Galois Theory 2: The General Quintic is Dead and Évariste Killed Him

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In this post, we continue our analysis of Galois Theory and dive further into the proof.

Why the preface about symmetric polynomials? It should be clear by now to readers of the previous post that demonstrating symmetry of polynomial roots and understanding what properties come with the various symmetries will be invaluable for further exploration of general polynomial solubility (or lack thereof).

Recall that f(x)=x22    α=±2f(x) = x^2 - 2 \implies \alpha = \pm \sqrt{2}. ±2\pm \sqrt{2} are symmetric roots of ff because α22=f\alpha^2 - 2 = f regardless of which root we use, and this is true for any polynomial with rational coefficients involving radicals, addition, and multiplication.

The Galois Theory states that if the Galois group of a polynomial ff is solvable, then ff itself is solvable in radicals. We can use the solvability of Galois groups of a polynomial to show that polynomials of degree n5n \geq 5 are not always solvable, whereas the Galois groups of polynomials of lower orders are always solvable.


In order to understand what the hell a Galois group of a polynomial is, let alone what it means to "solve" a group, we need to recap some algebraic and group theoretic definitions.

Radical Solubility

When a polynomial ff with rational coefficients in Q\mathbb Q can be solved using only rational numbers and the operations +,,÷,×+, -, \div, \times and nn-th roots xn\sqrt[n]{x}, we say that "f(x)f(x) is soluble by radicals"


A group is a collection of objects GG together with some operation \star which satisfy the following axioms: 1. For any x,yGx, y \in G, xyGx \star y \in G. That is, the result of the operation between two elements in our group is another element which is also in our collection; a group is closed under its operation. 2. ϵG  s.t.  xG,ϵx=x=xϵ\exists \epsilon \in G \; s.t. \; \forall x \in G, \epsilon \star x = x = x \star \epsilon. Every group must have a "no-op" element which is called the identity of the group GG. The identity of group formed by the integers with addition G(Z,+)G(\Z, +) is 00. 3. The group operation \star is associative. That is, x,y,zG,(zy)z=z(yz)\forall x,y,z \in G, (z \star y) \star z = z \star (y \star z) which allows us to write xyzx \star y \star z unambiguously 4. Every element in GG has an inverse: xGx1G  s.t.  xx1=ϵ=x1x\forall x \in G \exists x^{-1} \in G \; s.t. \; x \star x^{-1} = \epsilon = x^{-1} \star x

For example, (Z,+)(\Z, +) form a group, but (Z,×)(\Z, \times) does not, because ∄xZ  s.t.  2x=1\not\exists x \in Z \; s.t. \; 2x = 1

Abelian Groups

A group GG is said to be abelian if its operation is commutative:

x,yG();  xy=yx\forall x, y \in G(\star);\; x \star y = y \star x

Cyclic Groups

CpC_p is a cyclic group of order pp with elements {1,x,x2,...,xp1}\{1, x, x^2, ..., x^{p-1} \} with some operation \cdot which behaves similar to arithmetic multiplication: xn×xm=xm+nx^n \times x^m = x^{m+n} as well as xp=1x^p = 1.

For example, in C5C_5 the expression x2x4x^2 \cdot x^4 evaluates to x6=x5x1=xx^6 = x^5 \cdot x^1 = x. Hence, the inverse of xkx^k is xpkx^{p-k} for integer values of p,kp, k.

Symmetric Groups

Symmetric groups, denoted SnS_n and read "the symmetric group on nn elements" have objects which are themselves the permutations of the nn elements, with the binary operation of composition \circ. So, σS3\sigma \in S_3 might be the permutation which maps the indices of a triple (1,2,3)(2,3,1)(1, 2, 3) \rightarrow (2, 3, 1), and another element τS3\tau \in S_3 could be (1,2,3)(3,2,1)(1, 2, 3) \rightarrow (3, 2, 1). Their composition would then be:

στ=(1,2,3)(3,2,1)(1,3,2)\sigma \circ \tau = (1, 2, 3) \rightarrow (3, 2, 1) \rightarrow (1, 3, 2)

Alternating Groups and parity

Group parity: A permutation is even if it can be written as a product of an even number of transpositions, and odd if it can be written as an odd number of transpositions.

