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Évariste Galois

Évariste Galois

A course in two strands: the short, turbulent life of a republican student who died at twenty, and the theory he left behind, which explains when an equation can be solved by radicals and founded the study of groups.

Born 25 October 1811, Bourg-la-Reine Died 31 May 1832, Paris Published posthumously by Liouville, 1846 Level advanced undergraduate → graduate
The ten fields between ℚ and ℚ(α, i), α = ⁴√2. They match the ten subgroups of the dihedral group of order 8, with inclusions reversed. Highlighted: a tower of square roots.
Lesson 00

How to use this course

Évariste Galois wrote fewer than a hundred pages of mathematics. Most of them were rejected, lost or ignored during his lifetime, and he died after a duel at the age of twenty. Fourteen years later Joseph Liouville published them, and within a few decades they had changed algebra. Galois replaced the old search for formulas with a question about symmetry: which permutations of the roots of an equation respect every algebraic relation among them? The answer is a group, and the structure of that group decides whether the equation can be solved by radicals.

The legend of Galois is full of exaggeration: the night of frantic writing, the plot to kill him, the incompetent academicians. This course keeps to what the documents support, with Peter Neumann’s critical edition of the manuscripts as the main guide.

Prerequisites for Part II
  • Linear algebra and a first course in groups: subgroups, cosets, normal subgroups, quotients, Lagrange’s theorem.
  • Polynomials over a field: irreducibility, Eisenstein’s criterion, the division algorithm.
  • For Lessons 06–08, field extensions and degrees \([L:K]\) help but are reviewed.
  • For Lesson 10, some comfort with matrix groups over \(\mathbb Z/p\).

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern language, proof sketches and exercises with folded solutions. Where Galois’s own formulation differs from the modern one, the lesson says so. The lessons follow the order of his work: the theory of equations (1829–1831), equations of prime degree, finite fields (1830), and the final letter of May 1832.

Reference

Timeline

1811Born 25 October in Bourg-la-Reine, south of Paris, to Nicolas-Gabriel Galois and Adélaïde-Marie Demante. His mother teaches him at home until he is twelve.
1823October: enters the Lycée Louis-le-Grand in Paris as a boarder.
1827February: first mathematics class, with Hippolyte Vernier. Reads Legendre and Lagrange.
1828Fails the entrance examination of the École polytechnique. Joins the advanced class of Louis Richard.
1829April: first paper, on periodic continued fractions. 25 May and 1 June: two memoirs on equations to the Académie des sciences, with Cauchy as referee. 2 July: his father dies by suicide in Paris. Fails the Polytechnique examination again. November: enters the École préparatoire (the future École normale).
1830February: memoir submitted for the Academy’s Grand Prix; the secretary, Fourier, dies in May and the manuscript is lost. The prize goes to Abel (posthumously) and Jacobi. Three short papers in Férussac’s Bulletin, including the theory of finite fields. July Revolution. December: expelled from the school.
183117 January: third version of the memoir sent to the Academy. 9 May: the toast at a republican banquet; arrested. 15 June: acquitted. 4 July: Poisson and Lacroix reject the memoir. 14 July: arrested on the Pont Neuf in a banned uniform. 23 October: sentenced to six months.
1832March: moved from Sainte-Pélagie prison to a clinic because of cholera. 29 April: released. 29 May: letter to Auguste Chevalier summarizing his work. 30 May: wounded in a duel. 31 May: dies at the Hôpital Cochin, aged 20.
1832September: Chevalier publishes the letter in the Revue encyclopédique.
1843Liouville announces to the Academy that Galois’s papers contain a correct solution to the problem of equations of prime degree.
1846Liouville publishes the Œuvres mathématiques d’Évariste Galois in his Journal de mathématiques pures et appliquées.
1870Camille Jordan’s Traité des substitutions et des équations algébriques makes Galois theory a standard part of mathematics.
Part I · Formation
Lesson 01

Bourg-la-Reine and Louis-le-Grand, 1811–1828

Évariste Galois was born on 25 October 1811 in Bourg-la-Reine, a small town on the road south from Paris. His father, Nicolas-Gabriel Galois, ran a boarding school, held liberal and republican views, and became the town’s mayor. His mother, Adélaïde-Marie Demante, came from a family of lawyers and taught him Latin, Greek and religion until he was twelve. Nobody in the family had shown any interest in mathematics.

In October 1823 he entered the Lycée Louis-le-Grand in Paris as a boarder. The school was strict, royalist and turbulent; in his first term a protest by pupils led to around forty expulsions, though Galois was not involved. His early reports were good and he won prizes in Latin and Greek. In 1826 his teachers judged his rhetoric weak and made him repeat a year.

Discovering mathematics

The repeated year changed his life. In February 1827 he joined the mathematics class of Hippolyte Vernier and became absorbed. He read Adrien-Marie Legendre’s Éléments de géométrie, a book meant for two years of study, as quickly as a novel, then went on to Lagrange’s memoirs on the solution of equations. His other subjects collapsed. Teachers described him as original, odd, and lacking method.

In 1828 he tried the entrance examination of the École polytechnique a year early, without the usual preparation, and failed. He then entered the advanced mathematics class of Louis Paul Émile Richard, who recognized his talent at once and reported that the boy worked only in the highest regions of mathematics. Richard later taught Charles Hermite. By the end of 1828 Galois believed he could solve the general quintic equation by radicals. He soon found the mistake, and the attempt led him to Lagrange’s question of why the method fails.

Lesson 02

Failures and first memoirs, 1829–1830

In April 1829 Joseph Gergonne’s Annales de mathématiques printed Galois’s first paper, a proof of a theorem on periodic continued fractions (Lesson 09). On 25 May and 1 June he sent two memoirs on the algebraic solution of equations to the Académie des sciences. Augustin-Louis Cauchy and Louis Poinsot were named to examine them. Both manuscripts are lost, Cauchy never reported, and Galois seems to have withdrawn them in January 1830.

