---
title: Field Extensions and Algebraic Elements
module: Field Theory
moduleNumber: 10
lessonNumber: 1
order: 1001
summary: >
  A field extension makes a larger field K into a vector space over a smaller
  field F, and its degree [K:F] is that dimension. Adjoining a root of an
  irreducible polynomial builds a simple extension F(α) isomorphic to F[x]/(m),
  whose degree is the degree of the minimal polynomial. The tower law makes
  these degrees multiply, which turns algebra over fields into bookkeeping with
  integers.
topics: [Field Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 13 — Field Theory; §13.1 Basic Theory of Field Extensions, §13.2 Algebraic Extensions"
  - book: Judson
    ref: "Ch. 21 — Fields; §21.1 Extension Fields"
draft: false
---

A field is a commutative ring with identity in which every nonzero element has
an inverse: arithmetic in which division always works.
Two questions organize the subject. Given a polynomial with no root in $F$, like
$x^2 + 1$ over $\mathbb{R}$ or $x^2 - 2$ over $\mathbb{Q}$, is there a larger
field in which it has a root? And given a larger field $K \supseteq F$, how much
larger is it? The second question is answered by a single integer, the first by
a quotient of a polynomial ring.

## Characteristic and the prime subfield

Every field contains a smallest subfield. Repeatedly adding the identity
$1_F$ to itself gives elements
$1_F,\ 1_F + 1_F,\ 1_F + 1_F + 1_F, \dots$, which either are all distinct or
eventually return to $0$.

> **Definition (Characteristic).** The **characteristic** of a field $F$,
> written $\operatorname{ch}(F)$, is the smallest positive integer $p$ with
> $p \cdot 1_F = 0$ if such a $p$ exists, and $0$ otherwise.

The characteristic is $0$ or prime. If $n = ab$ were composite with
$n \cdot 1_F = 0$, then $(a \cdot 1_F)(b \cdot 1_F) = 0$, and because a field has
no zero divisors one of the factors already vanishes, contradicting minimality.
The map $\varphi : \mathbb{Z} \to F$, $n \mapsto n \cdot 1_F$, is a ring
homomorphism with kernel $\operatorname{ch}(F)\mathbb{Z}$, so its image is a copy
of either $\mathbb{Z}$ (characteristic $0$) or $\mathbb{Z}/p\mathbb{Z}$
(characteristic $p$). Since $F$ is a field it contains the fractions of that
image.

> **Definition (Prime subfield).** The **prime subfield** of $F$ is the subfield
> generated by $1_F$. It is isomorphic to $\mathbb{Q}$ when
> $\operatorname{ch}(F) = 0$ and to $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ when
> $\operatorname{ch}(F) = p$.

So $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ all have characteristic $0$ and
prime subfield $\mathbb{Q}$; the finite field $\mathbb{F}_p$ and the rational
function field $\mathbb{F}_p(x)$ both have characteristic $p$. Every field is an
extension of its prime subfield, so the subject reduces to what sits between
that subfield and a given field.

## Extensions and degree

> **Definition (Extension field).** If $K$ is a field containing a subfield $F$,
> then $K$ is an **extension** of $F$, written $K/F$. The multiplication of $K$
> makes it a vector space over $F$.

A field containing $F$ is in particular an $F$-vector space, and vector spaces
have dimensions.

> **Definition (Degree).** The **degree** of $K/F$, written $[K : F]$, is
> $\dim_F K$, the dimension of $K$ as a vector space over $F$. The extension is
> **finite** if $[K : F]$ is finite and **infinite** otherwise.

For example $\mathbb{C} = \mathbb{R} + \mathbb{R}i$ has $\{1, i\}$ as an
$\mathbb{R}$-basis, so $[\mathbb{C} : \mathbb{R}] = 2$. The notation $K/F$ is
read "$K$ over $F$" and is not a quotient; it only records that $F$ sits inside
$K$.

## Building an extension with a root

Given an irreducible $p(x) \in F[x]$, we want a field where it acquires a root.
The construction reuses the fact that $F[x]$ is a principal ideal domain in
which the ideal $(p(x))$ generated by an irreducible element is
maximal.[^df-polyring] Quotienting a ring by a maximal ideal produces a field,
so the following works.

> **Theorem (Existence of a root).** Let $p(x) \in F[x]$ be irreducible. Then
> $K = F[x]/(p(x))$ is a field containing an isomorphic copy of $F$ in which
> $p(x)$ has a root, namely the class $\theta = x + (p(x))$.