E.g., a permutation on a non-trivial group which is equivalent to the identity on that group can be expressed: σ=(1  2)(1  2)\sigma = (1 \; 2)( 1 \; 2), that is: we send the first element to the 2nd position, and the 2nd element to the first position, then repeat that same cycle, so for some tuple t=(a,b)t = (\color{red}a\color{black} , \color{blue}b\color{black}):

σ(t)=(abba)(baab)=(a,b)=ϵ(t)\begin{aligned} \sigma(t) = \begin{pmatrix} \color{red}a & \color{blue}b \\ \downarrow & \downarrow \\ \color{blue}b & \color{red}a \end{pmatrix} \begin{pmatrix} \color{blue}b & \color{red}a \\ \downarrow & \downarrow \\ \color{red}a & \color{blue}b \end{pmatrix} = (\color{red}a\color{black} , \color{blue}b\color{black}) = \epsilon(t) \end{aligned}

Like the set of real numbers, the parity of a group is multiplicative. That is, the product of an even permutation and an odd permutation is odd, the product of two even permutations or two odd permutations as even. The inverse of an even permutation is even, the inverse of an odd is odd, so we can define a subgroup of of the symmetric group consisting of all even permutations called the Alternating Group: AnSnA_n \subset S_n:

An={σϵ(σ)=1,σSn}A_n = \{ \sigma | \epsilon(\sigma) = 1, \sigma \in S_n \}

For more on the potential applications of alternating groups, check out this other post about solving Rubik's cubes with group theory.


A field is like a group F,+\mathbb F, + with an additional operation \cdot which also has identitive elements for both operations, such as 00 for the "additive" operation, and 11 for the "multiplicative" operation. Additionally, we specify the following constraints:

  1. (F,+)(\mathbb F, +) is a group, and (F\{0},)(\mathbb F \backslash \{ 0\}, \cdot) is a group (division by zero is verboten)
  2. The operation \cdot is distributive over ++ s.t. (x+y)z=xz+yz(x + y) \cdot z = x\cdot z + y \cdot z
  3. 0x=x0,xy=yx,x+y=y+x0 \cdot x = x \cdot 0, x \cdot y = y \cdot x, x + y = y + x, the multiplicative operation has an identity, and both operations are commutative (which is not always true for a group)

E.g. Q,R\mathbb {Q, R} are both fields.

Field Extensions

A field extension is a gerund insofar as we extend fields (verb), and the resultant extension is itself another field (noun). Extensions are accomplished by adjoining additional values to a base field. We write Q(2)\mathbb Q(\sqrt{2}) to denote field extension of the rationals with root two, implicitly adding all other numbers that can be formed by combining rational elements of the base field and 2\sqrt{2} via the operations of the base field. That is, all numbers that can be expressed in terms of a+b2a + b\sqrt{2} are in the field extension Q(2)\mathbb Q(\sqrt{2}) .

The only non-obvious axiom we might want to justify to ourselves is that there exists a multiplicative inverse of a number

x=1a+b2x = \frac{1}{a + b \sqrt{2}}

which can still be expressed in the form c+d2c + d\sqrt{2}. We can demonstrate that it exists by multiplying xx by its conjugate form of 11 and simplifying:

x=1(a+b2)(ab2)(ab2)=ab2(a2b2)=a(a2b2)b2(a2b2)\begin{aligned} x &= \frac{1 }{(a + b\sqrt{2})} \cdot \frac{(a - b\sqrt{2})}{(a - b\sqrt{2})} \\ &= \frac{a - b\sqrt{2}}{(a^2 - b^2)} \\ &= \frac{a}{(a^2 - b^2)} - \frac{b\sqrt{2}}{(a^2 - b^2)} \\ \end{aligned}

leaving us with an expression of the form c+d2c + d\sqrt{2} where

c=a(a2b2),d=b(a2b2)c = \frac{a}{(a^2 - b^2)},\quad d = - \frac{b}{(a^2 - b^2)} \\

As mentioned before, a field extension of FF is itself a field KK which contains FF. We write FKF \subseteq K, or more concretely, QR\mathbb Q \subseteq \mathbb R, or QQ(2)\mathbb Q \subseteq \mathbb Q(\sqrt{2}). (We didn't show that Q(2)\mathbb Q(\sqrt{2}) is a field, but it's plain to see that aQ,a+02Q(2))\forall a \in \mathbb Q, a + 0\sqrt{2} \in \mathbb Q(\sqrt{2}))

Radical Extensions

Because the rationals are a field Q\mathbb Q, but Q\mathbb Q is not closed under the nn-th root operation, oftentimes we will want to adjoin our roots to the rationals.