The death of his father

That summer brought disaster. A new parish priest at Bourg-la-Reine, a political enemy of the mayor, circulated malicious verses about members of the mayor’s family over Nicolas-Gabriel Galois’s forged signature. Galois’s father left for Paris and, on 2 July 1829, killed himself in an apartment close to Louis-le-Grand. His funeral at Bourg-la-Reine turned into a brawl. A few weeks later Galois sat the Polytechnique examination for the second and last time and failed again. Stories of a blackboard eraser thrown at the examiner date from long after his death and cannot be checked.

The École préparatoire

Galois turned to the less prestigious École préparatoire, the training school for teachers that soon resumed its old name of École normale. He entered in November 1829 and received the baccalaureate it required on 29 December. One examiner found him sometimes obscure in expressing his ideas but intelligent and with a remarkable spirit of research.

In February 1830 he submitted a new version of his theory to the Academy for the Grand Prix in mathematics. The permanent secretary, Joseph Fourier, took the manuscript home and died in May; it was never found. In June the prize went jointly to Niels Henrik Abel, who had died the year before, and Carl Gustav Jacob Jacobi, for their work on elliptic functions. Galois saw these losses as evidence of a hostile establishment. During 1830 he did publish three short notes in Baron de Férussac’s Bulletin des sciences mathématiques: an analysis of his memoir on equations, a note on numerical equations, and the paper Sur la théorie des nombres that introduced finite fields.

Lesson 03

Revolution and expulsion, 1830–1831

In the last days of July 1830 Paris rose against King Charles X. Students of the Polytechnique fought in the streets. The director of the École préparatoire, Joseph-Daniel Guigniaut, kept his students inside the school; Galois reportedly tried to climb the wall and failed. The revolution replaced Charles X with Louis-Philippe, a constitutional monarch, which disappointed the republicans. Galois joined the Société des amis du peuple, a republican society.

In December 1830 he wrote a letter to the Gazette des écoles attacking the director’s conduct during the revolution. He was expelled from the school; the decision was confirmed on 4 January 1831. He enlisted in the artillery of the National Guard, a unit with many republicans, which the king dissolved by decree on 31 December 1830.

The third memoir

Siméon-Denis Poisson invited him to submit his work once more. On 17 January 1831 Galois sent the Academy the Mémoire sur les conditions de résolubilité des équations par radicaux, now called the First Memoir. It is the founding document of Galois theory and the text studied in Lessons 05–08. The same month he announced a private course in advanced algebra at a bookshop; some forty people came to the first session, but attendance soon fell away. In April the mathematician Sophie Germain wrote to Guglielmo Libri that Galois had no money and was in a bad way since his expulsion.

Part II · The mathematics
Lesson 04

Before Galois: Lagrange, Ruffini, Abel

For the general reader. Formulas with square roots solve quadratic equations. In the sixteenth century Italian algebraists found formulas with cube and square roots for equations of degree three and four. For two and a half centuries nobody could do the same for degree five. Lagrange showed in 1770–71 that every known method worked for the same hidden reason, and that this reason seemed to fail for degree five. Galois read Lagrange as a schoolboy.

Goals
  • Solve the cubic with Lagrange resolvents.
  • State Lagrange’s theorem on rational functions of the roots.
  • Understand what Ruffini and Abel proved, and what question remained.

Lagrange resolvents for the cubic

Let \(x^3+px+q=0\) have roots \(x_1,x_2,x_3\), and let \(\omega=e^{2\pi i/3}\). Lagrange considered the resolvent \(r=x_1+\omega x_2+\omega^2x_3\). Under the six permutations of the roots, \(r\) takes six values, but \(r^3\) takes only two: the even permutations multiply \(r\) by a power of \(\omega\), and the odd ones exchange \(r^3\) with \(s^3\), where \(s=x_1+\omega^2x_2+\omega x_3\). Therefore \(r^3\) and \(s^3\) are the roots of a quadratic equation whose coefficients are symmetric in the roots, hence polynomials in \(p,q\). Solving it gives Cardano’s formula

\[ x=\sqrt[3]{-\frac q2+\sqrt{\Delta}}+\sqrt[3]{-\frac q2-\sqrt{\Delta}},\qquad \Delta=\frac{q^2}{4}+\frac{p^3}{27}, \]

with the two cube roots chosen so that their product is \(-p/3\). For the quartic, the function \(x_1x_2+x_3x_4\) takes three values and leads to a cubic resolvent. For the quintic, every function Lagrange tried led to an equation of degree at least five.

Theorem (Lagrange, 1771)

Let \(\varphi\) and \(\psi\) be rational functions of indeterminates \(x_1,\dots,x_n\) over a field \(K\). If every permutation of \(x_1,\dots,x_n\) that fixes \(\psi\) also fixes \(\varphi\), then \(\varphi\) is a rational function of \(\psi\) with coefficients in \(K(e_1,\dots,e_n)\), where the \(e_i\) are the elementary symmetric functions. Moreover the number of distinct values that \(\psi\) takes under \(S_n\) divides \(n!\).

In group language, the values of \(\psi\) correspond to the cosets of its stabilizer, so their number is an index in \(S_n\). This is the origin of the group-theoretic theorem that now carries Lagrange’s name.

Theorem (Ruffini 1799, Abel 1824–26)

For \(n\ge5\) the general equation \(x^n+a_1x^{n-1}+\cdots+a_n=0\), with \(a_1,\dots,a_n\) independent indeterminates, cannot be solved by radicals over \(\mathbb Q(a_1,\dots,a_n)\).

Paolo Ruffini’s proof had a gap; Abel’s was complete. Neither answered the question that matters for a given equation with numerical coefficients, such as \(x^5-x-1=0\): is this equation solvable? Some quintics are, for example \(x^5-2=0\). Galois gave a criterion that applies to every equation.

Exercise 4.1Use Cardano’s formula to solve \(x^3-6x-9=0\).

Show solution
Here \(p=-6,\ q=-9\), so \(\Delta=\frac{81}{4}-\frac{216}{27}=\frac{81}{4}-8=\frac{49}{4}\) and \(\sqrt\Delta=\frac72\). Then \(x=\sqrt[3]{\frac92+\frac72}+\sqrt[3]{\frac92-\frac72}=\sqrt[3]8+\sqrt[3]1=3\). Check: \(27-18-9=0\), and the product of the cube roots is \(2=-p/3\). The other roots solve \(x^2+3x+3=0\).