> **Proof.** The canonical projection $\pi : F[x] \to K$ restricts to a nonzero
> homomorphism on the constants $F$, and a nonzero field homomorphism is
> injective, so $F$ embeds in $K$. Identifying $F$ with its image,
> $p(\theta) = \overline{p(x)} = 0$ in the quotient, so $\theta$ is the promised
> root.[^df-thm3]

The elements of $K$ have a clean description.

> **Theorem (Basis of $F[x]/(p)$).** If $p(x)$ is irreducible of degree $n$ and
> $\theta = x + (p(x))$, then $1, \theta, \theta^2, \dots, \theta^{n-1}$ is an
> $F$-basis of $K = F[x]/(p(x))$. Hence $[K : F] = n$ and
> $$
> K = \{a_0 + a_1\theta + \cdots + a_{n-1}\theta^{n-1} : a_i \in F\}.
> $$

> **Proof.** Every polynomial $a(x)$ reduces modulo $p(x)$, by the division
> algorithm in $F[x]$, to a remainder of degree less than $n$, so the powers
> $1, \dots, \theta^{n-1}$ span. A nontrivial dependence among them would be a
> nonzero polynomial of degree less than $n$ divisible by $p(x)$, which is
> impossible on degree grounds; the powers are independent.[^df-thm4]

Addition in $K$ is polynomial addition; multiplication is polynomial
multiplication followed by reduction of the product modulo $p(x)$.

$$
% caption: The quotient F[x]/(p) collapses the polynomial ring onto the n-dimensional
% space of remainders modulo p, on which the class of x acts as a root of p.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=30mm, minimum height=13mm, align=center] (poly) at (0,0) {polynomials\\F[x]};
\node[draw=acc, thick, minimum width=44mm, minimum height=13mm, align=center, fill=acc!8] (quot) at (6.6,0) {remainders mod p\\basis 1, t, t squared, up to t to the n minus 1};
\draw[->, acc, very thick] (poly) -- (quot) node[midway, above, font=\footnotesize] {reduce mod p};
\node[font=\footnotesize, align=center, text=black] at (6.6,-1.5) {this space is F(t), where p(t) = 0};
\end{tikzpicture}
$$

For $F = \mathbb{R}$ and $p(x) = x^2 + 1$, the quotient
$\mathbb{R}[x]/(x^2 + 1)$ is a degree-$2$ extension with elements $a + b\theta$
and the rule $\theta^2 = -1$; the map $a + bx \mapsto a + bi$ is an isomorphism
onto $\mathbb{C}$. Taking $F = \mathbb{Q}$ and $p(x) = x^2 - 2$ gives a
degree-$2$ field whose root is written $\sqrt{2}$, with
$$
(a + b\sqrt 2)(c + d\sqrt 2) = (ac + 2bd) + (ad + bc)\sqrt 2.
$$
Over $\mathbb{F}_2$ the polynomial $x^2 + x + 1$ is irreducible, so
$\mathbb{F}_2[x]/(x^2 + x + 1)$ is a field of four elements
$\{0, 1, \theta, \theta + 1\}$ with $\theta^2 = \theta + 1$.

### Inverses by the Euclidean algorithm

Division in $F(\theta)$ looks harder than the other operations, but it reduces
to the extended Euclidean algorithm. If $a(\theta) \ne 0$ then
$\gcd(a(x), p(x)) = 1$, because $p$ is irreducible and does not divide the
lower-degree $a$. The extended algorithm produces $u(x), v(x)$ with
$u(x)a(x) + v(x)p(x) = 1$; reducing modulo $p$ leaves $u(\theta)a(\theta) = 1$,
so $u(\theta)$ is the inverse.

```algorithm
caption: $\textsc{Inverse}(a(x), p(x))$ — invert $a(\theta)$ in $F[x]/(p(x))$
require $p$ irreducible, $\deg a < \deg p$, $a \ne 0$
run the extended Euclidean algorithm on $a(x)$ and $p(x)$
obtain $u(x), v(x)$ with $u(x)\,a(x) + v(x)\,p(x) = 1$
return $u(x) \bmod p(x)$   // this is $a(\theta)^{-1}$
```