Abel-Ruffini Theorem

The Abel-Ruffini theorem –which predates Galois as hinted at in the previous post– states that some quintics have roots that are not expressible in terms of radicals.

Even if our root is some horrendous mess such as:

α=375+152378923\alpha = \sqrt[5]{\frac{3}{7}} + \frac{1}{5}\sqrt[23]{2 - \frac{3}{7}\sqrt{89}}

we can still feasibly construct a splitting field by adjoining

QQ(89)Q(89,28923)\mathbb Q \subseteq \mathbb Q(\sqrt{89}) \subseteq \mathbb Q(\sqrt{89}, \sqrt[23]{2 - \sqrt{89}}) \subseteq \cdots

and so on.

Splitting Fields

Given any polynomial f(x)f(x), the splitting field of ff is the smallest field extension of Q\mathbb Q that contains all the roots of ff.

For example, the splitting field of f(x)=x22f(x) = x^2 - 2 is Q(2)\mathbb Q(\sqrt{2}). This is obviously the minimal field since we can't remove any elements from it without either violating axioms of it being a field (closed under addition and multiplication) or removing one of the roots.

Another example of a splitting field is

f(x)=x45x2+6=(x23)(x22)    {α}={±3,±2,}\begin{aligned} f(x) &= x^4 - 5x^2 + 6 \\ &= (x^2 - 3)(x^2 - 2) \\ &\implies \{ \alpha \} = \{ \pm \sqrt{3}, \pm \sqrt{2}, \} \end{aligned}

Therefore the splitting field of ff is Q(3,2)\mathbb Q(\sqrt{3}, \sqrt{2}).


An automorphism is an invertible, one-to-one, structure-preserving map of a set (or a field) onto itself: σ:FF\sigma: F \rightarrow F which satisfies the following axioms for all x,yFx,y \in F:

  1. σ(x+y)=σ(x)+σ(y)\sigma(x + y) = \sigma(x) + \sigma(y)
  2. σ(xy)=σ(x)σ(y)\sigma(xy) = \sigma(x)\sigma(y)
  3. σ(1/x)=1/σ(x)\sigma(1/x) = 1/\sigma(x)

That is, applying the automorphism to a compound expression constructed from the operations {+,,×,÷}\{+, -, \times, \div \} is equivalent to to applying it to the individual terms of that expression.

For example, conjugation is an automorphism in Q(2)\mathbb Q(\sqrt{2}) given by:

σ:Q(2)Q(2)σ(a+b2)=ab2\begin{aligned} \sigma : \mathbb Q(\sqrt{2}) &\rightarrow \mathbb Q(\sqrt{2}) \\ \sigma(a + b\sqrt{2}) &= a - b\sqrt{2} \\ \end{aligned}


An FF-automorphism is an automorphism σ\sigma over a field extension K/FK/F s.t.

xF;σ(x)=x\forall x \in F; \quad \sigma(x) = x

That is, σ\sigma leaves elements of the base field unchanged. We can imagine how this might be useful when we want to describe permutations of just the roots of a polynomial which leaves all other rational elements of our field extension Q(α)/Q\mathbb Q(\alpha)/\mathbb Q unchanged.

As it turns out, conjugation is the only non-trivial Q\mathbb Q-automorphism for radical extensions Q(k)\mathbb Q(\sqrt{k}).

