Field Extensions

Introduction

A field extension is a pair of fields $F \subseteq K$; it is studied by forgetting, temporarily, that $K$ is a field and remembering only that $K$ is a vector space over $F$. The dimension of that vector space is the degree $[K:F]$, and almost everything in the theory of extensions is a computation with degrees. The further structure — which elements of $K$ satisfy a polynomial equation over $F$ — is what separates the algebraic from the transcendental case and produces the minimal polynomial of an element.

This article develops the calculus of degrees and minimal polynomials. It proves the tower law, which makes the degree multiplicative over intermediate fields, and it shows that an element is algebraic exactly when it generates a finite extension, in which case the extension is the quotient of a polynomial ring by the irreducible polynomial that vanishes at the element. These are the foundations on.

Throughout, $F \subseteq K$ are fields, so $\operatorname{char} F = \operatorname{char} K$ by Fields, §3, and $K$ is a vector space over $F$ in the sense. The prime subfield of $F$, the Frobenius map, and the basic field axioms are assumed from Fields. Where a result requires that $2$ be invertible, or that a polynomial be separable, this is flagged.

Extensions and Degree

The Vector-Space View

Definition. A field extension $K/F$ is a field $K$ together with a subfield $F \subseteq K$. It is written $K/F$ and read "$K$ over $F$". The field $F$ is the base field and $K$ the top field.

Since the field operations of $K$ are bilinear over $F$, indeed $F$-linear, the field $K$ is an $F$-vector space with the same addition and with scalar multiplication $F \times K \to K$ given by the multiplication of $K$. Its dimension is the degree:

Definition. The degree of $K/F$ is

$$ [K:F] = \dim_F K, $$

the cardinal number in $\{0, 1, 2, \ldots\} \cup \{\aleph_0, \aleph_1, \ldots\}$. The extension is finite if $[K:F]$ is a finite integer, and infinite otherwise.

The dimension is never $0$ for a field extension, since $1 \neq 0$ spans a one-dimensional subspace; so $[K:F] \geq 1$, with equality exactly when $K = F$, in which case $K/F$ is the trivial extension.

Proposition. Let $F \subseteq K$ be fields and let $E$ be an intermediate field, $F \subseteq E \subseteq K$. Then $E$ is an $F$-subspace of $K$.

Proof. $E$ is closed under addition and under multiplication by elements of $F \subseteq E$.

The Evaluation Homomorphism

For a field extension $K/F$ and an element $\alpha \in K$, define

$$ \varepsilon_\alpha : F[x] \longrightarrow K, \qquad \varepsilon_\alpha(p) = p(\alpha). $$

Proposition. $\varepsilon_\alpha$ is a unital ring homomorphism, and its image is the subring

$$ F[\alpha] = \{p(\alpha) : p \in F[x]\} \subseteq K, $$

the smallest subring of $K$ containing $F$ and $\alpha$.

Proof. Evaluation is additive and multiplicative by the ring axioms, and $\varepsilon_\alpha(1) = 1$. The image consists exactly of the polynomial expressions in $\alpha$ with coefficients in $F$.

Since $F[x]$ is a Euclidean domain, hence a PID, its ideals are principal, and the kernel of $\varepsilon_\alpha$ is an ideal. This single observation organises the whole theory: there are exactly two possibilities, whether that kernel is zero or not.

Definition of $F(\alpha)$ and $F(\alpha_1, \ldots, \alpha_n)$

Definition. For $\alpha_1, \ldots, \alpha_n \in K$, the field

$$ F(\alpha_1, \ldots, \alpha_n) $$

is the smallest subfield of $K$ containing $F$ and every $\alpha_i$, that is, the intersection of all such subfields. The extension $K/F$ is simple if $K = F(\alpha)$ for some $\alpha \in K$.

The subring $F[\alpha_1, \ldots, \alpha_n]$ is defined analogously, as the smallest subring of $K$ containing $F$ and the $\alpha_i$. In general $F[\alpha] \subseteq F(\alpha)$, and the inclusion is strict exactly in the transcendental case treated below.