Exercise 4.2How many values does \(\psi=x_1x_2+x_3x_4\) take under \(S_4\)? Identify its stabilizer.

Show solution
The values are \(x_1x_2+x_3x_4,\ x_1x_3+x_2x_4,\ x_1x_4+x_2x_3\): three, one for each way of splitting four roots into two pairs. The stabilizer of the first is generated by \((12)\), \((34)\) and \((13)(24)\), a dihedral group of order 8, and \(24/8=3\). The three values are the roots of Ferrari’s resolvent cubic.
Lesson 05

The group of an equation

For the general reader. The roots of an equation often satisfy hidden relations. The roots \(\pm\sqrt2\) of \(x^2-2\) add up to zero; the roots of \(x^4-2\) are \(\alpha, i\alpha, -\alpha, -i\alpha\), and opposite roots sum to zero. Galois looked at all the ways of shuffling the roots that keep every such relation true. Those shuffles form what he called the group of the equation.

Goals
  • Read Galois’s definition and its modern form as an automorphism group.
  • Understand his lemma on the primitive element.
  • Compute the Galois groups of \(x^3-2\), \(x^4-2\) and cyclotomic equations.

Galois’s definition

Galois fixed a field of “rational” quantities \(K\) (in his words, the quantities regarded as known) and an equation \(f(x)=0\) over \(K\) with distinct roots \(x_1,\dots,x_n\). His Proposition I states that there is a group of permutations of the roots with two properties: every rational function of the roots that is fixed by all its permutations lies in \(K\), and every rational function of the roots that lies in \(K\) is fixed by all its permutations. To build it he first proved a lemma.

Galois’s lemma (primitive element)

Let \(f\) have distinct roots \(x_1,\dots,x_n\) in characteristic 0. For suitable integers \(c_1,\dots,c_n\), the quantity \(V=c_1x_1+\cdots+c_nx_n\) takes \(n!\) distinct values under permutations of the roots, and every root \(x_i\) is a rational function of \(V\) over \(K\). Hence the splitting field is \(L=K(x_1,\dots,x_n)=K(V)\).

Let \(g\) be the minimal polynomial of \(V\) over \(K\), with roots \(V=V_1,V_2,\dots,V_m\). Writing \(x_i=\phi_i(V)\), Galois defined the group as the \(m\) permutations \(x_i=\phi_i(V)\mapsto\phi_i(V_k)\). In modern terms each \(V\mapsto V_k\) defines a field automorphism of \(L\) fixing \(K\).

Definition

Let \(L\) be the splitting field over \(K\) of a separable polynomial \(f\). The Galois group \(\mathrm{Gal}(L/K)\) is the group of field automorphisms \(\sigma:L\to L\) with \(\sigma(a)=a\) for all \(a\in K\). Each \(\sigma\) permutes the roots of \(f\), and \(\sigma\) is determined by that permutation, so \(\mathrm{Gal}(L/K)\) embeds in \(S_n\).

Theorem

With \(L\) as above, \(|\mathrm{Gal}(L/K)|=[L:K]\). The group acts transitively on the roots of \(f\) if and only if \(f\) is irreducible over \(K\).

Proof sketch. By the lemma \(L=K(V)\), and the automorphisms are the maps \(V\mapsto V_k\) for the roots \(V_k\) of the minimal polynomial \(g\) of \(V\). There are \(\deg g=[L:K]\) of them. If \(f\) is irreducible, any two roots \(x_i,x_j\) are conjugate, and the isomorphism \(K(x_i)\cong K(x_j)\) extends to an automorphism of \(L\). Conversely, an orbit of roots gives a factor of \(f\) with coefficients fixed by the group, hence in \(K\).

Exercise 5.1Show that the Galois group of \(x^3-2\) over \(\mathbb Q\) is \(S_3\).

Show solution
The roots are \(\theta,\omega\theta,\omega^2\theta\) with \(\theta=\sqrt[3]2\). The splitting field \(\mathbb Q(\theta,\omega)\) contains \(\mathbb Q(\theta)\), of degree 3 (Eisenstein at 2), and \(\omega\notin\mathbb Q(\theta)\subset\mathbb R\), so the degree is \(3\cdot2=6\). A subgroup of \(S_3\) of order 6 is \(S_3\) itself.

Exercise 5.2Let \(\alpha=\sqrt[4]2\), \(L=\mathbb Q(\alpha,i)\). Show \([L:\mathbb Q]=8\) and that \(\mathrm{Gal}(L/\mathbb Q)\) is dihedral of order 8, generated by \(\sigma:\alpha\mapsto i\alpha,\ i\mapsto i\) and complex conjugation \(\tau\).

Show solution
\(x^4-2\) is irreducible by Eisenstein, so \([\mathbb Q(\alpha):\mathbb Q]=4\); \(i\notin\mathbb Q(\alpha)\subset\mathbb R\), so \([L:\mathbb Q]=8\). An automorphism sends \(\alpha\) to one of four roots and \(i\) to \(\pm i\), giving at most 8 maps, so all 8 occur. Label the roots \(\alpha,i\alpha,-\alpha,-i\alpha\) as 1, 2, 3, 4: \(\sigma\) is the 4-cycle \((1234)\) and \(\tau=(24)\). They satisfy \(\sigma^4=\tau^2=1\), \(\tau\sigma\tau=\sigma^{-1}\), the symmetries of a square whose vertices are the roots.

Exercise 5.3Let \(p\) be prime and \(\zeta=e^{2\pi i/p}\). Show that \(\mathrm{Gal}(\mathbb Q(\zeta)/\mathbb Q)\cong(\mathbb Z/p)^\times\), a cyclic group of order \(p-1\).

Show solution
The minimal polynomial of \(\zeta\) is \(\Phi_p(x)=x^{p-1}+\cdots+x+1\) (irreducible by Eisenstein after \(x\mapsto x+1\)), with roots \(\zeta^a\), \(1\le a\le p-1\). Each automorphism is \(\sigma_a:\zeta\mapsto\zeta^a\), and \(\sigma_a\sigma_b=\sigma_{ab}\). The group \((\mathbb Z/p)^\times\) is cyclic because a primitive root mod \(p\) exists. This is the case Gauss had treated in 1801, which Galois generalized.
Lesson 06

The fundamental theorem

For the general reader. When you learn one new fact about the roots, say the value of \(\sqrt2\), fewer shuffles of the roots remain allowed. Galois saw that each piece of added knowledge corresponds exactly to a smaller group, and that the most useful pieces of knowledge correspond to special subgroups, now called normal. This dictionary between fields and groups is the heart of his theory.

