Évariste Galois

⚡ Key Takeaways

Évariste Galois (1811–1832) was a French mathematician who, before dying in a duel at age 20, founded what is now called Galois theory — the deep connection between field extensions and group theory that explains why the general polynomial equation of degree five or higher cannot be solved by radicals. His key result: a polynomial is solvable by radicals if and only if its associated symmetry group (the Galois group) is a solvable group. His work launched modern abstract algebra.

💡 The Core Insight

Symmetry controls solvability. To each polynomial equation, attach the group of all permutations of its roots that respect every polynomial relation with rational coefficients. Algebraic operations on the equation correspond to operations on this group. Solving "by radicals" (using $+,-,\times,\div$ and $n$ th roots) translates into building the group from successive abelian layers. Hence: solvable equation ⟺ solvable group. Because the symmetric group $S_5$ is not solvable, the general quintic is not solvable by radicals.

DefinitionField Extension and Galois Group

Let $K \subseteq L$ be a field extension. The Galois group $\mathrm{Gal}(L/K)$ is the group of all field automorphisms of $L$ that fix every element of $K$ :

1

\quad \mathrm{Gal}(L/K) = \{\sigma : L \to L \mid \sigma \text{ is a field automorphism},\ \sigma(a)=a\ \forall a \in K\}

Derivation hint: An automorphism must preserve $+$ and $\times$ ; fixing $K$ forces it to permute only the "new" elements, in particular the roots of any polynomial over $K$ . Here $K$ is the base field, $L$ is the extension field, and $\sigma$ ranges over the structure-preserving bijections of $L$ fixing $K$ pointwise. When $L$ is the splitting field of a separable polynomial over $K$ , the extension is called Galois, and we write $\mathrm{Gal}(L/K)$ for its Galois group.

DefinitionSolvable Group

A finite group $G$ is solvable if there exists a chain of subgroups (a subnormal series)

2

\quad \{e\} = G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n = G

Derivation hint: Each $G_i$ is normal in $G_{i+1}$ (the symbol $\triangleleft$ ), so the quotient $G_{i+1}/G_i$ is a well-defined group. such that every quotient $G_{i+1}/G_i$ is abelian (commutative). The integer $n$ measures how many abelian "layers" build the group. Every abelian group is solvable ( $n=1$ ); $S_2, S_3, S_4$ are solvable, but $S_5$ is not.

TheoremThe Fundamental Theorem of Galois Theory

Let $L/K$ be a finite Galois extension with group $G = \mathrm{Gal}(L/K)$ . There is an inclusion-reversing bijection between intermediate fields $K \subseteq M \subseteq L$ and subgroups $H \leq G$ , given by

3

\quad M \;\longmapsto\; \mathrm{Gal}(L/M), \qquad H \;\longmapsto\; L^{H} = \{x \in L \mid \sigma(x)=x\ \forall \sigma \in H\}

Derivation hint: Larger subgroups fix smaller subfields; the fixed-field operator and the Galois-group operator are mutually inverse on a Galois extension (Artin's theorem on linear independence of characters guarantees no collapse). Moreover $[L:M] = |\mathrm{Gal}(L/M)|$ and $[M:K] = [G : \mathrm{Gal}(L/M)]$ , and $M/K$ is Galois iff $\mathrm{Gal}(L/M)$ is normal in $G$ , in which case $\mathrm{Gal}(M/K) \cong G/\mathrm{Gal}(L/M)$ .

TheoremGalois's Solvability Criterion

A polynomial $f(x) \in K$ (with $K$ of characteristic $0$ ) is solvable by radicals if and only if its Galois group $\mathrm{Gal}(L/K)$ — where $L$ is the splitting field of $f$ — is a solvable group.

4

\quad f \text{ solvable by radicals} \iff \mathrm{Gal}(L/K) \text{ is solvable}

Derivation hint: A radical extension corresponds to adjoining roots of $x^n - a$ , whose Galois group (over a field with roots of unity) is cyclic, hence abelian; chaining radicals builds a solvable group, and conversely.

ProofSketch: Why

S_5

Being Non-Solvable Kills the General Quintic

Step 1. The general polynomial of degree $n$ over the field $K = \mathbb{Q}(a_1,\dots,a_n)$ of indeterminate coefficients has Galois group equal to the full symmetric group $S_n$ , because no nontrivial polynomial relation constrains the roots beyond the elementary symmetric ones.

Step 2. By the Solvability Criterion [4], the general degree- $n$ equation is solvable by radicals iff $S_n$ is a solvable group.