Algebraic and Transcendental Elements

Algebraic Elements

Definition. Let $K/F$ be an extension and $\alpha \in K$. Then $\alpha$ is algebraic over $F$ if $\alpha$ is a root of a nonzero polynomial $p \in F[x]$, that is, if $p(\alpha) = 0$ for some $p \neq 0$; otherwise $\alpha$ is transcendental over $F$. The extension $K/F$ is algebraic if every element of $K$ is algebraic over $F$.

Theorem. For $\alpha \in K$ the following are equivalent:

(a) $\alpha$ is algebraic over $F$;

(b) $F[\alpha] = F(\alpha)$, so the subring generated by $F$ and $\alpha$ is already a field;

(c) $[F(\alpha):F] < \infty$;

(d) the extension $F(\alpha)/F$ is algebraic.

Proof. (a) $\Rightarrow$ (b), (c): If $0 \neq p \in F[x]$ has $p(\alpha) = 0$, then $\ker \varepsilon_\alpha \neq 0$; as $F[x]$ is a PID, $\ker \varepsilon_\alpha = (m)$ for a nonzero $m$, which is therefore a nonzero prime ideal of $F[x]$ because $F[\alpha] \cong F[x]/(m)$ is a subring of the field $K$ and so is an integral domain. A nonzero prime ideal of the PID $F[x]$ is maximal, so $F[\alpha] \cong F[x]/(m)$ is a field. Hence $F[\alpha] = F(\alpha)$. Moreover $F[x]/(m)$ has $F$-basis the classes of $1, x, \ldots, x^{n-1}$ with $n = \deg m$, by the division algorithm, so $[F(\alpha):F] = n < \infty$.

(c) $\Rightarrow$ (a): if $[F(\alpha):F] = n < \infty$, then the $n+1$ elements $1, \alpha, \ldots, \alpha^n$ are linearly dependent over $F$, giving a nonzero polynomial of degree at most $n$ vanishing at $\alpha$.

(a) $\Rightarrow$ (d): every element of the field $F(\alpha) = F[\alpha]$ is a polynomial in $\alpha$; such an element lies in the finite extension $F(\alpha)$, so it is algebraic over $F$ by (c) $\Rightarrow$ (a). (d) $\Rightarrow$ (a) is the definition applied to $\alpha$ itself.

Transcendental Elements

Theorem. For $\alpha \in K$, the following are equivalent: (a) $\alpha$ is transcendental over $F$; (b) $\varepsilon_\alpha$ is injective; (c) $F[\alpha] \cong F[x]$ and $F(\alpha) \cong F(x)$; (d) no two distinct polynomials over $F$ take the same value at $\alpha$.

Proof. (a) $\Leftrightarrow$ (b) is the definition of the kernel, and (b) $\Leftrightarrow$ (d) is the definition of injectivity. Under (b), $\varepsilon_\alpha$ is an isomorphism onto $F[\alpha]$. Then $F[\alpha]$ is an integral domain that is not a field, since $x$ is not invertible in $F[x]$ and this property is preserved by isomorphism, so $F[\alpha] \subsetneq F(\alpha)$; the field of fractions of $F[\alpha] \cong F[x]$ is $F(x)$ by Localization and the Fraction Field, and it is the smallest field containing $F[\alpha]$, hence equals $F(\alpha)$.

Examples. The numbers $e$ and $\pi$ are transcendental over $\mathbb{Q}$, by theorems of Hermite (1873) and Lindemann (1882); a real number that is transcendental over $\mathbb{Q}$ is called a transcendental number, and since $\overline{\mathbb{Q}}$ is countable while $\mathbb{R}$ is not, most real numbers are transcendental. The element $x$ is transcendental over $F$ in the rational function field $F(x)$, by construction. The formal power series $e^{x} = \sum_{n \geq 0} x^n/n!$ is transcendental over $\mathbb{C}(x)$, so it generates a simple transcendental extension of $\mathbb{C}(x)$ inside $\mathbb{C}((x))$.