Goals
  • State the Galois correspondence precisely.
  • See how Galois reached normal subgroups through his “proper decomposition”.
  • Compute the full correspondence for \(x^4-2\).
Fundamental theorem of Galois theory

Let \(L/K\) be the splitting field of a separable polynomial, with group \(G=\mathrm{Gal}(L/K)\). For a subgroup \(H\le G\) let \(L^H=\{a\in L:\ h(a)=a\ \forall h\in H\}\). Then:

  1. \(H\mapsto L^H\) and \(E\mapsto\mathrm{Gal}(L/E)\) are mutually inverse, inclusion-reversing bijections between subgroups of \(G\) and fields \(K\subseteq E\subseteq L\);
  2. \([L:L^H]=|H|\) and \([L^H:K]=[G:H]\);
  3. \(E/K\) is itself the splitting field of a polynomial if and only if \(N=\mathrm{Gal}(L/E)\) is normal in \(G\), and then \(\mathrm{Gal}(E/K)\cong G/N\).

Proof sketch

The key inequality is \([L:L^H]\le|H|\), proved today by Emil Artin’s argument: if \(a_1,\dots,a_m\in L\) with \(m>|H|\), the linear system \(\sum_j h(a_j)y_j=0\) (\(h\in H\)) has a non-zero solution, and averaging over \(H\) produces a linear relation among the \(a_j\) over \(L^H\). Combined with \(|\mathrm{Gal}(L/E)|=[L:E]\) from Lesson 05, this gives the bijection. For part 3, \(\sigma\,\mathrm{Gal}(L/E)\,\sigma^{-1}=\mathrm{Gal}(L/\sigma E)\), so \(N\) is normal exactly when every \(\sigma\in G\) maps \(E\) to itself, and restriction gives a surjection \(G\to\mathrm{Gal}(E/K)\) with kernel \(N\).

Galois’s formulation

Galois worked with adjunctions. Adjoining to \(K\) a root of an auxiliary equation shrinks the group of \(f\) to a subgroup \(H\) (his Proposition II). He observed that if one adjoins all the roots of the auxiliary equation, the group \(G\) splits into cosets \(G=H\cup Hs\cup Hs'\cup\cdots\) that are also left cosets \(G=H\cup sH\cup s'H\cup\cdots\). He called such a splitting a décomposition propre. A subgroup whose left and right cosets coincide is exactly a normal subgroup, and Galois was the first to single out the notion.

Exercise 6.1In the notation of Exercise 5.2, find the fixed fields of \(\langle\sigma^2\rangle\), \(\langle\tau\rangle\) and \(\langle\sigma\tau\rangle\).

Show solution
\(\sigma^2:\alpha\mapsto-\alpha,\ i\mapsto i\) fixes \(\alpha^2=\sqrt2\) and \(i\); its fixed field has degree \(8/2=4\), so it is \(\mathbb Q(\sqrt2,i)\). \(\tau\) fixes \(\alpha\), giving \(\mathbb Q(\alpha)\). \(\sigma\tau\) (apply \(\tau\) first, then \(\sigma\)) sends \(\alpha\mapsto i\alpha\) and \(i\mapsto-i\). It maps \((1+i)\alpha\mapsto(1-i)\,i\alpha=(i+1)\alpha\), so the fixed field is \(\mathbb Q((1+i)\alpha)\), of degree 4.

Exercise 6.2Which of the five intermediate fields of degree 4 over \(\mathbb Q\) in the hero diagram are normal over \(\mathbb Q\)? Check with the group.

Show solution
The subgroups of order 2 in \(D_4\) are \(\langle\sigma^2\rangle\) (the centre, normal) and four reflections, none normal. So only \(\mathbb Q(\sqrt2,i)\), the fixed field of \(\langle\sigma^2\rangle\), is normal; it is the splitting field of \((x^2-2)(x^2+1)\). The field \(\mathbb Q(\alpha)\) is not normal: it contains the root \(\alpha\) of \(x^4-2\) but not \(i\alpha\).

Exercise 6.3Let \(\zeta=e^{2\pi i/7}\). List the subfields of \(\mathbb Q(\zeta)\) and give a generator of each.

Show solution
The group is cyclic of order 6, with one subgroup of each order 1, 2, 3, 6. The subgroup of order 2, \(\{\sigma_1,\sigma_6\}\), fixes \(\zeta+\zeta^{-1}=2\cos(2\pi/7)\), a cubic field. The subgroup of order 3, \(\{\sigma_1,\sigma_2,\sigma_4\}\) (the squares mod 7), fixes the Gauss period \(\eta=\zeta+\zeta^2+\zeta^4\), which satisfies \(\eta^2+\eta+2=0\), so the quadratic subfield is \(\mathbb Q(\sqrt{-7})\).
Lesson 07

Solvability by radicals

For the general reader. Taking a square root, a cube root or a fifth root is an operation with a simple symmetry: the possible answers differ by roots of unity. Galois showed that a formula with radicals exists exactly when the group of the equation can be broken down, step by step, into such simple symmetric pieces. The group of the general quintic cannot be broken down this way.

Goals
  • Define radical extensions and solvable groups.
  • Prove the criterion in the direction that matters for the quintic.
  • Exhibit an explicit quintic over \(\mathbb Q\) that is not solvable by radicals.
Definitions

A finite group \(G\) is solvable if there is a chain \(1=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_r=G\) with each \(G_{j}/G_{j-1}\) abelian (equivalently, cyclic of prime order). A polynomial \(f\in K[x]\) is solvable by radicals if its roots lie in a field \(K=K_0\subseteq K_1\subseteq\cdots\subseteq K_s\) with \(K_{j}=K_{j-1}(a_j)\) and \(a_j^{n_j}\in K_{j-1}\).