Step 3. Examine the derived series. For $n \le 4$ , $S_n$ is solvable:

5

\quad S_4 \triangleright A_4 \triangleright V_4 \triangleright \{e\}, \quad \text{quotients } \mathbb{Z}_2,\ \mathbb{Z}_3,\ \mathbb{Z}_2\times\mathbb{Z}_2

Derivation hint: $A_4/V_4 \cong \mathbb{Z}_3$ and $V_4$ (the Klein four-group) is abelian; all quotients are abelian, so the series witnesses solvability.

Step 4. For $n \ge 5$ , the alternating group $A_n$ is simple and non-abelian. In particular $A_5$ (order 60) has no nontrivial normal subgroups, so the only subnormal series of $S_5$ is $S_5 \triangleright A_5 \triangleright \{e\}$ , whose quotient $A_5/\{e\} = A_5$ is non-abelian.

Step 5. Therefore $S_5$ is not solvable, so by [4] the general quintic is not solvable by radicals. $\blacksquare$

∎

mathematician · Évariste Galois | 1811–1832

Born in Bourg-la-Reine near Paris to a politically liberal mayor father, Galois absorbed mathematics largely on his own, reading Legendre and Lagrange in the original. He twice failed the entrance exam to the École Polytechnique — once reportedly throwing an eraser at an examiner — and was expelled from the École Normale for a political letter. A passionate Republican during the turmoil following the 1830 Revolution, he was imprisoned twice. On 30 May 1832 he died of wounds from a duel whose cause remains obscure (a romantic entanglement, possibly engineered politically). The night before, he frantically annotated his manuscripts, writing "Je n'ai pas le temps" ("I have no time") in the margins. His complete mathematical output fills only about sixty pages, yet it reoriented all of algebra.

🧠 Intuition

Think of the roots of a polynomial as objects you can shuffle. Some shuffles are "invisible" from the rational numbers' point of view — if $\sqrt{2}$ and $-\sqrt{2}$ are roots, swapping them changes nothing a rational observer could detect, since both satisfy $x^2-2=0$ . The collection of all such invisible shuffles is the Galois group: it is the measured ambiguity in the roots. Solving by radicals means peeling that ambiguity away one clean layer at a time. Adjoining an $n$ th root resolves a cyclic (and so abelian) amount of ambiguity. The question "can I peel it all away?" becomes "can the group be torn down into abelian sheets?" For the quintic, the symmetry group $S_5$ contains the rigid, indivisible block $A_5$ — a knot of symmetry with no normal cuts. You can never unwind it abelian-layer-by-layer, so no radical formula exists.

ExampleGalois group of

x^2 - 2

over

\mathbb{Q}

Given: $f(x) = x^2 - 2$ , base field $K = \mathbb{Q}$ .

Step 1. Roots are $\pm\sqrt{2}$ , so the splitting field is $L = \mathbb{Q}(\sqrt 2)$ , of degree $[L:K]=2$ .

Step 2. An automorphism $\sigma \in \mathrm{Gal}(L/K)$ must send $\sqrt 2$ to a root of $x^2-2$ , i.e. $\sigma(\sqrt 2) \in \{\sqrt 2, -\sqrt 2\}$ .

Step 3. Two choices: identity and the conjugation $\sqrt 2 \mapsto -\sqrt 2$ . Both preserve field structure.

Result:

\boxed{\mathrm{Gal}(\mathbb{Q}(\sqrt2)/\mathbb{Q}) \cong \mathbb{Z}_2, \quad |G| = 2}

This group is abelian, hence solvable — consistent with $x^2-2$ being solvable by radicals ( $x = \pm\sqrt 2$ ).

ExampleGalois group of

x^3 - 2

over

\mathbb{Q}

Given: $f(x) = x^3 - 2$ , base field $K=\mathbb{Q}$ .

Step 1. Roots: $\sqrt[3]{2},\ \omega\sqrt[3]{2},\ \omega^2\sqrt[3]{2}$ , where $\omega = e^{2\pi i/3}$ . Splitting field $L = \mathbb{Q}(\sqrt[3]2,\ \omega)$ .

Step 2. Degree: $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt[3]2,\omega):\mathbb{Q}(\sqrt[3]2)]=2$ , so $[L:\mathbb{Q}] = 6$ .

Step 3. Thus $|G| = 6$ . Automorphisms permute the three roots and act faithfully, so $G$ embeds in $S_3$ ; since $|G|=6=|S_3|$ , we get $G \cong S_3$ .

Step 4. Check solvability via [5]: $S_3 \triangleright A_3 \triangleright \{e\}$ with quotients $\mathbb{Z}_2$ and $\mathbb{Z}_3$ , both abelian.