Remark. Whether an element is algebraic depends on the base field. The real number $\sqrt{2}$ is algebraic over $\mathbb{Q}$, with $\mathbb{Q}(\sqrt2)/\mathbb{Q}$ of degree $2$, but over $\mathbb{R}$ one has $\sqrt{2} \in \mathbb{R}$, so $\mathbb{R}(\sqrt2) = \mathbb{R}$ is trivial. Similarly $\pi$ is algebraic over $\mathbb{R}(\pi)$, indeed it lies in it.

Minimal Polynomials

Definition and Uniqueness

Definition. Let $\alpha \in K$ be algebraic over $F$. The minimal polynomial of $\alpha$ over $F$ is the monic polynomial $m_\alpha \in F[x]$ of least degree with $m_\alpha(\alpha) = 0$.

Theorem. Let $\alpha$ be algebraic over $F$ and let $m = m_\alpha$ have degree $n$.

(a) $m$ is unique and irreducible over $F$.

(b) For $p \in F[x]$, $p(\alpha) = 0$ if and only if $m \mid p$; equivalently $\ker \varepsilon_\alpha = (m)$.

(c) $F(\alpha) = F[\alpha] \cong F[x]/(m)$, and $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is an $F$-basis. Hence $[F(\alpha):F] = n = \deg m$.

(d) Every element of $F(\alpha)$ is uniquely of the form $c_0 + c_1 \alpha + \cdots + c_{n-1}\alpha^{n-1}$ with $c_i \in F$.

Proof. Uniqueness: two monic polynomials of least degree vanishing at $\alpha$ have a difference of smaller degree vanishing at $\alpha$, hence the difference is $0$. Irreducibility: if $m = gh$ with both factors of positive degree, then $0 = m(\alpha) = g(\alpha) h(\alpha)$ and $K$ is a field, so one factor vanishes at $\alpha$, contradicting the minimality of $\deg m$. For (b), the kernel is a nonzero ideal of $F[x]$, say $(g)$; minimality of degree forces $g$ and $m$ to be associates, and monicity forces $g = m$. Statement (c) follows from the division algorithm: for $p \in F[x]$, $p = qm + r$ with $\deg r < n$, so $p(\alpha) = r(\alpha)$, and $r(\alpha) = 0$ forces $r = 0$. Uniqueness of the coefficients in (d) is linear independence of the powers, which follows because a nontrivial dependence would give a nonzero polynomial of degree less than $n$ vanishing at $\alpha$.

Remark. The irreducibility of $m_\alpha$ makes the construction of extensions concrete: for any irreducible $m \in F[x]$, the quotient $F[x]/(m)$ is a field, and the class $\bar x$ is an element algebraic over $F$ with minimal polynomial $m$. Every finite extension is generated by elements obtained this way.

Examples of Minimal Polynomials

$\alpha$	Base	$m_\alpha$	$[F(\alpha):F]$
$\sqrt{2}$	$\mathbb{Q}$	$x^2 - 2$	$2$
$i$	$\mathbb{R}$	$x^2 + 1$	$2$
$\sqrt[3]{2}$	$\mathbb{Q}$	$x^3 - 2$	$3$
$\zeta_p$, $p$ prime	$\mathbb{Q}$	$x^{p-1} + \cdots + x + 1$	$p - 1$
$\sqrt{2} + \sqrt{3}$	$\mathbb{Q}$	$x^4 - 10x^2 + 1$	$4$
$\zeta_3 = \tfrac{-1+\sqrt{-3}}{2}$	$\mathbb{Q}$	$x^2 + x + 1$	$2$
$x$ (transcendental)	$\mathbb{C}$	none	infinite

The irreducibility of $x^2 - 2$ over $\mathbb{Q}$ is the irrationality of $\sqrt2$; that of $x^3 - 2$ and of $\Phi_p$ follows from Eisenstein's criterion as in Unique Factorisation Domains. For $\alpha = \sqrt2 + \sqrt3$ one computes

$$ \alpha^2 = 5 + 2\sqrt6, \qquad \alpha^4 = 49 + 20\sqrt6, \qquad \alpha^4 - 10\alpha^2 + 1 = 0, $$

so $\alpha$ satisfies $x^4 - 10x^2 + 1$; this polynomial is irreducible over $\mathbb{Q}$ because $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$, as shown under composites below.

