'\beta' = 0$ gives a contradiction), so the basis has $|\mathcal{A}||\mathcal{B}|$ elements, which proves 2 (with the convention that $\infty$ times anything nonzero is $\infty$). $\blacksquare$
Pick a tower below. The grid is the basis in part 1: each cell is a $K$-basis element (column) times an $F$-basis element (row). Then press Expand an element to watch the spanning half of the proof happen on an actual number.
The Tower Law has an immediate, very practical consequence: if $F \subseteq K \subseteq L$, then $[K:F]$ divides $[L:F]$. That turns "is this element in that field?" into arithmetic.
Show that $\sqrt3 \notin \mathbb{Q}(\sqrt2)$, and conclude $[\mathbb{Q}(\sqrt2, \sqrt3):\mathbb{Q}] = 4$.
Suppose $\sqrt3 = a + b\sqrt2$ with $a, b \in \mathbb{Q}$. Squaring, $3 = a^2 + 2b^2 + 2ab\sqrt2$. Since $\{1, \sqrt2\}$ is a basis, $2ab = 0$ and $a^2 + 2b^2 = 3$. If $b = 0$ then $a^2 = 3$; if $a = 0$ then $b^2 = \frac32$. Neither has a rational solution. So $\sqrt3 \notin \mathbb{Q}(\sqrt2)$.
Then $\{1, \sqrt3\}$ is a $\mathbb{Q}(\sqrt2)$-basis of $\mathbb{Q}(\sqrt2)(\sqrt3)$ (same argument as for $\mathbb{Q}(\sqrt2)/\mathbb{Q}$), so by the Tower Law, $[\mathbb{Q}(\sqrt2, \sqrt3):\mathbb{Q}] = 2 \cdot 2 = 4$, with basis $\{1, \sqrt2, \sqrt3, \sqrt6\}$.
The Tower Law needs a tower: one field inside the next. Two fields side by side, like $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt3)$, don't multiply on their own; you have to build the tower $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt2) \subseteq \mathbb{Q}(\sqrt2, \sqrt3)$ first and find the top degree separately. That's what the example above had to do.
Inside $\mathbb{C}$, every polynomial has roots. But what about over $\mathbb{F}_2$, where there's no "$\mathbb{C}$" lying around? Kronecker says you can always manufacture a root out of thin air.
If $F$ is a field and $f \in F[x]$ is irreducible, then there exists an extension of $F$ containing a root of $f$.
Proof. Set $K = F[x]/(f)$ (a field extension of $F$, as above), and define $\alpha := x + (f) \in K$. Writing $f(x) = \sum_{i=0}^n a_i x^i$ with $a_i \in F$, evaluate $f$ at $\alpha$:
$$\begin{aligned} f(\alpha) &= \sum_{i=0}^n a_i (x + (f))^i = \sum_{i=0}^n a_i (x^i + (f)) \\ &= \Big(\sum_{i=0}^n a_i x^i\Big) + (f) = f(x) + (f) = 0 + (f). \end{aligned}$$
Hence $\alpha$ is a root of $f$ in $K$. $\blacksquare$
The proof is short enough to feel like a trick. Run it on real polynomials: the new root is just "$x$, remembered modulo $f$."
Irreducibility is what makes $K$ a field. Kronecker's theorem still gives a root for any non-constant $f$: factor $f$ and apply the theorem to one irreducible factor. Just don't build $F[x]/(f)$ with a reducible $f$ and call it a field; $\mathbb{R}[x]/(x^2 - 1)$ has zero divisors ($(x-1)(x+1) = 0$).
Apply Kronecker over and over, one root at a time, and every polynomial eventually factors into linear pieces. The smallest field where that happens is the splitting field.
Every field has a "smallest piece" that all its extensions share. Find it by adding $1$ to itself.
The characteristic of a field $F$, denoted $\operatorname{char}(F)$, is the smallest positive integer $n$ such that $n \cdot 1_F = 0$, or $0$ if no such $n$ exists.
So $\operatorname{char}(\mathbb{Q}) = \operatorname{char}(\mathbb{C}) = 0$, and $\operatorname{char}(\mathbb{F}_{p^n}) = p$. You never see characteristic $6$:
For any field $F$, $\operatorname{char}(F)$ is either $0$ or a prime $p$. If $\operatorname{char}(F) = 0$, the prime subfield of $F$ is isomorphic to $\mathbb{Q}$. If $\operatorname{char}(F) = p$, the prime subfield is isomorphic to $\mathbb{F}_p \cong \mathbb{Z}/p\mathbb{Z}$.
Proof. The map $\mathbb{Z} \to F$, $m \mapsto m \cdot 1_F$, is a ring homomorphism; its image is an integral domain, so its kernel is a prime ideal $(n)$ of $\mathbb{Z}$, with $n = \operatorname{char}(F)$. Prime ideals of $\mathbb{Z}$ are $(0)$ and $(p)$, so $n = 0$ or $n = p$. If $n = p$, the image is $\mathbb{Z}/p\mathbb{Z}$, already a field, and it lies in every subfield (every subfield contains $1$). If $n = 0$, the map is injective, and every subfield contains the fractions $(a \cdot 1_F)(b \cdot 1_F)^{-1}$, a copy of $\mathbb{Q}$. $\blacksquare$
The Finite Fields page defined the prime subfield and found it inside finite fields. Now we know the full story: every field is an extension of either $\mathbb{Q}$ or some $\mathbb{F}_p$.
