Galois Theory of ℂ/ℝ

Introduction

This article applies the Galois theory of Fields to the extension $\mathbb{C}/\mathbb{R}$. The extension has degree $2$; it is normal and separable, hence Galois; its Galois group is $\mathbb{Z}/2\mathbb{Z}$, generated by complex conjugation; and the fundamental theorem then says that the only intermediate fields are $\mathbb{R}$ and $\mathbb{C}$.

The surrounding structure is the point. The base field $\mathbb{R}$ is real closed, so $\mathbb{C}$ is its algebraic closure and this quadratic step is the only one (the Artin–Schreier picture); the top field $\mathbb{C}$ is algebraically closed — the fundamental theorem of algebra — a statement equivalent to the real-closedness of $\mathbb{R}$. Orderability, the model theory of $\mathbb{R}$ and $\mathbb{C}$, the naive radical picture, the comparison with finite fields and with $\overline{\mathbb{Q}}$, and the automorphisms of $\mathbb{C}$ are companions of the same circle of ideas.

Notation is that of Fields and Complex Algebra: $i^2 = -1$, $\operatorname{Gal}(K/F)$ is the group of $F$-automorphisms of $K$, and $[K:F]$ is the degree. All statements are standard. The two real-analytic inputs needed for the fundamental theorem of algebra — the intermediate value theorem and the existence of real square roots of non-negative reals — are flagged where they are used.

1. The Extension and Its Degree

The imaginary unit satisfies $i^2 = -1$, so it is a root of $x^2+1 \in \mathbb{R}[x]$. That polynomial has no real root, since in the ordered field $\mathbb{R}$ squares are non-negative while $-1 < 0$, and a quadratic with no root is irreducible. Hence $x^2+1$ is the minimal polynomial of $i$ over $\mathbb{R}$, and

$$ \mathbb{C} = \mathbb{R}(i) = \mathbb{R}[i] = \{a+i a' : a,a' \in \mathbb{R}\}, \qquad [\mathbb{C}:\mathbb{R}] = 2, \qquad \mathbb{C} \cong \mathbb{R}[x]/(x^2+1), $$

with $\{1,i\}$ a basis over $\mathbb{R}$. The two roots of $x^2+1$ are $i$ and $-i$, the conjugates of $i$ over $\mathbb{R}$ in the sense of Fields, §8.

2. Normality and Separability

Recall (Fields, §§10, 12) that a finite extension is normal if it is a splitting field of a family of polynomials over the base, separable if all its elements are separable, and Galois if it is normal and separable, equivalently $|\operatorname{Gal}(K/F)| = [K:F]$.

$\mathbb{C}/\mathbb{R}$ is the splitting field of $x^2+1$, since $x^2+1 = (x-i)(x+i)$ in $\mathbb{C}[x]$ and $\mathbb{C} = \mathbb{R}(i,-i)$ is generated by the roots; hence it is normal. (Every degree-$2$ extension is normal: for $K = F(\alpha)$ the second root of the minimal polynomial is the trace minus $\alpha$, and the trace lies in $F$.) It is separable because $\mathbb{R}$ has characteristic $0$, so every irreducible polynomial over it is separable; concretely $\gcd(x^2+1,2x) = 1$, so $x^2+1$ has distinct roots. Therefore $\mathbb{C}/\mathbb{R}$ is Galois, with

$$ |\operatorname{Gal}(\mathbb{C}/\mathbb{R})| = [\mathbb{C}:\mathbb{R}] = 2. $$

3. The Galois Group: Conjugation Generates $\mathbb{Z}/2\mathbb{Z}$

An $\mathbb{R}$-automorphism $\sigma$ of $\mathbb{C}$ fixes $\mathbb{R}$ and is determined by $\sigma(i)$, because $\mathbb{C} = \mathbb{R}(i)$. Since $\sigma(i)^2 + 1 = \sigma(i^2+1) = 0$, the value $\sigma(i)$ must be a root of the minimal polynomial $x^2+1$, so

$$ \sigma(i) = i \quad \text{or} \quad \sigma(i) = -i. $$

There are at most two such automorphisms, and both occur: the identity, and complex conjugation

$$ \sigma(A) = \bar{A}, \qquad \sigma(a+i a') = a - i a', \qquad \sigma(i) = -i. $$

Conjugation is a field automorphism and an involution, with $\sigma^2 = \operatorname{id}$ and $\sigma \neq \operatorname{id}$. Hence

$$ \operatorname{Gal}(\mathbb{C}/\mathbb{R}) = \{\operatorname{id}, \sigma\} \cong \mathbb{Z}/2\mathbb{Z}, $$

with $\sigma$ corresponding to $-1$. So the Galois group is cyclic of order $2$, generated by complex conjugation: conjugation is its unique non-identity element, hence the unique nontrivial $\mathbb{R}$-automorphism of $\mathbb{C}$.