Result:

\boxed{\mathrm{Gal}(L/\mathbb{Q}) \cong S_3, \quad |G|=6, \text{ solvable}}

So $x^3-2$ is solvable by radicals (Cardano's cubic formula applies).

⚠️ Common mistake

"Galois proved the quintic has no solutions." False. Every quintic has five complex roots (Fundamental Theorem of Algebra). What Galois proved is that no general formula in radicals exists. Specific quintics like $x^5 - 1 = 0$ or $x^5 - 2 = 0$ are solvable by radicals — their Galois groups are solvable. The unsolvability is about the general equation with symbolic coefficients.

CounterexampleA solvable quintic

Consider $f(x) = x^5 - 2$ . Its splitting field is $\mathbb{Q}(\sqrt[5]2, \zeta_5)$ where $\zeta_5 = e^{2\pi i/5}$ . The Galois group has order 20 and is isomorphic to the Frobenius group $F_{20} = \mathbb{Z}_5 \rtimes \mathbb{Z}_4$ , which has the subnormal series $\{e\} \triangleleft \mathbb{Z}_5 \triangleleft F_{20}$ with abelian quotients $\mathbb{Z}_5$ and $\mathbb{Z}_4$ . Hence $F_{20}$ is solvable, and indeed $x = \sqrt[5]2\,\zeta_5^{\,k}$ are radical expressions. This single example refutes the belief that all degree-five equations resist radicals.

⚠️ Common mistake

"The Galois group is just the permutations of roots, $S_n$ ." Only for the generic polynomial. For a specific polynomial, hidden algebraic relations among roots shrink the group to a proper subgroup of $S_n$ . For instance $x^4 + 1$ over $\mathbb{Q}$ has Galois group $\mathbb{Z}_2 \times \mathbb{Z}_2$ , not $S_4$ , because the four roots satisfy extra rational relations.

connection

Galois's work is the keystone of modern algebra and reaches far beyond polynomials. It generalizes the impossibility results of Ruffini–Abel Theorem, which showed that the quintic is unsolvable; Galois explained why and exactly when. The machinery rests on Group Theory (he essentially invented normal subgroups and the word "group" in its modern sense) and Field Theory. The Fundamental Theorem of Galois Theory mirrors duality structures seen in Covering Space Theory (fundamental group ↔ covers) and in Galois Connections in order theory. It directly resolves the classical Compass and Straightedge Construction problems (trisecting the angle, doubling the cube) by translating constructibility into degree-power-of-two field extensions.

🔬 Where this appears

⏳ History

- 1770 — Lagrange studies resolvents and permutations of roots, planting the idea that symmetry governs solvability, but cannot resolve the quintic.

- 1799–1813 — Paolo Ruffini publishes incomplete proofs that the general quintic is unsolvable by radicals; largely ignored.

- 1824 — Niels Henrik Abel gives the first rigorous proof of quintic unsolvability (Ruffini–Abel Theorem), independent of Galois.

- 1829 — At age 17, Galois submits his first memoir on solvability to the Académie; Cauchy, the referee, loses or mislays it.

- 1830 — Galois publishes short notes in Férussac's Bulletin; his father commits suicide that year; he is admitted to the École Normale.

- 1831 — Galois submits a revised memoir; Poisson reviews it and declares it "incomprehensible," recommending fuller exposition. Galois is imprisoned for political activity.

- 30 May 1832 — Galois is mortally wounded in a duel; dies the next day, aged 20, after a night spent writing out his theory and a famous testamentary letter to Auguste Chevalier.

- 1843–1846 — Joseph Liouville studies the manuscripts, verifies their correctness, and publishes them in his Journal, finally bringing Galois's ideas to the mathematical world.

- 1870 — Camille Jordan's Traité des substitutions systematizes Galois theory into the foundation of modern group theory; later Dedekind, Weber, and Artin recast it in the field-automorphism language used today.

open-problem

The Inverse Galois Problem remains open: Is every finite group the Galois group of some Galois extension of $\mathbb{Q}$ ? It is known to hold for all solvable groups (Shafarevich's theorem), all symmetric and alternating groups, and 25 of the 26 sporadic simple groups, but the general case is unresolved. Related deep questions include the structure of the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ — a monstrous profinite object whose understanding is a central goal of the Langlands program and Grothendieck's theory of dessins d'enfants. Galois theory in positive characteristic and for inseparable extensions also requires careful generalization (Artin–Schreier theory, purely inseparable extensions).

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also