PMATH 348: Fields and Galois Theory

Introduction: Why Galois Theory?

The central question motivating this course is an old one: given a polynomial equation, can its roots be expressed using only the arithmetic operations of addition, subtraction, multiplication, and division together with the extraction of \(n\)th roots? An expression built from these five operations is called a radical. The quadratic formula shows that every quadratic equation is solvable by radicals, and analogous formulas — Cardano’s formula for cubics and Ferrari’s method for quartics — show the same is true for degrees three and four. The natural question is whether such a formula exists for degree five.

By changing variables, any cubic can be reduced to the depressed form \(x^3 + px = q\), and through the independent work of del Ferro, Tartaglia, and Fontana — with the solution published by Cardano in Ars Magna — the solution is

Ferrari extended this to quartic equations by reducing them to cubics. Since radical solutions exist for degrees up to four, it is natural to seek one for degree five. However, attempts by Euler, Bézout, and Lagrange all failed. Lagrange noticed that the roots of equations of degree at most four are preserved under certain permutations, but this property fails for quintics — a first hint that something fundamentally different happens at degree five. In 1799, Ruffini gave a 500-page proof of insolvability, though with a gap. In 1824, Abel filled the gap and completed the Abel-Ruffini theorem: the general quintic is not solvable by radicals.

But this raises a sharper question: given a specific polynomial of degree five, is it solvable by radicals? Galois theory answers this by reversing the question: suppose a radical solution exists — what must the polynomial look like?

The Galois approach proceeds in two steps. Given a root \(\alpha\) of a polynomial, we first pass from the element \(\alpha\) to the field \(\mathbb{Q}(\alpha)\), the smallest field containing \(\mathbb{Q}\) and \(\alpha\). A field has far more algebraic structure than a single element, but our knowledge of \(\mathbb{Q}(\alpha)\) is still limited — for instance, we may not know how many intermediate fields \(E\) sit between \(\mathbb{Q}\) and \(\mathbb{Q}(\alpha)\). We then take the second step: we associate \(\mathbb{Q}(\alpha)\) to a group, specifically the automorphism group \(\text{Aut}_{\mathbb{Q}}(\mathbb{Q}(\alpha))\) consisting of all field isomorphisms from \(\mathbb{Q}(\alpha)\) to itself that fix every element of \(\mathbb{Q}\). When \(\alpha\) is algebraic, this group is finite. The fundamental theorem of Galois theory then establishes a perfect correspondence between intermediate fields and subgroups of this automorphism group. The infinite problem of classifying field extensions is thus transformed into the finite problem of classifying subgroups of a group — and this simplification makes many ancient puzzles tractable.

In the words of Prof. Liu: Galois theory is an interplay between fields and groups. It transforms an infinite field question into a finite group problem. Welcome to PMATH 348 — a journey between finiteness and infiniteness.

Chapter 1: Ring Theory

This chapter reviews the ring theory needed throughout the course and proves Eisenstein’s criterion, a powerful tool for establishing irreducibility of polynomials.

1.1 Review of Ring Theory

A commutative ring with 1 (or simply ring in this course) is a set \(R\) with addition and multiplication such that \((R,+)\) is an abelian group with identity \(0\), multiplication is commutative and associative with identity \(1\), and the distributive law holds. A field is a ring in which every non-zero element has a multiplicative inverse. An integral domain is a ring in which the product of two non-zero elements is non-zero.

A key observation is that the only ideals of a field \(F\) are \(\{0\}\) and \(F\) itself. Consequently, if \(\phi: F \to S\) is a ring homomorphism from a field to a ring, then \(\phi\) is either injective or identically zero. An integral domain \(R\) is a principal ideal domain (PID) if every ideal is generated by a single element.

The polynomial ring \(F[x]\) and the integers \(\mathbb{Z}\) are closely parallel. Both are PIDs. The units of \(\mathbb{Z}\) are \(\{\pm 1\}\), while the units of \(F[x]\) are the non-zero constants \(F^* = F \setminus \{0\}\). The equivalence classes of non-zero elements modulo units are the positive integers in \(\mathbb{Z}\) and the monic polynomials in \(F[x]\). The field of fractions of \(\mathbb{Z}\) is \(\mathbb{Q}\), and the field of fractions of \(F[x]\) is the field of rational functions \(F(x) = \{f(x)/g(x) : f,g \in F[x],\, g \neq 0\}\).