> **Worked example.** Invert $1 + \theta$ in $\mathbb{Q}[x]/(x^3 - 2)$, where
> $\theta^3 = 2$. Divide $x^3 - 2$ by $x + 1$:
> $$
> x^3 - 2 = (x + 1)(x^2 - x + 1) - 3.
> $$
> The remainder $-3$ is a nonzero constant, confirming $\gcd(x + 1, x^3 - 2) = 1$.
> Rearranging isolates $1$:
> $$
> (x + 1)\cdot\tfrac13(x^2 - x + 1) - \tfrac13(x^3 - 2) = 1.
> $$
> Reducing modulo $x^3 - 2$ kills the second term, so
> $$
> (1 + \theta)^{-1} = \tfrac13(\theta^2 - \theta + 1).
> $$
> Check: $(1 + \theta)(\theta^2 - \theta + 1) = \theta^3 + 1 = 2 + 1 = 3$, and
> multiplying by $\tfrac13$ gives $1$.

## Generated subfields and simple extensions

Intersecting all subfields of $K$ that contain $F$ and a set of elements gives a
smallest such subfield.

> **Definition (Generated field, simple extension).** For $\alpha, \beta, \dots
> \in K$, the smallest subfield of $K$ containing $F$ and those elements is
> $F(\alpha, \beta, \dots)$. If $K = F(\alpha)$ for a single $\alpha$, the
> extension is **simple** and $\alpha$ is a **primitive element** for it.

The construction $F[x]/(p(x))$ and the concrete field $F(\alpha)$ are the same
object whenever $\alpha$ is a root of $p$.

> **Theorem.** Let $p(x) \in F[x]$ be irreducible and let $K \supseteq F$
> contain a root $\alpha$ of $p$. Then $F(\alpha) \cong F[x]/(p(x))$.

> **Proof.** The evaluation map $F[x] \to F(\alpha)$, $a(x) \mapsto a(\alpha)$,
> kills $p(x)$, so it factors through the field $F[x]/(p(x))$; the induced map
> is a nonzero homomorphism of fields, hence an isomorphism onto its image, and
> that image contains $F$ and $\alpha$, so it is all of $F(\alpha)$.[^df-thm6]

A useful consequence, proved the same way, is that the two roots of an
irreducible are algebraically interchangeable: an isomorphism
$\varphi : F \to F'$ carrying $p(x)$ to $p'(x)$ extends to
$F(\alpha) \cong F'(\beta)$ sending any root $\alpha$ of $p$ to any root $\beta$
of $p'$.[^df-thm8] The same lifting step drives the uniqueness of
[splitting fields](/abstract-algebra/field-theory/splitting-fields-and-algebraic-closure).

## Algebraic and transcendental elements

> **Definition (Algebraic, transcendental).** An element $\alpha \in K$ is
> **algebraic** over $F$ if it is a root of some nonzero polynomial in $F[x]$,
> and **transcendental** over $F$ otherwise. The extension $K/F$ is **algebraic**
> if every element of $K$ is algebraic over $F$.

$$
% caption: An element of K is either algebraic over F, satisfying a nonzero
% polynomial, or transcendental, satisfying none; the two cases split every element.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=26mm, minimum height=11mm] (root) at (0,0) {element a of K};
\node[draw=acc, thick, fill=acc!8, minimum width=40mm, minimum height=13mm, align=center] (alg) at (5.2,1.3) {algebraic over F\\a root of some nonzero f};
\node[draw, minimum width=40mm, minimum height=13mm, align=center] (tr) at (5.2,-1.3) {transcendental over F\\a root of no polynomial};
\draw[->, acc, thick] (root) -- (alg);
\draw[->, black, thick] (root) -- (tr);
\node[font=\footnotesize, text=black] at (5.2,0.05) {exactly one holds};
\end{tikzpicture}
$$

When $\alpha$ is algebraic, the polynomials it satisfies form a nonzero ideal of
$F[x]$; being a principal ideal domain, that ideal has a single monic generator.

> **Theorem (Minimal polynomial).** For $\alpha$ algebraic over $F$ there is a
> unique monic irreducible $m_{\alpha, F}(x) \in F[x]$ with $\alpha$ as a root.
> A polynomial $f(x) \in F[x]$ has $\alpha$ as a root if and only if
> $m_{\alpha, F}(x)$ divides $f(x)$.