Formally, we describe this relationship for any polynomial ff with rational coefficients, the field extension K/QK/\mathbb Q, and a Q\mathbb Q-automorphism σ\sigma of KK as:

σ(f(x))=f(σ(x))\sigma(f(x)) = f(\sigma(x))

This is useful since it shows that a Q\mathbb Q-automorphism of the splitting field KK of ff permutes just the roots:

f(x)=0    f(σ(x))=σ(f(x))=0    σ(α)=0\begin{aligned} f(x) = 0 &\implies f(\sigma(x)) = \sigma(f(x)) = 0 \\ &\implies \sigma(\alpha) = 0 \end{aligned}

Knowing how a single Q\mathbb Q-automorphism σ\sigma of a splitting field QK\mathbb Q \subseteq K rearranges the α\alpha roots of a polynomial ff is sufficient to derive how σ\sigma acts on all elements of KK. For example, for f(x)=x45x+6f(x) = x^4 - 5x + 6, with the splitting field K=Q(2,3)K = \mathbb Q(\sqrt{2},\sqrt{3}), and the roots {α}={±2,±3,}\{\alpha \} = \{ \pm \sqrt{2}, \pm \sqrt{3}, \}, there is no Q\mathbb Q-automorphism σ\sigma of KK where σ(2)=3\sigma(\sqrt{2}) = \sqrt{3}.

We can prove this by contradiction, supposing that there does exist such a permutation where σ(2)=3\sigma(\sqrt{2}) = \sqrt{3}. Then,

σ(2)2=σ(22)=σ(2)=2\begin{aligned} \sigma(\sqrt{2})^2 &= \sigma(\sqrt{2}^2) = \sigma(2) = 2 \end{aligned}

because σ\sigma must preserve multiplicative structure of KK, and σ(x)=x\sigma(x) = x for all xx in the base field Q\mathbb Q. But if σ(2)=3\sigma(\sqrt{2}) = \sqrt{3}, then σ(2)2=32\sigma(\sqrt{2})^2 = \sqrt{3}^2, or 2=32 = 3, which is not true.

Galois Groups

A Galois group of a field extension K/QK/\mathbb Q is the collection of all Q\mathbb Q-automorphisms where the group operation of composition \circ on permutations means:

στ    (στ)(x)=σ(τ(x))\sigma \circ \tau \implies (\sigma \circ \tau)(x) = \sigma(\tau(x))

Galois groups can be denoted Gal(K/Q)Gal(K/\mathbb Q) or just Gal(f)Gal(f) if KK is a splitting field of a polynomial ff.

For example, f(x)=x22f(x) = x^2 - 2 has the Galois group Gal(f)=(σ,τ)Gal(f) = (\sigma, \tau) with σ\sigma being the conjugate we've seen before defined by σ(a+b2)=ab2\sigma(a + b\sqrt{2}) = a - b\sqrt{2} and τ\tau is the identity: τ(x)=x=σσ\tau(x) = x = \sigma \circ \sigma.

This Galois group is the same as the cyclic group of order 2: C2C_2 (which is also the symmetric group on 2 elements: S2S_2). It's cyclic because it takes two applications of the non-identitive element to reach the identity on the group: σ2=τ\sigma^2 = \tau.

Begin the Problem

Galois Theory builds on the Abel-Ruffini Theorem claiming that if an equation ff is soluble by radicals, then we should be able to construct a sort of "tower" of radical extensions of the rationals with the roots.

Galois theory simplifies the problem of general radical solutions of higher order polynomials by translating it from the infinite domain of fields (ring theory) to the domain of groups (specifically, groups of finite permutations of the roots) making the problem infinitely easier. Specifically, we consider the special case of automorphisms acting on the splitting field of a polynomial.

Formally, for a polynomial ff of the form:

f(x)=anxn+an1xn1++a1x+a0f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_1x + a_0

with rational coefficients aiQa_i \in \mathbb Q, we consider the field extension containing all the roots of ff and denote L=Q(α1,α2,...,αn)L = \mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n).