For the quotient ring \(R/I\), elements have the form \(r + I\) with addition and multiplication inherited from \(R\). For \(n \in \mathbb{Z}\), we have \(\mathbb{Z}/\langle n \rangle = \{0, 1, \ldots, |n|-1\}\). The quotient \(R/I\) is a domain if and only if \(I\) is a prime ideal, and a field if and only if \(I\) is a maximal ideal.

1.2 Gauss’s Lemma and Eisenstein’s Criterion

A polynomial \(f(x) \in R[x]\) is irreducible over \(R\) if it is non-constant and cannot be written as a product of two polynomials of lower degree in \(R[x]\).

To sketch the proof: map \(f(x)\) to \(\bar{f}(x) \in \mathbb{Z}_p[x]\) by reducing coefficients modulo \(p\). By the conditions on coefficients, \(\bar{f}(x) = \bar{a}_n x^n\) — only the leading term survives. If \(f(x)\) were reducible in \(\mathbb{Q}[x]\), an application of Gauss’s Lemma would force \(p^2 \mid a_0\), contradicting the hypothesis. Eisenstein’s criterion generalises to unique factorization domains.

Chapter 2: Field Extensions

This chapter develops the theory of field extensions, with particular focus on distinguishing extensions generated by algebraic elements from those generated by transcendental elements.

2.1 Degree of Extensions

If \(E/F\) is a field extension, we can view \(E\) as a vector space over \(F\), where addition is that of \(E\) and scalar multiplication is the restriction of multiplication in \(E\) to \(F \times E\).

Proof sketch. Suppose \([E:K] = m\) and \([K:F] = n\), with bases \(\{a_1, \ldots, a_m\}\) for \(E/K\) and \(\{b_1, \ldots, b_n\}\) for \(K/F\). One verifies that \(\{a_i b_j : 1 \leq i \leq m,\ 1 \leq j \leq n\}\) is a basis of \(E/F\), giving \([E:F] = mn\). ∎

2.2 Algebraic and Transcendental Extensions

We also use the notion of an \(F\)-homomorphism: given rings \(R, R_1\) both containing \(F\), a ring homomorphism \(\psi: R \to R_1\) is an \(F\)-homomorphism if its restriction to \(F\) is the identity map.

Proof sketch. The \(F\)-homomorphism \(\psi: F(x) \to F(\alpha)\) mapping \(f(x)/g(x) \mapsto f(\alpha)/g(\alpha)\) is well-defined because \(\alpha\) is transcendental (so \(g(\alpha) \neq 0\) for \(g \neq 0\). One shows it is an isomorphism. ∎

Proof sketch. Consider the \(F\)-homomorphism \(\psi: F[x] \to F(\alpha)\) mapping \(x \mapsto \alpha\). Since \(\alpha\) is algebraic, \(\ker(\psi) \neq 0\). Since \(F[x]/\ker(\psi) \cong \text{Im}(\psi)\) is a subring of a field, \(\ker(\psi)\) is a prime ideal. Since \(F[x]\) is a PID, the kernel is generated by an irreducible polynomial \(p(x)\); taking \(p(x)\) monic ensures uniqueness. ∎

Corollary. If \(p(x)\) is the minimal polynomial of \(\alpha\) over \(F\) with \(\deg(p) = n\), then \([F(\alpha):F] = n\). This explains why the degree of \(F(\alpha)/F\) equals the degree of the minimal polynomial — and why we call the vector space dimension the “degree” of the extension.

Furthermore, if \(E/F\) is a finite extension, then by induction there exist \(\alpha_1, \ldots, \alpha_n \in E\) such that \(E = F(\alpha_1, \ldots, \alpha_n)\), and every finite extension is a tower of simple extensions. A field extension \(E/F\) is algebraic if every element of \(E\) is algebraic over \(F\); it is transcendental otherwise.

Proof sketch. If \([E:F] = n\) and \(\alpha \in E\), then the \(n+1\) elements \(1, \alpha, \alpha^2, \ldots, \alpha^n\) cannot be linearly independent over \(F\), so there exist coefficients giving a polynomial with \(\alpha\) as a root. ∎

Chapter 3: Splitting Fields

This chapter introduces splitting fields — the natural home for all roots of a given polynomial — and proves their existence and uniqueness.