Theorem (Galois, 1831)

Let \(K\) have characteristic 0 and let \(f\in K[x]\) have splitting field \(L\). Then \(f\) is solvable by radicals if and only if \(\mathrm{Gal}(L/K)\) is a solvable group.

Proof sketch

Two facts carry the proof. First, if \(K\) contains the \(n\)-th roots of unity and \(a^n\in K\), then \(K(a)/K\) has cyclic Galois group, since every automorphism sends \(a\) to \(\zeta a\) for an \(n\)-th root of unity \(\zeta\). Also \(K(\zeta)/K\) is abelian. Second, conversely, if \(K\) contains the \(p\)-th roots of unity and \(E/K\) is cyclic of prime degree \(p\) with generator \(\sigma\), then a Lagrange resolvent \(r=\sum_{k=0}^{p-1}\zeta^{-k}\sigma^k(b)\) is non-zero for suitable \(b\), satisfies \(\sigma(r)=\zeta r\), and so \(r^p\in K\) and \(E=K(r)\).

(⇒) Adjoin enough roots of unity first. A radical tower is then a chain of abelian extensions, and by the fundamental theorem the group of the top field has a chain of normal subgroups with abelian quotients. The group of \(f\) is a quotient of it, and quotients of solvable groups are solvable. (⇐) Run the chain of subgroups backwards, adjoining roots of unity first and using resolvents at each step. Galois’s own proof follows the same plan, adjoining at each step a root of an auxiliary equation of prime degree.

Corollary

For \(n\ge5\), \(S_n\) is not solvable, because its only normal subgroups are \(1, A_n, S_n\) and \(A_n\) is simple and non-abelian. Therefore any polynomial of degree \(n\ge5\) with group \(S_n\) is not solvable by radicals.

An explicit example

Let \(f=x^5-4x+2\). By Eisenstein at 2 it is irreducible, so its group \(G\le S_5\) is transitive and \(5\mid|G|\); by Cauchy’s theorem \(G\) contains a 5-cycle. The graph of \(f\) crosses the axis three times, so \(f\) has three real roots and two complex conjugate ones. Complex conjugation therefore acts as a transposition. A 5-cycle and a transposition generate \(S_5\).

Exercise 7.1Verify that \(x^5-4x+2\) has exactly three real roots.

Show solution
\(f(-2)=-22,\ f(-1)=5,\ f(0)=2,\ f(1)=-1,\ f(2)=26\): three sign changes give at least three real roots. \(f'(x)=5x^4-4\) has exactly two real zeros, \(\pm(4/5)^{1/4}\), so by Rolle’s theorem \(f\) has at most three real roots.

Exercise 7.2Show that \(S_4\) is solvable by writing a chain with abelian quotients. Why does this match the existence of Ferrari’s formula?

Show solution
\(1\trianglelefteq V_4\trianglelefteq A_4\trianglelefteq S_4\), with \(V_4=\{1,(12)(34),(13)(24),(14)(23)\}\). The quotients are \(V_4\cong(\mathbb Z/2)^2\), \(A_4/V_4\cong\mathbb Z/3\), \(S_4/A_4\cong\mathbb Z/2\). Ferrari’s method mirrors this chain: a square root (the discriminant, for \(A_4\)), a cubic resolvent (for \(A_4/V_4\)) and square roots (for \(V_4\)).

Exercise 7.3Let \(p\) be prime. Show that a \(p\)-cycle and any transposition generate \(S_p\).

Show solution
Relabel so the transposition is \((1\,2)\) and, replacing the \(p\)-cycle by a suitable power (possible because \(p\) is prime), the cycle is \(c=(1\,2\,\cdots\,p)\). Conjugating \((1\,2)\) by powers of \(c\) gives \((2\,3),(3\,4),\dots\), and adjacent transpositions generate \(S_p\).
Lesson 08

Equations of prime degree

For the general reader. Galois’s criterion says “look at the group”, but in 1831 nobody knew how to compute groups. So Galois added a test that needs no group at all for equations of prime degree: such an equation is solvable exactly when, knowing any two of its roots, you can compute all the others rationally. Liouville singled out this result when he announced Galois’s work in 1843.

Goals
  • State Propositions VII and VIII of the First Memoir.
  • Identify the solvable transitive groups of prime degree with subgroups of the affine group.
  • Use the criterion to show that real quintics with three real roots are not solvable.
Theorem (Galois, 1831)

Let \(K\) have characteristic 0 and let \(f\in K[x]\) be irreducible of prime degree \(p\), with roots indexed as \(x_0,x_1,\dots,x_{p-1}\) by \(\mathbb Z/p\). The following are equivalent:

  1. \(f\) is solvable by radicals;
  2. for a suitable indexing, the group of \(f\) consists of permutations \(x_k\mapsto x_{ak+b}\) with \(a\in(\mathbb Z/p)^\times,\ b\in\mathbb Z/p\);
  3. every root is a rational function over \(K\) of any two of the roots: \(L=K(x_i,x_j)\) for all \(i\ne j\).

Proof sketch

(1⇒2) The group \(G\) is solvable and transitive on \(p\) points. Take the last non-trivial term \(N\) of a chain as in Lesson 07, refined so that \(N\) is a minimal normal subgroup. A normal subgroup of a transitive group has orbits of equal size, which must divide \(p\), so \(N\) is transitive; being abelian and transitive, it is regular, of order \(p\), generated by a \(p\)-cycle. Index the roots so that this cycle is \(k\mapsto k+1\). Every element of \(G\) normalizes \(N\), and the normalizer of \(\langle k\mapsto k+1\rangle\) in \(S_p\) is the affine group \(\{k\mapsto ak+b\}\) of order \(p(p-1)\).

(2⇒3) A non-identity map \(k\mapsto ak+b\) has at most one fixed point. So the only element of \(G\) fixing both \(x_i\) and \(x_j\) is the identity, and by the fundamental theorem \(K(x_i,x_j)=L\).

(2⇒1) The affine group has the chain \(1\trianglelefteq\{k\mapsto k+b\}\trianglelefteq\mathrm{AGL}(1,p)\) with cyclic quotients, so \(G\) is solvable. (3⇒2) is the delicate direction; it uses the fact that a transitive group of prime degree in which only the identity fixes two points is affine. Galois’s argument here was terse, and later writers filled it in.