Conjugates

Definition. Let $\alpha$ be algebraic over $F$ with minimal polynomial $m_\alpha$, and let $L$ be a field containing $F(\alpha)$ in which $m_\alpha$ splits into linear factors. The roots of $m_\alpha$ in $L$ are the conjugates of $\alpha$ over $F$.

Proposition. The conjugates of $\alpha$ are exactly the elements $\sigma(\alpha)$ as $\sigma$ ranges over the $F$-embeddings $F(\alpha) \to L$; if $m_\alpha$ has distinct roots then there are $\deg m_\alpha$ of them.

Proof. If $\sigma$ is an $F$-embedding then $m_\alpha(\sigma(\alpha)) = \sigma(m_\alpha(\alpha)) = 0$, so $\sigma(\alpha)$ is a conjugate. Conversely every root $\beta$ of $m_\alpha$ admits an $F$-embedding $F(\alpha) \to L$ sending $\alpha$ to $\beta$, because $F(\alpha) \cong F[x]/(m_\alpha)$ and evaluation of $x$ at $\beta$ factors through this quotient.

Thus the number of conjugates of $\alpha$ equals $\deg m_\alpha$ exactly when $m_\alpha$ is separable, which is the case in characteristic $0$ , more generally, for every perfect field; the inseparable case is treated. In $\mathbb{C}$ the conjugates of $\sqrt2$ over $\mathbb{Q}$ are $\pm\sqrt2$, and the conjugates of a primitive $n$-th root of unity $\zeta_n$ over $\mathbb{Q}$ are the $\zeta_n^{k}$ with $\gcd(k,n) = 1$.

The Tower Law

Statement

Theorem (tower law). Let $F \subseteq E \subseteq K$ be fields. Then

$$ [K:F] = [K:E]\,[E:F], $$

with the convention that the product is infinite if either factor is infinite.

Proof. Let $\{e_i\}_{i \in I}$ be a basis of $E$ over $F$ and $\{k_j\}_{j \in J}$ a basis of $K$ over $E$. The claim is that the products $e_i k_j$ form a basis of $K$ over $F$.

(a) Spanning. Given $x \in K$, write $x = \sum_j a_j k_j$ with $a_j \in E$. Each $a_j$ is a finite $F$-linear combination of the $e_i$, so $x$ is a finite $F$-linear combination of the $e_i k_j$.

(b) Independence. Suppose $\sum_{i,j} c_{ij} e_i k_j = 0$ with $c_{ij} \in F$ and only finitely many nonzero. Collect the terms by $j$: $\sum_j \left(\sum_i c_{ij} e_i\right) k_j = 0$, and each coefficient $\sum_i c_{ij} e_i$ lies in $E$. Since the $k_j$ are $E$-independent, $\sum_i c_{ij} e_i = 0$ for every $j$, and since the $e_i$ are $F$-independent, $c_{ij} = 0$ for every $i, j$.

Note on infinite degrees. The same argument shows that the set of products is a basis when either index set is infinite, and then the cardinal arithmetic gives $[K:F] = [K:E]\,[E:F]$. In particular $[\mathbb{R}:\mathbb{Q}]$ is infinite, since $\mathbb{R}$ is uncountable while a finite-dimensional $\mathbb{Q}$-vector space is countable.

Consequences

Corollary (divisibility). If $K/F$ is finite and $E$ is intermediate, then $[E:F]$ and $[K:E]$ both divide $[K:F]$, and if $[K:F]$ is prime then the only intermediate fields are $F$ and $K$.