> **Proof.** A monic generator $g$ of minimal degree cannot factor as $g = ab$
> with both factors of smaller degree, since $g(\alpha) = a(\alpha)b(\alpha) = 0$
> would force one factor to vanish at $\alpha$, contradicting minimality; so $g$
> is irreducible. Any $f$ with $f(\alpha) = 0$ leaves remainder $r$ on division
> by $g$, and $r(\alpha) = 0$ with $\deg r < \deg g$ forces $r = 0$.[^df-prop9]

> **Definition (Minimal polynomial and degree).** The polynomial
> $m_{\alpha, F}(x)$ is the **minimal polynomial** of $\alpha$ over $F$; its
> degree is the **degree of $\alpha$** over $F$.

Combining the two theorems, the simple extension a single algebraic element
generates has degree equal to the degree of its minimal polynomial:
$$
F(\alpha) \cong F[x]/(m_{\alpha, F}(x)),
\qquad
[F(\alpha) : F] = \deg m_{\alpha, F}(x) = \deg \alpha.
$$

| Element $\alpha$ | base field $F$ | minimal polynomial | $[F(\alpha):F]$ |
| --- | --- | --- | --- |
| $\sqrt{2}$ | $\mathbb{Q}$ | $x^2 - 2$ | $2$ |
| $\sqrt[3]{2}$ | $\mathbb{Q}$ | $x^3 - 2$ | $3$ |
| $\sqrt[n]{2}$ | $\mathbb{Q}$ | $x^n - 2$ (Eisenstein) | $n$ |
| $i$ | $\mathbb{Q}$ | $x^2 + 1$ | $2$ |
| $\sqrt[3]{2}$ | $\mathbb{R}$ | $x - \sqrt[3]{2}$ | $1$ |

The last row shows that degree depends on the base field: over
$\mathbb{R}$ the cube root of $2$ already lives in the field, so its minimal
polynomial is linear. The irreducibility claims for $x^n - 2$ come from
[Eisenstein's criterion](/abstract-algebra/factorization-and-polynomials/irreducibility-criteria-and-groebner).

Finiteness and algebraicity of a single element are the same condition.

> **Theorem.** An element $\alpha$ is algebraic over $F$ if and only if
> $F(\alpha)/F$ is finite. Consequently, every finite extension is algebraic.

> **Proof.** If $\alpha$ is algebraic, $[F(\alpha) : F] = \deg m_\alpha$ is
> finite. Conversely if $[F(\alpha) : F] = n$, then the $n + 1$ powers
> $1, \alpha, \dots, \alpha^n$ are linearly dependent over $F$, and any
> nontrivial dependence is a nonzero polynomial vanishing at $\alpha$.[^df-prop12]
> For the corollary, if $\alpha$ lies in a finite extension $K$, then $F(\alpha)$
> is a subspace of $K$, so $[F(\alpha) : F] \le [K : F] < \infty$.

## The tower law

Degrees multiply through a tower of fields, exactly as index does through a chain
of subgroups.

> **Theorem (Tower law).** For fields $F \subseteq K \subseteq L$,
> $$
> [L : F] = [L : K]\,[K : F],
> $$
> where the equation holds in the extended sense that one side is infinite
> precisely when the other is.

> **Proof.** If $\{\alpha_i\}$ is a $K$-basis of $L$ and $\{\beta_j\}$ an
> $F$-basis of $K$, then every element of $L$ is a $K$-combination of the
> $\alpha_i$, and each coefficient in turn is an $F$-combination of the
> $\beta_j$; substituting shows the products $\alpha_i \beta_j$ span $L$ over
> $F$. A dependence among them, read back through the two bases, forces all
> coefficients to zero, so they are a basis and there are exactly
> $[L:K][K:F]$ of them.[^df-thm14]

$$
% caption: Extension degrees multiply along a tower; the composite degree is the
% product of the two step degrees, so an F-basis of L has that many elements.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (F) at (0,0) {F};
\node (K) at (0,1.7) {K};
\node (L) at (0,3.4) {L};
\draw[thick] (F) -- (K) node[midway, right=2pt, font=\footnotesize] {2};
\draw[thick] (K) -- (L) node[midway, right=2pt, font=\footnotesize] {3};
\draw[->, acc, thick] (0.9,0) .. controls (1.9,1.7) .. (0.9,3.4);
\node[acc, font=\footnotesize] at (2.35,1.7) {6 = 2 times 3};
\node[font=\footnotesize, align=left, text=black] at (5.6,1.7)
  {basis of L over F:\\products of a K-basis of L\\with an F-basis of K};
\end{tikzpicture}
$$