4. The Fundamental Theorem in This Case

For $K/F$ finite Galois with group $G$, the fundamental theorem (Fields, §16) gives an inclusion-reversing bijection between intermediate fields $F \subseteq E \subseteq K$ and subgroups $H \leq G$, via $E \mapsto \operatorname{Gal}(K/E)$ and $H \mapsto K^H$, with $[K:E] = |H|$ and $[E:F] = [G:H]$; normal intermediate extensions correspond to normal subgroups. Here $G = \mathbb{Z}/2\mathbb{Z}$ has only the subgroups $\{e\}$ and $G$:

$H$	$K^H$	$[K^H:\mathbb{R}]$
$\{e\}$	$\mathbb{C}$	$2$
$G$	$\mathbb{R}$	$1$

The fixed field of the full group is the fixed field of conjugation, $\mathbb{C}^G = \{A \in \mathbb{C} : \bar{A} = A\} = \mathbb{R}$, as in Complex Algebra: $a+i a'$ is fixed precisely when $a' = 0$. Hence the only intermediate fields of $\mathbb{C}/\mathbb{R}$ are $\mathbb{R}$ and $\mathbb{C}$. Both intermediate extensions are Galois, consistent with $G$ being abelian so that every subgroup is normal. Since every degree-$2$ extension of a field of characteristic not $2$ is normal and separable, this is the simplest nontrivial Galois extension.

5. Real Closed Fields and the Artin–Schreier Characterization

A field is formally real if it admits an ordering, equivalently if $-1$ is not a sum of squares; it is real closed if it is formally real and has no proper algebraic extension that is formally real.

Theorem (Artin–Schreier). For a field $F$ the following are equivalent:

(a) $F$ is real closed;

(b) $F$ is not algebraically closed and $F(i) = F[x]/(x^2+1)$ is algebraically closed;

(c) $F$ has an ordering in which every positive element is a square and every polynomial of odd degree has a root in $F$.

Every formally real field has a real closure, unique up to an isomorphism fixing the field. Statement (b) says that the algebraic closure of a real closed field is obtained by adjoining one square root of $-1$, so that $[\overline{F}:F] = 2$; the converse is the Artin–Schreier theorem.

$\mathbb{R}$ is real closed: it is not algebraically closed ($x^2+1$ has no real root), and $\mathbb{R}(i) = \mathbb{C}$ is algebraically closed by the fundamental theorem of algebra. Directly, every positive real is a square, $-1$ is not a sum of squares, and every odd-degree real polynomial has a real root by the intermediate value theorem. Hence

$$ \overline{\mathbb{R}} = \mathbb{C}, \qquad \operatorname{Gal}(\overline{\mathbb{R}}/\mathbb{R}) = \mathbb{Z}/2\mathbb{Z}. $$

The degree $2$ is the minimum for the algebraic closure of a field that is not itself algebraically closed, and a field with finite absolute Galois group is either algebraically closed (trivial group) or real closed (group $\mathbb{Z}/2\mathbb{Z}$); so $\mathbb{R}$ realizes the smallest nontrivial absolute Galois group.

6. The Fundamental Theorem of Algebra and Equivalent Formulations

Theorem. $\mathbb{C}$ is algebraically closed.

For a field $K$ the following are equivalent, and any may serve as the definition (Fields, §11): (a) every nonconstant polynomial over $K$ has a root in $K$; (b) every nonconstant polynomial factors into linear factors; (c) every irreducible polynomial over $K$ is linear; (d) $K$ has no nontrivial finite (equivalently algebraic) extension; (e) $K = \overline{K}$. For $K = \mathbb{C}$, a nontrivial finite extension would contain a simple extension $\mathbb{C}(\alpha)$ whose minimal polynomial has no root, so (a) and (d) agree; and because $[\mathbb{C}:\mathbb{R}] = 2$, (a) is equivalent to (f) $\mathbb{R}$ is real closed, by Artin–Schreier (b). Thus the fundamental theorem of algebra and the real-closedness of $\mathbb{R}$ are one statement seen from the two ends of the extension.