3.1 Splitting Fields: Definitions and Existence

To prove existence, we first need to find a field extension containing at least one root of a given irreducible polynomial.

Proof sketch. Since \(p(x)\) is irreducible, the ideal \(I = \langle p(x) \rangle\) is maximal, so \(E = F[x]/I\) is a field. The map \(a \mapsto a + I\) is an injective ring homomorphism from \(F\) to \(E\), so \(F\) embeds as a subfield. Since \(I = \langle p(x) \rangle\), the element \(\alpha = x + I\) satisfies \(p(\alpha) = 0\). ∎

By applying Kronecker’s theorem repeatedly (inductively), we obtain:

Corollary. Every polynomial \(f(x) \in F[x]\) has a splitting field, which is a finite extension of \(F\). Explicitly, if \(\alpha_1, \ldots, \alpha_n\) are the roots of \(f(x)\) in some extension, then \(F(\alpha_1, \ldots, \alpha_n)\) is the splitting field.

3.2 Uniqueness of Splitting Fields

To prove uniqueness, we need to understand how field isomorphisms extend to polynomial rings.

Corollary (Uniqueness). Taking \(\phi = \text{id}_F\), any two splitting fields of \(f(x) \in F[x]\) over \(F\) are \(F\)-isomorphic.

3.3 Degree of Splitting Fields

Proof sketch. By induction on \(n = \deg(f)\). If \(f(x)\) is irreducible and \(\alpha \in E\) is a root, then \([F(\alpha):F] = n\). The splitting field of \(f(x)/(x - \alpha)\) over \(F(\alpha)\) has degree dividing \((n-1)!\) by induction, so \([E:F] = n \cdot [E:F(\alpha)]\) divides \(n!\). If \(f = gh\) with \(\deg(g) = m\) and \(\deg(h) = k\) where \(m + k = n\), then \([E:F]\) divides \(m! \cdot k!\), which divides \(n!\). ∎

Chapter 4: Characteristic and Prime Fields

This chapter introduces the characteristic of a field, prime fields, derivatives in positive characteristic, finite fields, and separable polynomials.

4.1 Prime Fields and Characteristic

Proof sketch. Let \(F_1\) be a subfield of \(F\). Consider the ring map \(\chi: \mathbb{Z} \to F_1\) sending \(n \mapsto n \cdot 1_{F_1}\). The image is a subring of a field, hence an integral domain, so \(I = \ker(\chi)\) is a prime ideal of \(\mathbb{Z}\). If \(I = 0\), then \(\mathbb{Z} \hookrightarrow F_1\) and \(\mathbb{Q} \subseteq F_1\). If \(I = \langle p \rangle\) for a prime \(p\), then \(\mathbb{Z}_p \subseteq F_1\). ∎

4.2 Derivatives and Repeated Roots

Taking formal derivatives is straightforward in characteristic zero, but behaves differently in characteristic \(p\).

The key is that \((ax^n)' = nax^{n-1}\): in characteristic \(p\), the term \(nax^{n-1}\) vanishes precisely when \(p \mid n\), which means \(f(x)\) involves only powers \(x^p, x^{2p}, \ldots\) — that is, \(f(x) = g(x^p)\).

An element \(\alpha\) in an extension field is a repeated root of \(f(x)\) if \(f(x) = (x - \alpha)^2 g(x)\) for some \(g(x)\).

Note the important distinction: the condition of having a repeated root depends on which extension we work in, but the gcd condition involves only polynomials over \(F\).

4.3 Finite Fields

Proof sketch. The multiplicative group \(F^* = F \setminus \{0\}\) has order \(p^n - 1\). One can show \(F^*\) is cyclic, so every \(a \in F^*\) satisfies \(a^{p^n - 1} = 1\), meaning every element of \(F\) is a root of \(x^{p^n} - x\). ∎

4.4 Separable Polynomials

Chapter 5: Sylow Theorems

This chapter develops the group theory needed for Galois theory, culminating in the three Sylow theorems about the structure of finite groups.