Proof. Both degrees are positive integers whose product is $[K:F]$. If $[K:F]$ is prime and $E \neq K$, then $[K:E] > 1$ and $[E:F] = [K:F]/[K:E] < [K:F]$, so $[E:F] = 1$ and $E = F$.

Corollary (finite implies algebraic). If $[K:F] < \infty$ then $K/F$ is algebraic.

Proof. For $\alpha \in K$ the subfield $F(\alpha)$ is intermediate, so $[F(\alpha):F] \le [K:F] < \infty$, and $\alpha$ is algebraic by the theorem above.

Corollary (transitivity). For $F \subseteq E \subseteq K$, the extension $K/F$ is algebraic if and only if both $K/E$ and $E/F$ are algebraic. If both are finite then $K/F$ is finite.

Proof. If $K/F$ is algebraic then $K/E$ is algebraic (fewer equations are available) and $E/F$ is algebraic. Conversely, let $\alpha \in K$ be algebraic over $E$, with minimal polynomial $m \in E[x]$ of degree $n$ and coefficients $c_0, \ldots, c_{n-1} \in E$, which are algebraic over $F$ and therefore generate a finite extension $E' = F(c_0, \ldots, c_{n-1})$ of $F$, by the corollary on finitely generated algebraic extensions below; then $F(\alpha)$ lies in $E'(\alpha)$, whose degree over $E'$ is at most $n$, so $\alpha$ is algebraic over $F$ by the tower law. The finite statement is the tower law.

Composites and an Inequality

Definition. Let $E$ and $L$ be subfields of a common field $K$, both containing $F$. The compositum $EL$ is the smallest subfield of $K$ containing both, that is, $EL = E(L)$.

Theorem. For finite extensions $E/F$ and $L/F$ inside a common field,

$$ [EL:F] \le [E:F]\,[L:F]. $$

Proof. Choose an $F$-basis $\{e_1, \ldots, e_m\}$ of $E$ and an $F$-basis $\{l_1, \ldots, l_n\}$ of $L$, with $m = [E:F]$ and $n = [L:F]$, arranged so that $e_1 = l_1 = 1$. Let $S$ be the $F$-linear span in $K$ of the $mn$ products $e_i l_j$. Then $S \subseteq EL$, and $S$ contains $E$ and $L$ and is closed under multiplication, because $$ (e_i l_j)(e_{i'} l_{j'}) = (e_i e_{i'})\,(l_j l_{j'}), $$ where $e_i e_{i'}$ is an $F$-linear combination of the $e$'s and $l_j l_{j'}$ is one of the $l$'s. So $S$ is a subring of the field $K$, of dimension at most $mn$ over $F$. A finite-dimensional domain over a field is a field: multiplication by a nonzero element is an injective $F$-linear endomorphism of $S$, hence surjective, so the element is invertible. Therefore $S$ is a field containing $E$ and $L$, so $EL \subseteq S$, whence $S = EL$ and $[EL:F] = \dim_F EL = \dim_F S \le mn = [E:F]\,[L:F]$.

Example. $E = \mathbb{Q}(\sqrt2)$, $L = \mathbb{Q}(\sqrt3)$. Then $[E:\mathbb{Q}] = [L:\mathbb{Q}] = 2$, and the inequality gives $[EL:\mathbb{Q}] \le 4$. It is exactly $4$: the field $EL = \mathbb{Q}(\sqrt2, \sqrt3)$ contains $\sqrt2 \sqrt3 = \sqrt6$, and $\mathbb{Q}(\sqrt6)$ is a subfield of degree $2$ over $\mathbb{Q}$ distinct from both $E$ and $L$, since $\sqrt6 \notin \mathbb{Q}(\sqrt2)$ (if $\sqrt6 = a + b\sqrt2$ then $6 = a^2 + 2b^2 + 2ab\sqrt2$, so $ab = 0$, and neither alternative is consistent with $a, b \in \mathbb{Q}$). Hence $EL \supsetneq E$, so $[EL:E] = 2$ and $[EL:\mathbb{Q}] = 4$. The element $\sqrt2 + \sqrt3$ has $\mathbb{Q}(\sqrt2 + \sqrt3) = \mathbb{Q}(\sqrt2, \sqrt3)$: indeed $(\sqrt2+\sqrt3)^3 = 11\sqrt2 + 9\sqrt3$, so $\sqrt2$ and $\sqrt3$ are rational combinations of $\alpha = \sqrt2+\sqrt3$ and $\alpha^3$, and $\alpha$ generates the compositum. This is the computation behind the minimal polynomial $x^4 - 10x^2 + 1$ of the table.