An immediate corollary: if $L/F$ is finite and $F \subseteq K \subseteq L$, then
$[K : F]$ divides $[L : F]$. This settles containment questions cheaply. Because
$[\mathbb{Q}(\sqrt 2) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\alpha) : \mathbb{Q}] =
3$ for $\alpha$ a root of $x^3 - 3x - 1$, and $2 \nmid 3$, the number $\sqrt 2$
cannot lie in $\mathbb{Q}(\alpha)$ — a fact awkward to check by hand. For a
positive use, $\sqrt[6]{2}$ has degree $6$, and since $(\sqrt[6]{2})^3 = \sqrt 2$,
the field $\mathbb{Q}(\sqrt 2)$ sits inside $\mathbb{Q}(\sqrt[6]{2})$ with
$[\mathbb{Q}(\sqrt[6]{2}) : \mathbb{Q}(\sqrt 2)] = 3$.

$$
% caption: The degree-6 extension Q(2^(1/6)) factors through Q(2^(1/2)) with step
% degrees 2 and 3, so the sixth root of 2 has degree 3 over the square root of 2.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (Q) at (0,0) {Q};
\node (Q2) at (0,1.7) {Q(a)};
\node (Q6) at (0,3.4) {Q(b)};
\draw[thick] (Q) -- (Q2) node[midway, right=2pt, font=\footnotesize] {2};
\draw[thick] (Q2) -- (Q6) node[midway, right=2pt, font=\footnotesize] {3};
\node[font=\footnotesize, align=left, text=black] at (3.6,1.7)
  {a is the square root of 2\\b is the sixth root of 2\\b cubed equals a};
\end{tikzpicture}
$$

## Finitely generated algebraic extensions

Adjoining several algebraic elements one at a time keeps degrees finite. Writing
$F(\alpha, \beta) = (F(\alpha))(\beta)$ realizes $F(\alpha_1, \dots, \alpha_k)$
as a chain of simple extensions $F = F_0 \subseteq F_1 \subseteq \cdots
\subseteq F_k$, and each step degree is at most the degree of the next generator
over $F$.

> **Theorem.** An extension $K/F$ is finite if and only if $K$ is generated by
> finitely many algebraic elements. A field generated by algebraic elements of
> degrees $n_1, \dots, n_k$ is algebraic of degree at most $n_1 n_2 \cdots n_k$.

The bound can be strict, since a later generator may already lie in the field a
previous one built. For example $\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{2}) =
\mathbb{Q}(\sqrt[3]{2})$ because the second generator adds nothing, so the naive
product bound of $9$ collapses to $3$.

> **Worked example.** Compute $[\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}]$, an
> $F$-basis, and a primitive element. First, $\sqrt 3 \notin \mathbb{Q}(\sqrt 2)$:
> if $\sqrt 3 = a + b\sqrt 2$ with $a, b \in \mathbb{Q}$, squaring gives
> $3 = a^2 + 2b^2 + 2ab\sqrt 2$, so $ab = 0$. Then $b = 0$ makes $\sqrt 3$
> rational, and $a = 0$ makes $b^2 = \tfrac32$ with no rational solution, both
> impossible. So $x^2 - 3$ stays irreducible over $\mathbb{Q}(\sqrt 2)$, and the
> tower law gives
> $$
> [\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}]
> = [\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}(\sqrt 2)]\,
>   [\mathbb{Q}(\sqrt 2) : \mathbb{Q}] = 2 \cdot 2 = 4,
> $$
> with $F$-basis $\{1, \sqrt 2, \sqrt 3, \sqrt 6\}$. A single element generates
> the whole field: put $\gamma = \sqrt 2 + \sqrt 3$. Then
> $\gamma^3 = 11\sqrt 2 + 9\sqrt 3$, so
> $$
> \sqrt 2 = \tfrac12(\gamma^3 - 9\gamma), \qquad \sqrt 3 = \gamma - \sqrt 2,
> $$
> placing both radicals in $\mathbb{Q}(\gamma)$; hence
> $\mathbb{Q}(\gamma) = \mathbb{Q}(\sqrt 2, \sqrt 3)$. Its minimal polynomial
> follows from $\gamma^2 = 5 + 2\sqrt 6$: rearranged, $(\gamma^2 - 5)^2 = 24$, so
> $$
> \gamma^4 - 10\gamma^2 + 1 = 0,
> $$
> a degree-$4$ polynomial matching $[\mathbb{Q}(\gamma) : \mathbb{Q}] = 4$, hence
> the minimal polynomial of $\sqrt 2 + \sqrt 3$.