Exercise 8.1Let \(f\in\mathbb Q[x]\) be irreducible of prime degree \(p\ge5\) with at least two but not all roots real. Show that \(f\) is not solvable by radicals.

Show solution
Complex conjugation restricts to an automorphism of the splitting field. It fixes the two real roots \(x_i,x_j\) but moves a non-real root, so it is a non-identity element fixing two roots. This contradicts condition 3. The quintic of Lesson 07 is a case: three real roots.

Exercise 8.2Find the order of the largest possible group of a solvable irreducible quintic, and show that \(x^5-2\) over \(\mathbb Q\) attains it.

Show solution
\(|\mathrm{AGL}(1,5)|=5\cdot4=20\). For \(x^5-2\) the splitting field is \(\mathbb Q(\sqrt[5]2,\zeta_5)\), of degree \(5\cdot4=20\) because the degrees 5 and 4 of the two subfields are coprime. Indexing the roots \(x_k=\zeta^k\sqrt[5]2\), the automorphism \(\sqrt[5]2\mapsto\zeta^b\sqrt[5]2,\ \zeta\mapsto\zeta^a\) sends \(x_k\mapsto x_{ak+b}\), exactly Galois’s form.
Lesson 09

Galois fields and number theory

For the general reader. Arithmetic modulo a prime \(p\) gives a number system with \(p\) elements. In June 1830 Galois showed how to build number systems with \(p^2\), \(p^3\) and in general \(p^\nu\) elements by adjoining “imaginary” roots of polynomials that have no solution mod \(p\). These finite fields now carry his name and run through coding theory and cryptography.

Goals
  • Construct the fields with \(p^\nu\) elements as Galois did.
  • Prove that their multiplicative groups are cyclic and their Galois groups are generated by Frobenius.
  • See how finite fields enter his classification of primitive solvable equations, and his first paper on continued fractions.

Galois imaginaries

In Sur la théorie des nombres Galois took a polynomial \(F(x)\) of degree \(\nu\) that is irreducible modulo \(p\) and introduced a symbol \(i\) for one of its roots, as \(\sqrt{-1}\) is introduced for \(x^2+1\). The expressions \(a_0+a_1i+\cdots+a_{\nu-1}i^{\nu-1}\) with \(a_j\in\mathbb Z/p\) number \(p^\nu\), and they form a field. He proved that every element \(\alpha\) satisfies \(\alpha^{p^\nu}=\alpha\), and that some element generates all the non-zero ones as powers.

Theorem (finite fields)

For every prime \(p\) and integer \(\nu\ge1\) there is a field \(\mathbb F_{p^\nu}\) with \(p^\nu\) elements, unique up to isomorphism: the splitting field of \(x^{p^\nu}-x\) over \(\mathbb F_p\), whose elements are exactly the roots of that polynomial. Its multiplicative group is cyclic of order \(p^\nu-1\), and \(\mathrm{Gal}(\mathbb F_{p^\nu}/\mathbb F_p)\) is cyclic of order \(\nu\), generated by the Frobenius map \(x\mapsto x^p\).

Proof sketch. In characteristic \(p\), \((a+b)^p=a^p+b^p\), so the roots of \(x^{p^\nu}-x\) form a field; the derivative is \(-1\), so there are \(p^\nu\) distinct roots. Any field \(F\) with \(q=p^\nu\) elements satisfies \(a^{q-1}=1\) for \(a\ne0\) by Lagrange’s theorem, so \(F\) is that splitting field. A finite abelian group in which \(x^d=1\) has at most \(d\) solutions for each \(d\) is cyclic. Frobenius fixes exactly \(\mathbb F_p\), so it generates a group of order \(\nu\), which is all of the Galois group.

Primitive solvable equations

An equation is primitive if its group does not preserve any partition of the roots into blocks of equal size larger than 1. In the Second Memoir, written in 1830, Galois used his imaginaries to describe the solvable ones.

Theorem (Galois 1830; completed by Jordan)

If an irreducible primitive equation is solvable by radicals, its degree is a prime power \(p^\nu\). Identifying the roots with the points of \(\mathbb F_p^{\,\nu}\), its group is contained in the affine group \(\mathrm{AGL}(\nu,p)=\{v\mapsto Av+b\}\), \(A\in\mathrm{GL}(\nu,p)\).

The first paper: periodic continued fractions

A quadratic irrational \(x\) has an eventually periodic continued fraction, by Lagrange. Galois’s paper of 1829 characterized the purely periodic ones.

Theorem (Galois, 1829)

Let \(x\) be a real quadratic irrational with conjugate \(x'\). Then \(x\) has a purely periodic continued fraction \(x=[\overline{a_0;a_1,\dots,a_{n-1}}]\) if and only if \(x>1\) and \(-1<x'<0\). In that case \(-1/x'=[\overline{a_{n-1};a_{n-2},\dots,a_0}]\).

Exercise 9.1Build \(\mathbb F_9\) as \(\mathbb F_3[i]\) with \(i^2=-1\). Show that \(1+i\) generates the multiplicative group.

Show solution
\(x^2+1\) has no root mod 3 (squares are 0 and 1), so it is irreducible. \((1+i)^2=1+2i+i^2=2i\), \((1+i)^4=(2i)^2=4i^2=-4\equiv2=-1\), so \((1+i)^8=1\) and \((1+i)^4\ne1\). The order divides 8 but not 4, so it is 8: a generator.

Exercise 9.2Show that \(x^4+x+1\) is irreducible over \(\mathbb F_2\) and that a root \(\theta\) generates \(\mathbb F_{16}^\times\).

Show solution
It has no roots in \(\mathbb F_2\) (\(f(0)=f(1)=1\)). The only irreducible quadratic over \(\mathbb F_2\) is \(x^2+x+1\), and \((x^2+x+1)^2=x^4+x^2+1\ne f\). So \(f\) is irreducible and \(\mathbb F_2[\theta]\) has 16 elements. The order of \(\theta\) divides 15. \(\theta^3\ne1\) since \(1,\theta,\theta^2,\theta^3\) are independent, and \(\theta^5=\theta\cdot\theta^4=\theta(\theta+1)=\theta^2+\theta\ne1\). So the order is 15.