Corollary. If $E$ and $L$ are finite extensions of $F$ with $[EL:F] = [E:F][L:F]$, then for any basis $\{e_i\}$ of $E$ over $F$ and any basis $\{l_j\}$ of $L$ over $F$ the products $e_i l_j$ form a basis of $EL$ over $F$; the extensions are then called linearly disjoint. Without the equality the products still span but are dependent.

Simple Extensions and Generators

Degrees of Successive Simple Extensions

Theorem. Let $K = F(\alpha_1, \ldots, \alpha_n)$ and let $K_i = F(\alpha_1, \ldots, \alpha_i)$, with $K_0 = F$. If every $\alpha_i$ is algebraic over $K_{i-1}$, then

$$ [K:F] = \prod_{i=1}^{n} [K_i : K_{i-1}], $$

and each factor is the degree of the minimal polynomial of $\alpha_i$ over $K_{i-1}$.

Proof. Immediate from the tower law and the description of a simple algebraic extension.

Corollary. A finitely generated algebraic extension is finite. In particular, if $\alpha_1, \ldots, \alpha_n$ are algebraic over $F$, then $[F(\alpha_1, \ldots, \alpha_n):F] \le \prod_i [F(\alpha_i):F]$, and an algebraic extension is finitely generated if and only if it is finite.

Proof. Each $[K_i:K_{i-1}] \le [F(\alpha_i):F]$, because the minimal polynomial of $\alpha_i$ over $F$ also annihilates $\alpha_i$ over $K_{i-1}$. The last statement is the finite case, together with the observation that a finite extension is generated by a finite basis.

The Algebraic Elements Form a Subfield

Theorem. Let $K/F$ be an extension. The set

$$ E = \{\alpha \in K : \alpha \text{ is algebraic over } F\} $$

is a subfield of $K$, called the algebraic closure of $F$ in $K$.

Proof. Let $\alpha, \beta \in E$. Then $F(\alpha)$ and $F(\beta)$ are finite over $F$, so the compositum $F(\alpha, \beta)$ is finite over $F$ by the theorem on composites, hence algebraic over $F$. It contains $\alpha \pm \beta$, $\alpha \beta$, and, when $\beta \neq 0$, the element $\alpha/\beta$. Hence these lie in $E$, and $E$ is a field.

Corollary. Every element of a finite extension is algebraic over the base field, and the algebraic closure of $F$ in $K$ equals $K$ precisely when $K/F$ is algebraic.

The corollary distinguishes the algebraic from the transcendental part of an extension: for $K = \mathbb{C}$ and $F = \mathbb{Q}$, the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$ is the countable field $\overline{\mathbb{Q}}$, and the complementary part consists of the transcendental numbers. This relative algebraic closure is not to be confused with the absolute algebraic closure $\overline{F}$ constructed; it is algebraic over $F$ by construction, but it need not be algebraically closed.

Simple Extensions and the Primitive Element Theorem

A finite extension that equals $F(\alpha)$ is called simple, and by the theorem on minimal polynomials such an extension is completely described by the irreducible polynomial $m_\alpha$. Not every finite extension is visibly simple, but over a large class of fields every finite extension is simple.

Theorem (primitive element theorem). Every finite separable extension $K/F$ is simple: there exists $\alpha \in K$ with $K = F(\alpha)$.