$$
% caption: The subfield lattice of Q(root 2, root 3): three quadratic subfields
% sit between Q and the degree-4 top, one for each of root 2, root 3, root 6.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (Q) at (0,0) {Q};
\node (a) at (-2.6,1.7) {Q(root 2)};
\node (b) at (0,1.7) {Q(root 3)};
\node (c) at (2.6,1.7) {Q(root 6)};
\node[draw=acc, thick, fill=acc!8] (K) at (0,3.4) {Q(root 2, root 3)};
\draw[thick] (Q) -- (a) node[midway, left=2pt, font=\footnotesize] {2};
\draw[thick] (Q) -- (b);
\draw[thick] (Q) -- (c) node[midway, right=2pt, font=\footnotesize] {2};
\draw[thick] (a) -- (K);
\draw[thick] (b) -- (K);
\draw[thick] (c) -- (K) node[midway, right=2pt, font=\footnotesize] {2};
\end{tikzpicture}
$$

In this genuinely independent case the product bound is attained.

Two closure statements follow. First, sums, products, and quotients of algebraic
elements are algebraic, because they all live in the finite extension
$F(\alpha, \beta)$; so the elements of any extension $L/F$ that are algebraic
over $F$ form a subfield of $L$. Second, algebraicity is transitive.

> **Theorem (Transitivity of algebraic extensions).** If $K/F$ is algebraic and
> $L/K$ is algebraic, then $L/F$ is algebraic.

> **Proof.** Take $\alpha \in L$; it satisfies a polynomial with coefficients
> $a_0, \dots, a_n \in K$, each algebraic over $F$. The field
> $F(a_0, \dots, a_n)$ is finite over $F$, and $\alpha$ generates a finite
> extension of it, so by the tower law $F(\alpha, a_0, \dots, a_n)$ is finite
> over $F$, making $\alpha$ algebraic over $F$.[^df-thm20]

Applied to $\mathbb{C}/\mathbb{Q}$, the elements of $\mathbb{C}$ algebraic over
$\mathbb{Q}$
form the field $\overline{\mathbb{Q}}$ of **algebraic numbers**, an infinite
algebraic extension since it contains $\sqrt[n]{2}$ for every $n$. A counting
argument shows $\overline{\mathbb{Q}}$ is countable while $\mathbb{R}$ is not, so
transcendental reals exist; $\pi$ and $e$ are the standard examples, though
proving it for either is hard.

[^df-polyring]: **Dummit & Foote**, §7.4; §9.2 — Properties of Ideals and polynomial rings over fields: $F[x]$ is a principal ideal domain and $(p(x))$ is maximal exactly when $p$ is irreducible, so the quotient is a field. Covered in [ideals and quotient rings](/abstract-algebra/ring-theory/ideals-quotients-and-homomorphisms).
[^df-thm3]: **Dummit & Foote**, §13.1, Theorem 3 — for irreducible $p(x)$ the quotient $F[x]/(p(x))$ is an extension of $F$ containing a root of $p$.
[^df-thm4]: **Dummit & Foote**, §13.1, Theorem 4 and Corollary 5 — the powers $1, \theta, \dots, \theta^{n-1}$ are an $F$-basis, and multiplication is polynomial multiplication reduced modulo $p$.
[^df-thm6]: **Dummit & Foote**, §13.1, Theorem 6 and Corollary 7 — $F(\alpha) \cong F[x]/(p(x))$ when $\alpha$ is a root of the irreducible $p$.
[^df-thm8]: **Dummit & Foote**, §13.1, Theorem 8 — an isomorphism of base fields extends to an isomorphism of the simple extensions obtained by adjoining corresponding roots.
[^df-prop9]: **Dummit & Foote**, §13.2, Proposition 9 — existence and uniqueness of the minimal polynomial and its divisibility characterization.
[^df-prop12]: **Dummit & Foote**, §13.2, Proposition 12 and Corollary 13 — $\alpha$ algebraic iff $F(\alpha)/F$ finite; finite extensions are algebraic.
[^df-thm14]: **Dummit & Foote**, §13.2, Theorem 14 and Corollary 15 — multiplicativity of degrees in a tower and the divisibility of intermediate degrees.
[^df-thm20]: **Dummit & Foote**, §13.2, Theorems 17 and 20, Corollaries 18–19 — finite equals finitely generated algebraic, algebraic elements form a subfield, and algebraicity is transitive.