By definition of a root, f(αk)=0f(\alpha_k) = 0 for all values of αk\alpha_k we've adjoined to the rationals (I'll omit the subscript on α\alpha going forward since we cease caring about specific roots, but rather whole permutations of roots). But what happens when we apply a Q\mathbb Q-automorphism to the polynomial?

f(x)=anxn+an1xn1++a1x+a0f(α)=anαn+an1αn1++a1α+a0=0σ(0)=σ(anαn+an1αn1++a1α+a0)=σ(anαn)+σ(an1αn1)++σ(a1α)+σ(a0)=σ(an)σ(αn)+=σ(an)σ(α)n+=anσ(α)n+an1σ(α)na1σ(α)+a0\begin{aligned} f(x) &= a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \\ f(\alpha) &= a_{n}\alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0 \\ \\ \sigma(0) &= \sigma(a_{n}\alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0) \\ &= \sigma(a_{n}\alpha^{n}) + \sigma(a_{n-1}\alpha^{n-1}) + \cdots + \sigma(a_1\alpha) + \sigma(a_0) \\ &= \sigma(a_{n})\sigma(\alpha^{n}) + \cdots \\ &= \sigma(a_{n})\sigma(\alpha)^n + \cdots \\ &= a_{n}\sigma(\alpha)^n + a_{n-1}\sigma(\alpha)^{n-} \cdots a_1\sigma(\alpha) + a_0 \\ \end{aligned}

We're left with just the roots being expressed in terms of σ\sigma, since by definition of σ\sigma a Q\mathbb Q-automorphism of the splitting field of ff, it leaves elements of Q\mathbb Q unchanged.

If we fairly assume all roots of our polynomial have single multiplicity,1 then according to our properties of automorphisms, σ\sigma just permutes the roots leaving the rest of the expression unchanged.

We call this minimal extension Q(α)\mathbb Q(\alpha) a Galois Extension. The collection of all automorphisms of this extension (sometimes denoted Aut(L/Q)Aut(L/\mathbb Q)) is the Galois group of ff.

More Extensions

Two additional types of field extensions omitted from the front-loaded glossary are cyclotomic extensions and Kummer extensions.

Cyclotomic Extensions

Consider all polynomials of the form f(x)=xn1f(x) = x^n - 1. For n=5n=5 We can plot the roots of unity2 on the complex plane which are evenly distributed along the unit circle:

Where, in general, {α1,α2,...,αn}\{ \alpha_1, \alpha_2, ..., \alpha_n\} are 11 and the powers of the nn-th roots of unity ζ\zeta.

To analyze the Galois Group of ff, we extend the rationals to include our roots of unity: Q(ζ)\mathbb Q(\zeta).

Again, it is sufficient to adjoin just ζ\zeta since we know the other roots are its powers up to ζn1\zeta^{n-1}. So, what automorphisms are in this Galois Group? 11 needs to remain unchanged as it is in the base field Q\mathbb Q, but starting with the first root of unity ζ\zeta, we can consider any automorphism sending this to any other root.

Let's consider the automorphism σ:Q(ζ)Q(ζ)\sigma: \mathbb Q(\zeta) \rightarrow \mathbb Q(\zeta) which permutes ζ\zeta by sending it to ζ3\zeta^3. As mentioned earlier, we can derive the rest of σ\sigma from just this relation. Moving to the next root of unity ζ2\zeta^2, we apply the definition of σ\sigma:

σ(ζ2)=σ(ζ)2=σ(ζ)σ(ζ)=ζ3ζ3=ζ6=ζ5ζ=1ζ=ζ\begin{aligned} \sigma(\zeta^2) = \sigma(\zeta)^2 &= \sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 = \zeta^6 \\ &= \zeta^5 \zeta = 1 \cdot \zeta \\ &= \zeta \end{aligned}

Similarly for the other roots we get

σ(ζ3)=σ(ζ)3=σ(ζ)σ(ζ)σ(ζ)=ζ3ζ3ζ3=ζ9=ζ5ζ4=1ζ4=ζ4σ(ζ4)=σ(ζ)4=σ(ζ)σ(ζ)σ(ζ)σ(ζ)=ζ3ζ3ζ3ζ3=ζ12=ζ5ζ5ζ2=11ζ2=ζ2\begin{aligned} \sigma(\zeta^3) = \sigma(\zeta)^3 &= \sigma(\zeta)\sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 \zeta^3 = \zeta^9 \\ &= \zeta^5 \zeta^4 = 1 \cdot \zeta^4 \\ &= \zeta^4 \\ \sigma(\zeta^4) = \sigma(\zeta)^4 &= \sigma(\zeta)\sigma(\zeta)\sigma(\zeta)\sigma(\zeta) \\ &= \zeta^3 \zeta^3 \zeta^3 \zeta^3 = \zeta^{12} \\ &= \zeta^5 \zeta^5 \zeta^2 = 1 \cdot 1 \cdot \zeta^2 \\ &= \zeta^2 \\ \end{aligned}