Proof sketch. Two real-analytic inputs are used: (i) every odd-degree real polynomial has a real root (intermediate value theorem); (ii) every non-negative real has a square root, so every $A = a+i a'$ has a square root, namely $\pm\bigl(\sqrt{(|A|+a)/2} + i\,\operatorname{sgn}(a')\sqrt{(|A|-a)/2}\bigr)$ with $|A|^2 = a^2+a'^2$.

Suppose $\mathbb{C}$ is not algebraically closed. The normal closure over $\mathbb{R}$ of a proper finite extension is a finite Galois extension $L/\mathbb{R}$ with $L \supseteq \mathbb{C}$ and $L \neq \mathbb{C}$, of group $G$ with $|G| = [L:\mathbb{R}] = 2[L:\mathbb{C}]$. Write $|G| = 2^s m$ with $m$ odd. The fixed field of a Sylow $2$-subgroup has degree $m$ over $\mathbb{R}$, and $\mathbb{R}$ has no odd-degree extension by (i); so $m = 1$ and $G$ is a $2$-group. Since $L \neq \mathbb{C}$ we have $|G| \geq 4$, so $G$ has a subgroup $H$ of index $2$, whose fixed field $M$ satisfies $[M:\mathbb{R}] = 2$. A degree-$2$ extension of $\mathbb{R}$ is $\mathbb{R}(\sqrt{d})$ with $d$ a non-square; since $M \neq \mathbb{R}$, $d < 0$, so $M$ contains a square root of $-1$, necessarily $\pm i$, and hence $M = \mathbb{C}$. Then $M = \mathbb{C}$, so $H = \operatorname{Gal}(L/\mathbb{C})$ is a $2$-group of order $|G|/2$. If $|H| > 1$ it has a subgroup $N$ of index $2$, whose fixed field $M'$ satisfies $[M':\mathbb{C}] = [H:N] = 2$, and by (ii) every quadratic over $\mathbb{C}$ splits, so $M' = \mathbb{C}$, a contradiction; hence $|H| = 1$, $|G| = 2$, and $[L:\mathbb{C}] = 1$, contradicting $L \neq \mathbb{C}$. Hence $\mathbb{C}$ is algebraically closed.

Every proof of the fundamental theorem of algebra uses some completeness or ordering property of $\mathbb{R}$; for a general real closed field the facts (i) and (ii) are taken as axioms.

7. Ordering Consequences

In any ordered field, squares are non-negative and $1 > 0$ (Fields, §18).

$\mathbb{R}$ has a unique ordering. Let $P$ be the positive cone of an ordering of $\mathbb{R}$. A nonzero square is positive, so every positive element lies in $P$: if $a > 0$ then $a$ is a nonzero square, since positive reals have square roots. Hence $P = \{a \in \mathbb{R} : a > 0\}$, the usual positive cone. More generally a real closed field has a unique ordering, determined by its squares: $a \geq 0$ if and only if $a$ is a square, a definable condition.

$\mathbb{C}$ has no ordering. Otherwise $-1 = i^2$ would be a square, hence non-negative, contradicting $-1 < 0$. Equivalently $\mathbb{C}$ is not formally real, since $-1$ is a sum of squares. In fact no algebraically closed field is formally real: there every element is a square. So passing from $\mathbb{R}$ to $\mathbb{C}$ makes $-1$ a square and destroys the ordering. This is general: a real closed field is exactly a field whose algebraic closure has degree $2$, and its orderability, its unique ordering, and the finiteness of its absolute Galois group are three aspects of one condition.

8. Model-Theoretic Consequences

The real closed fields form an elementary class $\mathrm{RCF}$, axiomatized in the language of ordered fields by the field and order axioms, the statement that every positive element is a square, and, for each odd $n$, that every monic polynomial of degree $n$ has a root. Let $\mathrm{ACF}_0$ be the theory of algebraically closed fields of characteristic $0$.

Tarski's theorem. $\mathrm{RCF}$ admits quantifier elimination, is complete, decidable and model complete, and $(\mathbb{R},+,\cdot,<)$ is o-minimal. Hence every real closed field is elementarily equivalent to $\mathbb{R}$: the real algebraic numbers and any non-Archimedean real closed field satisfy exactly the same first-order sentences. First-order logic therefore cannot pin down $\mathbb{R}$ — the least upper bound property for arbitrary subsets is second-order — and a proper elementary extension of $\mathbb{R}$ of cardinality $\mathrm{C}$, obtained by adjoining a constant exceeding every real and applying compactness and downward Löwenheim–Skolem, is real closed and elementarily equivalent to $\mathbb{R}$ but non-Archimedean, hence not isomorphic to it. So $\mathrm{RCF}$ is not $\mathrm{C}$-categorical, nor $\aleph_0$-categorical, since the countable real algebraic numbers and the real closure of $\mathbb{Q}(t)$ with $t$ infinite are non-isomorphic.