5.1 Group Actions and the Class Equation

Let \(G\) be a group acting on a set \(S\). For \(x \in S\):

• The orbit of \(x\) is \(Gx = \{gx : g \in G\}\).
• The stabiliser of \(x\) is \(G_x = \{g \in G : gx = x\}\), which is a subgroup of \(G\).

When \(G\) acts on itself by conjugation (i.e., \(g \cdot x = gxg^{-1}\), the stabiliser of \(x\) is the centraliser \(C_G(x) = \{g \in G : gx = xg\}\). The orbit \(Gx\) is a singleton precisely when \(x\) is in the centre \(Z(G)\).

5.2 Sylow Theorems

A \(p\)-group is a group in which every element has order a power of \(p\). By Cauchy’s theorem, a finite group is a \(p\)-group if and only if its order is a power of \(p\). If \(H\) is a subgroup of \(G\), the normaliser of \(H\) in \(G\) is \(N_G(H) = \{g \in G : gHg^{-1} = H\}\). Note that \(H \trianglelefteq N_G(H)\).

A Sylow \(p\)-subgroup of \(G\) is a maximal \(p\)-subgroup. By the first Sylow theorem, if \(|G| = p^n m\) with \(\gcd(p,m) = 1\), then every Sylow \(p\)-subgroup has order exactly \(p^n\).

Proof sketch of Third Sylow. Let \(S\) be the set of all Sylow \(p\)-subgroups and \(P \in S\). By the second theorem, \(|S| = [G : N_G(P)]\), which divides \(|G|\). Let \(P\) act on \(S\) by conjugation; orbits of size 1 correspond to elements of \(S\) normalised by \(P\). One shows the only such element is \(P\) itself, giving \(|S| \equiv 1 \pmod{p}\). ∎

Chapter 6: Solvable Groups

This chapter introduces solvable groups, which are the group-theoretic key to understanding solvability of polynomial equations by radicals.

6.1 Definition and Basic Properties

Proof of (1). Let \(H_i = H \cap G_i\). Then \(H = H_0 \trianglerighteq H_1 \trianglerighteq \cdots \trianglerighteq H_m = \{1\}\), and by the second isomorphism theorem \(H_i/H_{i+1} \hookrightarrow G_i/G_{i+1}\), which is abelian.

Proof of (2). Consider the expanded tower \(G_0 N \trianglerighteq G_1 N \trianglerighteq \cdots \trianglerighteq G_m N = N\). Taking quotients by \(N\) and applying the second and third isomorphism theorems shows each factor is abelian. ∎

6.2 Simple Groups and Failure of Solvability

The alternating group \(A_5\) is simple. Since the only possible tower for \(A_5\) is \(A_5 \trianglerighteq \{1\}\), and \(A_5/\{1\} \cong A_5\) is not abelian, \(A_5\) is not solvable. By Theorem 6.0.1, \(S_5\) is not solvable either (since \(A_5 \leq S_5\). More generally, \(S_n\) is not solvable for all \(n \geq 5\), since each such \(S_n\) contains a copy of \(A_5\).

This connection between the symmetric group and solvability is precisely what makes the Abel-Ruffini theorem work.

Chapter 7: Automorphism Groups of Field Extensions

This chapter associates a group to each field extension: the group of field automorphisms that fix the base field.

7.1 Automorphism Groups

Two fundamental lemmas govern these automorphisms:

Lemma 7.1.1. If \(\psi \in \text{Aut}_F(E)\) and \(\alpha \in E\) is a root of \(f(x) \in F[x]\), then \(\psi(\alpha)\) is also a root of \(f(x)\). (Automorphisms permute the roots of polynomials with coefficients in the base field.)

Lemma 7.1.2. If \(E = F(\alpha_1, \ldots, \alpha_n)\) and two maps \(\psi_1, \psi_2 \in \text{Aut}_F(E)\) agree on each generator \(\alpha_i\), then \(\psi_1 = \psi_2\).