Characteristic $p$ does not mean finite. The field of rational functions $\mathbb{F}_p(t)$ (fractions of polynomials with $\mathbb{F}_p$ coefficients) is infinite, yet $p \cdot 1 = 0$ in it.
The simplest extensions are the ones you get by adjoining a single element.
An extension $K/F$ is a simple extension if $K = F(\alpha)$ for some $\alpha \in K$.
If $K = F(\alpha)$, the generator $\alpha$ is called a primitive element.
"Simple" is about how the field can be described, not how it was built. $\mathbb{Q}(\sqrt2, \sqrt3)$ was built with two generators, but it turns out one is enough (first exercise below).
The Tower Law does most of the work in each of these.
Show that $\mathbb{Q}(\sqrt2, \sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt3)$, so $\mathbb{Q}(\sqrt2, \sqrt3)/\mathbb{Q}$ is a simple extension.
Let $\gamma = \sqrt2 + \sqrt3$. Clearly $\gamma \in \mathbb{Q}(\sqrt2, \sqrt3)$, so $\mathbb{Q}(\gamma) \subseteq \mathbb{Q}(\sqrt2, \sqrt3)$.
For the other inclusion, $\gamma^{-1} = \dfrac{1}{\sqrt3 + \sqrt2} = \dfrac{\sqrt3 - \sqrt2}{3 - 2} = \sqrt3 - \sqrt2 \in \mathbb{Q}(\gamma)$. So $\mathbb{Q}(\gamma)$ contains $\frac12(\gamma - \gamma^{-1}) = \sqrt2$ and $\frac12(\gamma + \gamma^{-1}) = \sqrt3$, hence $\mathbb{Q}(\sqrt2, \sqrt3) \subseteq \mathbb{Q}(\gamma)$. $\blacksquare$
Given that $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}] = 3$, show that $\sqrt2 \notin \mathbb{Q}(\sqrt[3]2)$ and that $[\mathbb{Q}(\sqrt[3]2, \sqrt2):\mathbb{Q}] = 6$.
If $\sqrt2 \in \mathbb{Q}(\sqrt[3]2)$, then $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt2) \subseteq \mathbb{Q}(\sqrt[3]2)$, and the Tower Law says $2 = [\mathbb{Q}(\sqrt2):\mathbb{Q}]$ divides $3$. It doesn't, so $\sqrt2 \notin \mathbb{Q}(\sqrt[3]2)$.
Let $L = \mathbb{Q}(\sqrt[3]2, \sqrt2)$. Both $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt[3]2)$ sit inside $L$, so $2 \mid [L:\mathbb{Q}]$ and $3 \mid [L:\mathbb{Q}]$, hence $6 \mid [L:\mathbb{Q}]$. On the other hand $L = \mathbb{Q}(\sqrt[3]2)(\sqrt2)$, and $\{1, \sqrt2\}$ spans it over $\mathbb{Q}(\sqrt[3]2)$ (since $\sqrt2^2 = 2$), so $[L:\mathbb{Q}(\sqrt[3]2)] \le 2$ and $[L:\mathbb{Q}] \le 6$. So $[L:\mathbb{Q}] = 6$.
Suppose $[K:F] = p$ is prime. Show that $K = F(\alpha)$ for every $\alpha \in K \setminus F$.
$F \subseteq F(\alpha) \subseteq K$, so by the Tower Law $[F(\alpha):F]$ divides $p$: it is $1$ or $p$. It isn't $1$, because then $F(\alpha) = F$ (a $1$-dimensional $F$-space containing $1$ is $F$), but $\alpha \notin F$. So $[F(\alpha):F] = p = [K:F]$, and then $[K:F(\alpha)] = 1$, i.e. $K = F(\alpha)$. $\blacksquare$
Two fields can have the same degree over $\mathbb{Q}$ and still be completely different.
Prove that $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt3)$ are not isomorphic as fields, even though both have degree $2$ over $\mathbb{Q}$.
Proof. Suppose $\sigma\colon \mathbb{Q}(\sqrt3) \to \mathbb{Q}(\sqrt2)$ is a field isomorphism. Then $\sigma(1) = 1$, so $\sigma(n) = n$ for every integer $n$, and $\sigma(a/b) = \sigma(a)\sigma(b)^{-1} = a/b$: $\sigma$ fixes $\mathbb{Q}$ pointwise.
Let $\sigma(\sqrt3) = a + b\sqrt2$ with $a, b \in \mathbb{Q}$. Then $3 = \sigma(3) = \sigma(\sqrt3)^2 = a^2 + 2b^2 + 2ab\sqrt2$. As in the example above, this forces $ab = 0$ and $a^2 + 2b^2 = 3$, which has no rational solutions ($a^2 = 3$ or $b^2 = \frac32$). Contradiction. $\blacksquare$
(The point: an isomorphism must fix the prime subfield, so it has to send a square root of $3$ to a square root of $3$, and $\mathbb{Q}(\sqrt2)$ doesn't have one.)