By contrast $\mathrm{ACF}_0$ is complete, decidable, model complete and, by Steinitz's theorem, categorical in every uncountable cardinal, so $\mathbb{C}$, of cardinality $\mathrm{C} = 2^{\aleph_0}$, is the unique such field of that cardinality up to isomorphism. The Galois-theoretic reading: a real closed $F$ has absolute Galois group $\mathbb{Z}/2\mathbb{Z}$, while a model of $\mathrm{ACF}_0$ has no nontrivial algebraic extension, its structure lying in the transcendence degree over the prime field.

9. Radical Extensions and Solvability

A radical extension of $F$ is built by a tower $F = K_0 \subseteq \cdots \subseteq K_m$ with $K_{j+1} = K_j(\alpha_j)$ and $\alpha_j^{n_j} \in K_j$. By Galois's criterion a polynomial is solvable by radicals if and only if its Galois group is solvable.

The polynomial $x^2+1$ over $\mathbb{R}$ has splitting field $\mathbb{C}$ and group $\mathbb{Z}/2\mathbb{Z}$, which is abelian, hence solvable; the quadratic formula gives the roots $\pm\sqrt{-1} = \pm i$. The extension is itself radical in one step,

$$ \mathbb{R} \subseteq \mathbb{C} = \mathbb{R}(i), \qquad i^2 = -1 \in \mathbb{R}, $$

a degree-$2$ radical (indeed cyclic) extension, with no solvability obstruction.

The naive reading that "$\mathbb{C}$ cannot be reached from $\mathbb{R}$ by radicals because $-1$ has no real square root" confuses a radical extension with a tower of real extensions. If every step must remain inside $\mathbb{R}$ — for instance if only radicals $\sqrt[n]{a}$ with $a > 0$ are allowed — the tower never leaves $\mathbb{R}$, and a subfield of $\mathbb{R}$ cannot be $\mathbb{C}$; but a radical extension allows a root of $-1$, and the tower $\mathbb{R} \subseteq \mathbb{C}$ with $\alpha^2 = -1$ has degree $2$. Consistently, $\mathbb{R}$ has no proper finite extension other than $\mathbb{C}$, so this is the only nontrivial radical tower over $\mathbb{R}$.

10. Comparison with Finite Fields

For prime $p$ and $n \geq 1$ the extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is Galois of degree $n$, with

$$ \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \langle \varphi \rangle \cong \mathbb{Z}/n\mathbb{Z}, \qquad \varphi(a) = a^p, $$

and its subfields are the fields $\mathbb{F}_{p^m}$ with $m \mid n$ (Fields, §§13–14). In both this and the complex case the group is cyclic, hence abelian, every subgroup is normal and every intermediate extension is Galois; both base fields are perfect.

Extension	Degree	Group	Generator
$\mathbb{C}/\mathbb{R}$	$2$	$\mathbb{Z}/2\mathbb{Z}$	complex conjugation
$\mathbb{F}_{p^n}/\mathbb{F}_p$	$n$	$\mathbb{Z}/n\mathbb{Z}$	Frobenius $a \mapsto a^p$

The differences are instructive. No finite field is algebraically closed, and $\overline{\mathbb{F}_p}$ is infinite over $\mathbb{F}_p$, with $\operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p) \cong \widehat{\mathbb{Z}} = \varprojlim_n \mathbb{Z}/n\mathbb{Z}$ topologically generated by Frobenius; by contrast $\operatorname{Gal}(\overline{\mathbb{R}}/\mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$. The analogue of adjoining $i$ depends on $p$: $-1$ is a square in $\mathbb{F}_p$ exactly when $p = 2$ or $p \equiv 1 \pmod 4$. Hence for $p \equiv 1 \pmod 4$ one has $\mathbb{F}_p(i) = \mathbb{F}_p$; for $p \equiv 3 \pmod 4$ one gets the quadratic extension $\mathbb{F}_{p^2}$; and for $p = 2$ the polynomial $x^2+1 = (x+1)^2$ is a square and is not separable. So $\mathbb{C}/\mathbb{R}$ behaves like the case $p \equiv 3 \pmod 4$, except that over $\mathbb{R}$ the element $-1$ is never a square because $\mathbb{R}$ is ordered.