Proof sketch. Since \(E/F\) is finite, write \(E = F(\alpha_1, \ldots, \alpha_n)\). Any \(\psi \in \text{Aut}_F(E)\) is determined by where it sends each \(\alpha_i\). By Lemma 7.1.1, each \(\psi(\alpha_i)\) is a root of the minimal polynomial of \(\alpha_i\), giving finitely many choices. So \(|\text{Aut}_F(E)| < \infty\). ∎

7.2 Automorphism Groups of Splitting Fields

7.3 Fixed Fields

Proof sketch. Let \(L = E^G\). By definition of \(L\), every \(F\)-automorphism of \(E\) also fixes \(L\), so \(\text{Aut}_L(E) = \text{Aut}_F(E)\). Since \(f(x)\) is separable, \(|\text{Aut}_F(E)| = [E:F]\) and \(|\text{Aut}_L(E)| = [E:L]\). Thus \([E:F] = [E:L]\), and since \(F \subseteq L\), we get \(L = F\). ∎

Chapter 8: Separable and Normal Extensions

These two properties together define Galois extensions, which are the central objects of the theory.

8.1 Separable Extensions

Proof sketch. For \(\alpha \in E\) with minimal polynomial \(p(x)\), consider the distinct roots \(\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_n\) of \(p(x)\) in \(E\) and set \(\tilde{p}(x) = \prod_{i=1}^n (x - \alpha_i)\). By Theorem 7.3.1, the coefficients of \(\tilde{p}(x)\) lie in \(F\), and so \(\tilde{p}(x) = p(x)\). Thus \(p(x)\) has distinct roots and is separable. ∎

8.2 The Primitive Element Theorem

Proof sketch. Assume \(F\) is infinite (the finite field case is handled separately). It suffices to show that any extension generated by two algebraic elements is simple. Let \(E = F(\alpha, \beta)\) with minimal polynomials \(a(x)\) and \(b(x)\). Choose \(\lambda \in F\) such that \(\lambda \neq (\tilde{\alpha} - \alpha)/(\beta - \tilde{\beta})\) for all roots \(\tilde{\alpha}\) of \(a(x)\) and all roots \(\tilde{\beta} \neq \beta\) of \(b(x)\). Then \(F(\alpha, \beta) = F(\alpha + \lambda \beta)\). The condition on \(\lambda\) can always be satisfied since \(F\) is infinite and there are only finitely many excluded values. ∎

8.3 Normal Extensions

Forward direction. If \(E = F(\alpha_1, \ldots, \alpha_n)\) and \(p_i(x)\) is the minimal polynomial of \(\alpha_i\), then \(E\) is the splitting field of \(p_1(x) \cdots p_n(x)\).

Backward direction. Let \(E/F\) be the splitting field of \(f(x)\), and \(p(x) \in F[x]\) irreducible with root \(\alpha_1 \in E\). For any other root \(\alpha_2 \neq \alpha_1\) in the splitting field of \(p(x)\) over \(E\), Theorem 3.2.1 gives an automorphism \(\psi\) of the splitting field mapping \(\alpha_1 \mapsto \alpha_2\). Since \(\psi(E) = E\), we get \(\alpha_2 \in E\). ∎

Proof. Let \(E = F(\alpha_1, \ldots, \alpha_n)\) with minimal polynomial \(p_i(x)\) for \(\alpha_i\). Take \(N\) to be the splitting field of \(p_1(x) \cdots p_n(x)\) over \(E\). ∎

Chapter 9: Galois Theory

This chapter is the heart of the course: the definition of Galois extensions, Artin’s theorem, and the Fundamental Theorem of Galois Theory.

9.1 Galois Extensions

By Theorems 8.1.1 and 8.2.1, a finite Galois extension is equivalent to the splitting field of a separable polynomial. If \(f(x)\) has degree \(n\), then \(\text{Gal}_F(E) \leq S_n\) and \(|\text{Gal}_F(E)| = [E:F]\).