The complete permutation derived from ζζ3\zeta \rightarrow \zeta^3 is then:

We can denote this unique mapping as σ3\sigma_3 since it sends the first non-fixed element (since σ(1)=1\sigma(1) = 1) ζ\zeta to ζ3\zeta^3. In general σl=ζl\sigma_l = \zeta^{l}. Note that for cyclotomic automorphisms we have the following property:

σlσm(ζ)=σl(ζm)=(σl(ζ))m=ζlm\sigma_l\sigma_m(\zeta) = \sigma_l(\zeta^m) = (\sigma_l(\zeta))^m = \zeta^{lm}

Furthermore, this also implies commutativity of such permutations: σlσm=σmσl=ζlm\sigma_l\sigma_m = \sigma_m\sigma_l = \zeta^{lm}. Only some Galois groups will have this abelian property.

Recall that A group GG is said to be abelian if its operation is commutative. Here, the elements of our Galois group are permutations σ,τ\sigma, \tau, and the operation is composition (implicitly, \circ):

σ,τG;  στ=τσ\forall \sigma, \tau \in G;\; \sigma \tau = \tau \sigma

So, the Galois group of f(x)=xn1f(x) = x^n - 1 is abelian.

Kummer Extensions

Consider the generalization of the simple polynomial from the previous example f(x)=xnθf(x) = x^n - \theta. Again, the nn-th roots of θ\theta fall along a circle on the complex plane (just not necessarily the unit circle).

Our first root becomes α\alpha, rather than 11, and the rest of the roots take the form αζ,αζ2,...,αζn\alpha\zeta, \alpha\zeta^2, ..., \alpha\zeta^{n}. To find the Galois group, our base field is no longer simply the rationals, but instead it's some field extension KK. We'll define KK to be the field containing ζ\zeta, though by default it doesn't matter since we can simply adjoin ζ\zeta to an existing field and then call that sumbitch KK without loss of generality. We also have θK\theta \in K, and next we adjoin the roots of ff to KK yielding the splitting field over ff denoted K(α)K(\alpha).

In this case, unlike the Cyclotomic extension, the first element of our root α\alpha is not constrained by the KK-automorphism over K(α)K(\alpha), so σ(α)\sigma(\alpha) is not fixed like σ(1)=1\sigma(1) = 1 was. We can permute α\alpha anywhere, so lets choose the automorphism σ2\sigma_2 which sends ααζ2\alpha \rightarrow \alpha\zeta^2.

Continuing derivation of the next element in our group αζ\alpha\zeta, according to the preservation of algebraic structure, we get:

σ2(αζ)=σ2(α)σ2(ζ)=σ2(α)ζ=αζ2ζ=αζ3\begin{aligned} \sigma_2(\alpha\zeta) &= \sigma_2(\alpha)\sigma_2(\zeta) \\ &= \sigma_2(\alpha)\zeta \\ &= \alpha\zeta^2\zeta \\ &= \alpha\zeta^3 \end{aligned}

Note that since ζ\zeta is a member of the base group KK, σ\sigma leaves it unchanged and instead effectively only permutes α\alpha. Similarly, we get