11. Comparison with the Algebraic Closure of $\mathbb{Q}$

$\overline{\mathbb{Q}}$ is algebraically closed of characteristic $0$, but $\overline{\mathbb{Q}}/\mathbb{Q}$ is infinite, and $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is the absolute Galois group of $\mathbb{Q}$: an infinite profinite group, not abelian and not explicitly known. It has many finite quotients (the inverse Galois problem asks which finite groups occur, and remains open, though all solvable groups and the symmetric groups do). This is the sharpest contrast with the group $\mathbb{Z}/2\mathbb{Z}$ of $\mathbb{C}/\mathbb{R}$.

The two examples are related by real closure. The real algebraic numbers

$$ \overline{\mathbb{Q}} \cap \mathbb{R} = \{\alpha \in \overline{\mathbb{Q}} : \alpha \text{ real}\} $$

form a real closed field, the real closure of $\mathbb{Q}$; it is countable, and

$$ \overline{\mathbb{Q}} = (\overline{\mathbb{Q}} \cap \mathbb{R})(i), \qquad [\overline{\mathbb{Q}} : \overline{\mathbb{Q}} \cap \mathbb{R}] = 2, $$

the same quadratic pattern as $\mathbb{C} = \mathbb{R}(i)$. The difference is entirely in the base: $\mathbb{R}$ is real closed, so $[\mathbb{C}:\mathbb{R}] = 2$, while $\mathbb{Q}$ is not, so $[\overline{\mathbb{Q}}:\mathbb{Q}]$ is infinite. Thus $\mathbb{C}/\mathbb{R}$ is the real-closed case of the algebraic-closure phenomenon, with the smallest nontrivial absolute Galois group.

Note also that $\mathbb{C}$ is an algebraic closure of $\mathbb{R}$ but not of $\mathbb{Q}$: $\mathbb{C}/\mathbb{Q}$ is transcendental. Its algebraic part over $\mathbb{Q}$ is the countable field $\overline{\mathbb{Q}}$, and $\mathbb{C}$ is obtained from it by adjoining a transcendence basis of cardinality $\mathrm{C} = 2^{\aleph_0}$.

12. The Automorphism Group of $\mathbb{C}$

Three nested groups must be distinguished: the Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$, the group of continuous automorphisms, and the full automorphism group $\operatorname{Aut}(\mathbb{C})$ of $\mathbb{C}$ as a field.

$\mathbb{R}$-linear automorphisms. A field automorphism is $\mathbb{R}$-linear precisely when it fixes $\mathbb{R}$, that is, precisely when it lies in $\operatorname{Gal}(\mathbb{C}/\mathbb{R}) = \{\operatorname{id}, \text{conjugation}\}$ by the computation of the Galois group above.

Continuous automorphisms. Give $\mathbb{C}$ its usual topology. A continuous field automorphism $\sigma$ fixes $\mathbb{Q}$, hence, by continuity and the density of $\mathbb{Q}$ in $\mathbb{R}$,

$$ \sigma(a) = \sigma(\lim q_n) = \lim \sigma(q_n) = \lim q_n = a \qquad (a \in \mathbb{R}). $$

Then $\sigma(i)^2 = \sigma(-1) = -1$, so $\sigma(i) = \pm i$, and $\sigma$ is determined by $\sigma(i)$ since $\mathbb{C} = \mathbb{R}(i)$. Hence the continuous automorphisms of $\mathbb{C}$ are exactly the identity and conjugation; in particular every continuous automorphism fixes $\mathbb{R}$.

The full group. As an abstract group, $\operatorname{Aut}(\mathbb{C})$ is far larger:

$$ |\operatorname{Aut}(\mathbb{C})| = 2^{\mathrm{C}} = 2^{2^{\aleph_0}}. $$

The upper bound is $|\mathbb{C}|^{|\mathbb{C}|} = \mathrm{C}^{\mathrm{C}} = 2^{\mathrm{C}}$. For the lower bound, let $B$ be a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$, so $|B| = \mathrm{C}$ and $\mathbb{C}$ is an algebraic closure of $\mathbb{Q}(B)$. Every permutation of $B$ extends to an automorphism of $\mathbb{Q}(B)$ and then to one of $\mathbb{C}$; distinct permutations differ on $B$, and $|\operatorname{Sym}(B)| = 2^{\mathrm{C}}$. Hence $|\operatorname{Aut}(\mathbb{C})| = 2^{\mathrm{C}} > \mathrm{C} = |\mathbb{C}|$.