9.2 Artin’s Theorem

Proof sketch. Let \(n = |G|\) and \(F = E^G\). For any \(\alpha \in E\), consider its \(G\)-orbit \(\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_m\). Set \(f(x) = \prod_{i=1}^m (x - \alpha_i)\). Since every \(\psi \in G\) permutes the \(\alpha_i\)’s, the coefficients of \(f(x)\) are symmetric in the \(\alpha_i\)’s, hence fixed by all of \(G\) and so lie in \(F\). One shows \(f(x)\) is the minimal polynomial of \(\alpha\) over \(F\) — it is separable and splits over \(E\), so \(E/F\) is Galois. The bound \([E:F] \leq n\) follows from a linear algebra argument: if \([E:F] > n\), we could find a linear dependence contradicting the minimality hypothesis. Combined with \(|G| \leq |\text{Gal}_F(E)| = [E:F] \leq n = |G|\), we get equality throughout. ∎

9.3 The Fundamental Theorem of Galois Theory

Proof. The maps are mutually inverse: if \(L \in \text{Int}(E/F)\), then \((L^*)^* = E^{\text{Gal}_L(E)} = L\) by Theorem 7.3.1. If \(H \in \text{Sub}(G)\), then \((H^*)^* = \text{Gal}_{E^H}(E) = H\) by Artin’s theorem. ∎

The correspondence can be visualised as an inverted diagram and is illustrated below:

Proof. One shows: \(L/F\) is normal \(\Leftrightarrow\) \(\psi(L) = L\) for all \(\psi \in G\) \(\Leftrightarrow\) \(\psi \text{Gal}_L(E) \psi^{-1} = \text{Gal}_L(E)\) for all \(\psi \in G\) \(\Leftrightarrow\) \(L^* \trianglelefteq G\). The restriction map \(G \to \text{Gal}_F(L)\), \(\psi \mapsto \psi|_L\), is surjective with kernel \(L^*\). ∎

Chapter 10: Cyclic Extensions

Cyclic extensions — Galois extensions with cyclic Galois group — are the building blocks for understanding solvability by radicals.

10.1 Dedekind’s Lemma

In other words: distinct field homomorphisms are linearly independent over any field. This is the key tool in proving that cyclic extensions are generated by elements whose powers lie in the base field.

Proof sketch. Suppose the minimal number of non-zero \(c_i\) giving a relation is \(m \geq 2\). Choose \(\beta \in L\) with \(\psi_1(\beta) \neq \psi_2(\beta)\). By substituting \(\alpha \beta\) and dividing by \(\psi_1(\beta)\), one obtains a shorter relation — a contradiction with minimality of \(m\). ∎

10.2 Kummer Theory: Cyclic Extensions by Radicals

One checks \(\sigma(\alpha) = \zeta_n \alpha\), so \(\alpha, \alpha\zeta_n, \ldots, \alpha\zeta_n^{n-1}\) are the conjugates of \(\alpha\). Their product gives \(\alpha^n \in F\) and \(E = F(\alpha)\). ∎

10.3 Cyclic Extensions in Characteristic p

When the degree equals the characteristic, a different criterion applies.

Chapter 11: Solvability by Radicals and the Abel-Ruffini Theorem

This final chapter brings together all the preceding theory to prove the Abel-Ruffini theorem: the general polynomial of degree \(\geq 5\) is not solvable by radicals.

11.1 Radical Extensions

11.2 Solvability by Radicals ↔ Solvable Galois Group

Since each successive quotient is cyclic (by Theorem 9.2.3), \(G\) is solvable. Since \(\text{Gal}(f)\) is a quotient of the solvable group \(G\), it is solvable.

Proof sketch (\(\Leftarrow\). If \(\text{Gal}(f)\) is solvable, adjoin \(\zeta_n\) where \(n = |\text{Gal}(f)|\). The solvability tower of \text{Gal}(L/K)) (which is a subgroup of \text{Gal}(f)) gives, via Kummer theory, a corresponding radical tower: each cyclic step \(H_{i-1}/H_i \cong C_{d_i}\) contributes a field extension \(K_i = K_{i-1}(\alpha_i)\) with \(\alpha_i^{d_i} \in K_{i-1}\). ∎

11.3 The Abel-Ruffini Theorem

Proof. Since \(S_p\) is generated by a 2-cycle and a \(p\)-cycle, it suffices to find both in \(\text{Gal}(f) \leq S_p\). Since \(f(x)\) is irreducible of degree \(p\), we have \(p \mid |\text{Gal}(f)|\), so by Cauchy’s theorem there exists a \(p\)-cycle. The complex conjugation map \(a + bi \mapsto a - bi\) interchanges the two non-real roots and fixes all real roots — this is a 2-cycle in \(\text{Gal}(f)\). ∎