σ2(αζ2)=σ2(α)σ2(ζ2)=σ2(α)ζ2=αζ2ζ2=αζ4σ2(αζ3)=σ2(α)σ2(ζ3)=σ2(α)ζ3=αζ2ζ3=αζ5=αζσ2(αζ4)=σ2(α)σ2(ζ4)=σ2(α)ζ4=αζ2ζ4=αζ6=αζ2\begin{aligned} \sigma_2(\alpha\zeta^2) &= \sigma_2(\alpha)\sigma_2(\zeta^2) \\ &= \sigma_2(\alpha)\zeta^2 \\ &= \alpha\zeta^2\zeta^2 \\ &= \alpha\zeta^4\\ \\ \sigma_2(\alpha\zeta^3) &= \sigma_2(\alpha)\sigma_2(\zeta^3) \\ &= \sigma_2(\alpha)\zeta^3 \\ &= \alpha\zeta^2\zeta^3 \\ &= \alpha\zeta^5 \\ &= \alpha\zeta \\ \\ \sigma_2(\alpha\zeta^4) &= \sigma_2(\alpha)\sigma_2(\zeta^4) \\ &= \sigma_2(\alpha)\zeta^4 \\ &= \alpha\zeta^2\zeta^4 \\ &= \alpha\zeta^6 \\ &= \alpha\zeta^2 \end{aligned}

So, all together, σ2\sigma_2 on Gal(f)Gal(f) permutes the roots in this starshape:

And, similar to the unitary expansion, we can label an automorphism σl(α)=αζl\sigma_l(\alpha) = \alpha\zeta^l according to where σ\sigma sends the first root α\alpha, and we once again can try to determine whether these groups of automorphisms are abelian:

σlσm(α)=σl(αζm)=αζl+m\sigma_l\sigma_m(\alpha) = \sigma_l(\alpha\zeta^m) = \alpha\zeta^{l + m}

Though the rules of composition result in summing the powers rather than multiplying them, these powers still commute:

σlσm=σmσl=αζl+m\sigma_l\sigma_m = \sigma_m\sigma_l = \alpha\zeta^{l+m}

so the group is still abelian. Note, however, that since the automorphisms over this kind of field extension are all determined by where α\alpha goes, it eliminates several possibilities of permutations. The only automorphism sending ααζ2\alpha \rightarrow \alpha\zeta^2 permutes the roots in the starshape illustrated above, meaning that a permutation

ααζ2αζ4αζαζ3α\alpha \rightarrow \alpha\zeta^2 \rightarrow \alpha\zeta^4 \rightarrow \alpha\zeta \rightarrow \alpha\zeta^3 \rightarrow \alpha

like this is not possible:

In this sense, the Galois group, or rather the size of the Galois group is a measure of how related the roots of a polynomial are. In our example, the roots differ by powers of ζ\zeta, so there really aren't that many automorphisms.

Naturally, we'll want to compose these extensions for some polynomial of the form f=xnθf = x^n - \theta with roots α\alpha.

But what about a direct extension from Q\mathbb Q to the largest field extension Q(ζ,α)\mathbb Q(\zeta, \alpha)?

The result would still be a splitting field over the same polynomial ff with roots in α\alpha (assuming θQ\theta \in \mathbb Q) to begin with. So it would still be a Galois extension and we can then consider the Galois group of this extension.

Towers of Galois

Now we can begin to show how analysis of the Galois groups of field extensions can tell us meaningful things about the general solubility of polynomials. Ultimately, we want to devise a sequence of field extensions resembling the following:

where the β\beta terms are whatever unholy perversions like those

28923\sqrt[23]{2 - \sqrt{89}}

terms we ran into earlier. First, we'll consider a generic, simple case involving 3 field extensions to formalize the problem of Group Solubility.

We're most interested in the Galois group of the Galois extension M/KM/K containing the KK-automorphisms of MM as well as the LL-automorphisms of MM. Note that because

KLMK \subseteq L \subseteq M

The LL-automorphisms of MM, in addition to leaving elements of LL unchanged by definition, also leave KK unchanged!

On the other hand, the Galois group Aut(M/K)Aut(M/K) containing automorphisms of the field extension from KK straight to MM implies the existence of a broader Galois group Aut(M/K)Aut(M/K) which also contains Aut(M/L)Aut(M/L) since they both leave elements of KK unchanged, and every σAut(M/L)\sigma \in Aut(M/L) necessarily leaves elements of KK unchanged too, so we can also say:

Aut(M/L)Aut(M/K)Aut(M/L) \subseteq Aut(M/K)

Furthermore, because L/KL/K is a Galois extension, we can say something more. Our prior constraint of automorphisms –that they are transformations σ:LL\sigma: L \rightarrow L where LL is the splitting field of ff– can be loosened. The only requirement for this tower we've constructed is that αL\alpha \in L which needs to be true so that we can apply σ(α)\sigma(\alpha). In fact, LL can be a bigger field so long as it still contains the splitting field over FF.