The proof requires separability, which is developed, and is given there. Two cases make the theorem concrete: every finite extension of a field of characteristic $0$ is separable, so every finite extension of $\mathbb{Q}$ is simple; and every finite extension of a finite field is separable, so every finite field extension is simple. Separability is genuinely needed: the extension $\mathbb{F}_p(x, y)/\mathbb{F}_p(x^p, y^p)$ has degree $p^2$ but every element outside the base field has degree at most $p$ over it, so it is not simple.

Summary

A field extension $K/F$ is a field $K$ containing $F$, regarded as an $F$-vector space; its degree $[K:F] = \dim_F K$ is finite exactly for the finite extensions. The evaluation map $\varepsilon_\alpha : F[x] \to K$, $p \mapsto p(\alpha)$, is a ring homomorphism, and an element $\alpha$ is algebraic exactly when its kernel is nonzero, equivalently $F[\alpha] = F(\alpha)$, equivalently $[F(\alpha):F] < \infty$. In that case the kernel is generated by the unique monic minimal polynomial $m_\alpha$, which is irreducible and divides every polynomial over $F$ vanishing at $\alpha$; moreover $F(\alpha) = F[\alpha] \cong F[x]/(m_\alpha)$ and $[F(\alpha):F] = \deg m_\alpha$, with $\{1, \alpha, \ldots, \alpha^{\deg m_\alpha - 1}\}$ a basis. An element is transcendental exactly when $\varepsilon_\alpha$ is injective, in which case $F(\alpha) \cong F(x)$ and the extension is infinite.

The tower law $[K:F] = [K:E][E:F]$ makes degree multiplicative over intermediate fields; consequently degrees of intermediate fields divide the total degree, a finite extension is algebraic, and algebraicness is transitive. The elements of $K$ algebraic over $F$ form a subfield of $K$, the algebraic closure of $F$ in $K$, and a finitely generated algebraic extension is finite. Composites satisfy $[EL:F] \le [E:F][L:F]$, with equality for linearly disjoint extensions, and every finite separable extension is simple.

Notion	Definition	Equivalent form	Criterion
Algebraic $\alpha$	$p(\alpha) = 0$ for some $0 \neq p \in F[x]$	$F[\alpha] = F(\alpha)$	$[F(\alpha):F] < \infty$
Transcendental $\alpha$	no such $p$	$\varepsilon_\alpha$ injective	$F(\alpha) \cong F(x)$
Degree $[K:F]$	$\dim_F K$	$[K:E][E:F]$	multiplicative by the tower law
Minimal polynomial $m_\alpha$	monic of least degree with $m_\alpha(\alpha) = 0$	generator of $\ker\varepsilon_\alpha$	irreducible, $\deg m_\alpha = [F(\alpha):F]$

Summary of Notation

Symbol	Meaning
$F$, $K$, $E$, $L$	Fields; $F \subseteq K$ a field extension
$K/F$	Field extension, $K$ over $F$
$[K:F]$	Degree, $\dim_F K$
$\operatorname{char} F$	Characteristic of the field $F$
$F(\alpha_1, \ldots, \alpha_n)$	Smallest subfield containing $F$ and the $\alpha_i$
$F[\alpha_1, \ldots, \alpha_n]$	Smallest subring containing $F$ and the $\alpha_i$
$\varepsilon_\alpha$	Evaluation homomorphism $F[x] \to K$, $p \mapsto p(\alpha)$
$m_\alpha$	Minimal polynomial of $\alpha$ over $F$
$\deg m_\alpha$	$[F(\alpha):F]$ for algebraic $\alpha$
$\zeta_n$	Primitive $n$-th root of unity
$\overline{\mathbb{Q}}$	Algebraic numbers, the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$
$F(x)$	Rational function field $\operatorname{Frac}(F[x])$
$EL$	Compositum of subfields $E$ and $L$
$\operatorname{Frac}(R)$	Fraction field of a domain $R$