Wild automorphisms. Every field automorphism of $\mathbb{R}$ is the identity: it preserves squares, hence the unique ordering, hence is order-preserving, and it fixes the dense subfield $\mathbb{Q}$. Therefore if $\sigma(\mathbb{R}) = \mathbb{R}$ then $\sigma|_{\mathbb{R}} = \operatorname{id}$ and $\sigma \in \{\operatorname{id}, \text{conjugation}\}$. Any other automorphism, called wild, moves $\mathbb{R}$: the image $\sigma(\mathbb{R})$ is a real-closed subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$ but distinct from it, and $\sigma$ is not continuous.

Two cautions. First, not every automorphism of $\mathbb{C}$ is continuous, and not every automorphism fixes $\mathbb{R}$; conjugation is the unique nontrivial $\mathbb{R}$-linear automorphism, not the unique automorphism. Second, wild automorphisms require the axiom of choice (a transcendence basis, and extension of isomorphisms to algebraic closures): it is consistent with $\mathsf{ZF} + \mathsf{DC}$ that every automorphism of $\mathbb{C}$ is continuous, so that identity and conjugation are the only ones.

Finally, restriction gives a surjection $\operatorname{Aut}(\mathbb{C}) \to \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ (again using choice), because any automorphism of $\overline{\mathbb{Q}}$ extends to one of $\mathbb{C}$; so the wild automorphisms combine the absolute Galois group of $\mathbb{Q}$ with a large kernel fixing $\overline{\mathbb{Q}}$.

Summary

This article applies the Galois theory of Fields to the extension $\mathbb{C}/\mathbb{R}$. The extension has degree $2$: the imaginary unit is a root of $x^2 + 1$, which has no real root because $\mathbb{R}$ is ordered, and it generates $\mathbb{C}$ over $\mathbb{R}$. The extension is normal and separable, hence Galois.

The Galois group is generated by complex conjugation, which fixes $\mathbb{R}$ and is determined by its value on $i$; it is cyclic of order two, so $\operatorname{Gal}(\mathbb{C}/\mathbb{R}) \cong \mathbb{Z}/2\mathbb{Z}$. The fundamental theorem of Galois theory then gives the inclusion-reversing bijection between the intermediate fields and the subgroups, and in this case the only intermediate fields are $\mathbb{R}$ and $\mathbb{C}$ themselves.

The article places the extension in its wider setting. Real closed fields are characterised by the Artin–Schreier theorem, the fundamental theorem of algebra is stated with its equivalent formulations, and the ordering of $\mathbb{R}$ is shown to be unique. The model-theoretic consequences follow, the real closed fields being an elementary class, together with the radical extensions and solvability, and the comparisons with the finite fields and with the algebraic closure of $\mathbb{Q}$. The final section distinguishes the three nested groups of automorphisms of $\mathbb{C}$: the Galois group, the group of continuous automorphisms, and the full automorphism group.

Summary of Notation

Symbol	Meaning
$F$, $K$, $E$	Base field, extension field, intermediate field
$\mathbb{R}, \mathbb{C}$	Real and complex fields
$i$, $\bar{A}$	Imaginary unit $i^2 = -1$; complex conjugation
$[K:F]$, $\operatorname{Gal}(K/F)$	Degree; group of $F$-automorphisms
$G$, $H$, $K^H$	Galois group, a subgroup, its fixed field
$\overline{F}$	Algebraic closure of $F$
$\overline{\mathbb{Q}}$, $\overline{\mathbb{Q}} \cap \mathbb{R}$	Algebraic closure of $\mathbb{Q}$; real algebraic numbers
$\mathbb{F}_{p^n}$, $\varphi(a) = a^p$	Finite field; Frobenius automorphism
$\widehat{\mathbb{Z}} = \varprojlim_n \mathbb{Z}/n\mathbb{Z}$	Absolute Galois group of $\mathbb{F}_p$
$\operatorname{Aut}(\mathbb{C})$	All field automorphisms of $\mathbb{C}$
$P$	Positive cone of an ordering of a field
$\mathrm{RCF}$, $\mathrm{ACF}_0$	Theories of real closed fields; of algebraically closed fields of characteristic $0$
$B$, $\operatorname{Sym}(B)$, $\mathrm{C}$	Transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$, its symmetry group, the cardinality $2^{\aleph_0}$