Exercise 9.3Check Galois’s continued-fraction theorem for \(x=1+\sqrt2\).

Show solution
\(x=2+(\sqrt2-1)\) and \(1/(\sqrt2-1)=\sqrt2+1=x\), so \(x=[\overline{2}]\). The conjugate \(x'=1-\sqrt2\approx-0.414\) lies in \((-1,0)\), and \(-1/x'=1/(\sqrt2-1)=1+\sqrt2=[\overline 2]\), the reversed period.
Lesson 10

The last letter: PSL(2, p) and modular equations

For the general reader. On 29 May 1832, the night before the duel, Galois wrote to his friend Auguste Chevalier, summarizing his memoirs and stating new results without proof. One concerned the groups of the modular equations of elliptic functions, a topic then at the frontier. Galois found that these groups are what we now call \(\mathrm{PSL}(2,p)\), that they cannot be broken down, and that an exceptional reduction happens only for the primes 5, 7 and 11. The claims were correct.

Goals
  • Compute the orders of \(\mathrm{GL}(\nu,p)\) and \(\mathrm{PSL}(2,p)\).
  • State Galois’s claims about modular equations in group-theoretic form.
  • See why 5, 7 and 11 are special.

The letter

The letter describes three memoirs: the theory of equations, the primitive equations of Lesson 09, and integrals of algebraic functions, where Galois sketched a classification of abelian integrals into three kinds. It asks Chevalier to have Jacobi or Gauss give their opinion, which Galois pointedly said should concern the importance of the theorems rather than their truth. It also contains the first computation of the order of the general linear group.

Proposition (order of the general linear group)

\(|\mathrm{GL}(\nu,p)|=(p^\nu-1)(p^\nu-p)(p^\nu-p^2)\cdots(p^\nu-p^{\nu-1})\). Consequently \(|\mathrm{PSL}(2,p)|=\tfrac12p(p^2-1)\) for odd \(p\).

Modular equations

Transforming an elliptic function by a prime order \(p\) leads to a modular equation of degree \(p+1\), whose roots correspond to the \(p+1\) points of the projective line over \(\mathbb F_p\). Galois identified its group with the fractional linear maps \(k\mapsto\frac{ak+b}{ck+d}\) of this line, and, after adjoining a square root, with \(\mathrm{PSL}(2,p)\). He observed that for \(p>3\) this group has no proper decomposition, so it is simple in modern language. He then asked when the degree can be lowered from \(p+1\) to \(p\).

Theorem (claimed by Galois, 1832; proved later)

For a prime \(p\ge5\), \(\mathrm{PSL}(2,p)\) is simple. It has a subgroup of index \(p\) if and only if \(p\in\{5,7,11\}\). Accordingly, the modular equation of degree \(p+1\) can be reduced to an equation of degree \(p\) exactly for \(p=5,7,11\).

Why 5, 7 and 11. A subgroup of index \(p\) would have order \(\tfrac12(p^2-1)\). By the classification of subgroups of \(\mathrm{PSL}(2,p)\) (completed by L. E. Dickson in 1901), every proper subgroup is contained in a Borel subgroup of order \(\tfrac12p(p-1)\), a dihedral group of order \(p\pm1\), or a copy of \(A_4\), \(S_4\) or \(A_5\) (orders 12, 24, 60). The first two are too small for \(p\ge5\), and \(\tfrac12(p^2-1)\) equals 12, 24, 60 for \(p=5,7,11\) respectively. Indeed \(\mathrm{PSL}(2,5)\cong A_5\) contains \(A_4\), \(\mathrm{PSL}(2,7)\) of order 168 contains \(S_4\), and \(\mathrm{PSL}(2,11)\) of order 660 contains \(A_5\). The case \(p=5\) is the bridge that Hermite and Kronecker crossed in 1858, when they solved the general quintic with elliptic modular functions.

Exercise 10.1Prove the formula for \(|\mathrm{GL}(\nu,p)|\) and deduce \(|\mathrm{PSL}(2,p)|\).

Show solution
An invertible matrix is a sequence of \(\nu\) linearly independent columns: \(p^\nu-1\) choices for the first (non-zero), \(p^\nu-p\) for the second (outside the line spanned), and so on, with \(p^\nu-p^{k}\) for the \((k+1)\)-st. For \(\nu=2\): \((p^2-1)(p^2-p)\). The determinant maps onto \(\mathbb F_p^\times\), so \(|\mathrm{SL}(2,p)|=p(p^2-1)\). The centre \(\{\pm I\}\) has order 2 for odd \(p\), giving \(\tfrac12p(p^2-1)\).

Exercise 10.2Compute \(|\mathrm{PSL}(2,p)|\) and \(\tfrac12(p^2-1)\) for \(p=5,7,11,13\), and explain why \(p=13\) fails.

Show solution
Orders: 60, 168, 660, 1092. The would-be index-\(p\) subgroups have orders 12, 24, 60, 84. For \(p=13\), 84 is not the order of \(A_4\), \(S_4\) or \(A_5\), is not a divisor of a dihedral order \(p\pm1\in\{12,14\}\), and is not a divisor of the Borel order \(78\). So no subgroup of index 13 exists.

Exercise 10.3Show that \(\mathrm{PSL}(2,5)\) acts on the 6 points of the projective line over \(\mathbb F_5\), and that its order matches \(A_5\).

Show solution
The line has \(5+1=6\) points \(0,1,2,3,4,\infty\), permuted by \(k\mapsto\frac{ak+b}{ck+d}\). Its order is \(\tfrac12\cdot5\cdot24=60=|A_5|\). (In fact \(\mathrm{PSL}(2,5)\cong A_5\), acting on 6 points as \(A_5\) acts on the 6 axes through opposite vertices of an icosahedron.)
Part III · Life and legacy
Lesson 11

Prison, the duel, death, 1831–1832

On 9 May 1831 about two hundred republicans held a banquet to celebrate the acquittal of nineteen artillery officers accused of conspiracy. Galois rose with a glass in one hand and an open knife in the other and toasted King Louis-Philippe. The gesture was read as a threat, and he was arrested the next day and held at Sainte-Pélagie prison. At his trial on 15 June Galois said he had added the words “if he betrays”, lost in the noise, and the jury acquitted him.

On 4 July Poisson and Sylvestre-François Lacroix reported on the First Memoir. They had tried to follow the proofs and could not, and they noted that the criterion did not give a practical way to decide whether a given equation was solvable. They suggested he publish a fuller account. On 14 July, Bastille Day, Galois was arrested on the Pont Neuf wearing the uniform of the dissolved artillery and carrying weapons. He was held for months before being sentenced on 23 October to six months in prison. At Sainte-Pélagie the chemist François-Vincent Raspail, a fellow prisoner, recorded his bitterness and drinking. Galois learned of the rejection there and decided to publish his work privately with the help of his friend Auguste Chevalier; he wrote a sharp preface against the Academy.

The duel

In March 1832, during a cholera epidemic, he was moved to a clinic, the maison de santé of Sieur Faultrier, where he met Stéphanie-Félicie Poterin du Motel, the daughter of a resident physician. He was released on 29 April. Letters show an unhappy attachment that she ended in May. Shortly afterwards he was challenged to a duel. Its cause and his opponent remain uncertain: Alexandre Dumas named Pescheux d’Herbinville, and other evidence points to Ernest Duchâtelet, a republican imprisoned with Galois after the arrests of July 1831. Theories of a police plot have no documentary support.

On 29 May he wrote letters to his republican friends and the mathematical letter to Chevalier (Lesson 10). In the margin of one manuscript he noted that a proof needed completing and he had no time. The account that he wrote all his mathematics in that last night is a myth: the memoirs already existed. Early on 30 May he was shot in the abdomen and left on the ground; a passer-by took him to the Hôpital Cochin. His brother Alfred came to his bedside. He refused a priest and died the next morning, 31 May 1832, probably of peritonitis, aged twenty. He was buried on 2 June in a common grave in the Montparnasse cemetery; its exact location is unknown. A cenotaph stands in the cemetery of Bourg-la-Reine.

DiscussionPoisson’s report is often cited as a failure of the Academy. Read a translation of it (Neumann’s edition) and of the First Memoir’s proofs. Was the criticism fair given the standards of 1831? What would have helped the referees?

Lesson 12

Rediscovery and legacy

Chevalier and Alfred Galois copied the manuscripts and published the testamentary letter in the Revue encyclopédique in September 1832. They sent copies to Gauss and Jacobi, as Galois had asked; no reply is known. Around 1842 the papers reached Joseph Liouville, who studied them with care. In 1843 he told the Academy that he had found in them a solution of the problem of deciding when an irreducible equation of prime degree is solvable by radicals, as rigorous as it was deep. In 1846 he published the Œuvres mathématiques d’Évariste Galois in his journal, about sixty pages.

Understanding came slowly. Enrico Betti in Italy wrote proofs of Galois’s main claims in the early 1850s. Richard Dedekind lectured on Galois theory at Göttingen in the late 1850s, among the first university lectures on the subject, and pushed it toward the language of fields. Joseph-Alfred Serret included it in later editions of his Cours d’algèbre supérieure. Camille Jordan’s Traité des substitutions et des équations algébriques (1870) turned Galois’s ideas into a systematic theory of permutation groups and finally proved many of his claims. Emil Artin’s lectures of the 1920s and 1930s gave the linear-algebra presentation used in most textbooks today.

Legacy

  • Group theory: the word groupe in its technical sense, normal subgroups, simple groups, and the first infinite family of non-abelian simple groups after the alternating groups, \(\mathrm{PSL}(2,p)\).
  • Algebra and number theory: the Galois correspondence, Galois fields, Galois cohomology and Galois representations; the inverse Galois problem remains open.
  • Beyond algebra: Galois connections in order theory, differential Galois theory, the Galois theory of covering spaces, and Grothendieck’s Galois categories all extend his idea of matching structures to symmetry groups.
  • Applications: finite fields underlie error-correcting codes, elliptic-curve cryptography and the Advanced Encryption Standard.

The legend has often outgrown the man. Eric Temple Bell’s Men of Mathematics (1937) popularized a heroic and partly invented version; historians from René Taton to Peter Neumann have corrected it from the archives. The mathematics needs no embellishment.

Final projectTake an irreducible quartic over \(\mathbb Q\) of your choice. Compute its Galois group using the resolvent cubic and the discriminant, draw the lattice of intermediate fields as in the hero figure, and write the solution by radicals that the chain of subgroups predicts.

Reference

Connections

People linked to Évariste Galois across this collection. A name without a link has no page here yet. See Évariste Galois in the network of all 337 lives →

Teachers and advisors

  • Hippolyte Vernier 1827-1828
  • Louis Paul Émile Richard 1828-1829

Influenced by

Influenced

Colleagues and collaborators

Friends

  • Auguste Chevalier

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Évariste Galois, MacTutor History of Mathematics.
  2. Évariste Galois, Wikipedia (with extensive references).
  3. P. M. Neumann, The memoirs and legacy of Évariste Galois, Gresham College lecture, 2011.
  4. MacTutor, Évariste Galois’ family.
  5. R. Taton, Les relations d’Évariste Galois avec les mathématiciens de son temps, Revue d’histoire des sciences 1 (1947).
  6. T. Rothman, Genius and biographers: the fictionalization of Évariste Galois, American Mathematical Monthly 89 (1982).

Mathematics

  1. É. Galois, Œuvres mathématiques, ed. J. Liouville, Journal de mathématiques pures et appliquées 11 (1846). Gallica holds scans of the journal.
  2. P. M. Neumann, The Mathematical Writings of Évariste Galois, European Mathematical Society, 2011 (critical edition with translation).
  3. H. M. Edwards, Galois Theory, Springer GTM 101, 1984 (follows Galois’s own route).
  4. H. M. Edwards, Galois for 21st-century readers, Bull. AMS 51 (2014).
  5. C. Jordan, Traité des substitutions et des équations algébriques, Gauthier-Villars, 1870.
  6. E. Artin, Galois Theory, Notre Dame, 1942.
  7. J.-P. Tignol, Galois’ Theory of Algebraic Equations, World Scientific, 2nd ed. 2016.
  8. B. Kiernan, The development of Galois theory from Lagrange to Artin, Archive for History of Exact Sciences 8 (1971).