Returning to our tower, we fix LL to in fact be the splitting field of ff over KK. Aut(M/KAut(M/K) then will still contain permutations of the roots of ff. So, even though we're only fixing our smaller field KK, permutations σL:αiαj\sigma_L: \alpha_i \rightarrow \alpha_j are still transformations entirely within LL, that is:

L=K(α1,α2,...,αn)L = K(\alpha_1, \alpha_2, ..., \alpha_n)

Next we consider a new pair of automorphisms: σAut(M/K)\sigma \in Aut(M/K) is a KK-automorphism of MM insofar as it leaves all elements of KK unchanged. Additionally, we can consider an LL-automorphism of MM denoted τAut(M/L)\tau \in Aut(M/L), and observe that the composition of these two automorphisms on some element L\ell \in L:


And, without knowing anything else about σ,τ\sigma, \tau or \ell other than these properties of groups, we can infer that:

  • σ1()=L\sigma^{-1}(\ell) = \ell' \in L because L\ell \in L
  • τ()=L\tau(\ell') = \ell'' \in L leaves \ell unchanged, so it is also in LL
  • σ()\sigma(\ell'') is effectively just σσ1=\sigma\sigma^{-1} = \ell''

So the whole chain of permutations is itself an automorphism in Aut(M/L)Aut(M/L). This kind of relationship allows us to say that Aut(M/L)Aut(M/L) is a normal subgroup of Aut(M/K)Aut(M/K).

In general, we say that NN is a normal subgroup of GG (denoted NGN \vartriangleleft G) iff:

γνγ1(x)=(x)νN,γG\begin{aligned} \gamma\nu\gamma^{-1}(x) = (x) \\ \nu \in N, \gamma \in G \end{aligned}

Like the characters from Lost and the island, we return to the tower once again we can label the Galois group over the field extension from our base group KK to MM as G=Aut(M/K)G = Aut(M/K) and the Galois group of the extension from LL to MM as N=Aut(M/L)N = Aut(M/L) which we recognize to be a normal subgroup of GG.

Considering finally the third Galois group of the lowest order field extension L/KL/K, elements of this group HH are the KK-automorphisms over LL. And the last insight we'll draw from this simplest-form of a Galois tower is the consistent behavior of automorphisms of a subgroup HH within a broader group NN.

To illustrate, we consider automorphisms νN\nu \in N again as one-to-one mappings:

Straightforwardly, ν\nu sends elements of LL to itself. ν\nu on LL behaves like an automorphism of LL.

It's also possible to have another automorphism νN\nu' \in N which preserves the transformations on elements of LL while doing something completely different to the elements in MM:

The composition νν\nu \circ \nu' of any two LL-automorphisms ν,νAut(M/L)\nu, \nu' \in Aut(M/L) on the subfield LL is effectively the same as σσ1\sigma \circ \sigma^{-1}.

Automorphic Difference and Quotient Groups

We say that the result of ν11ν2\nu_1^{-1}\nu_2 (or sometimes this composite permutation itself) is the difference between automorphisms. If ν11,ν2\nu_1^{-1}, \nu_2 both do the same thing to the elements of the smaller field LL, then their composition leaves LL fixed (ν2\nu_2 sends an element of LL from left to right, and ν11\nu_1^{-1} sends an element of LL back from right to left).

This observation is restricted to elements in LL, we can make no inferences about the nature of ν\nu on MM. This difference itself though is another LL-automorphism of the group Aut(M/L)Aut(M/L) leaving LL unchanged. When ν11,ν2N\nu_1^{-1}, \nu_2 \in N we say that ν11ν2\nu_1^{-1}\nu_2 is a quotient group G/NG/N. Two automorphisms of GG are in G/NG/N if they differ by an element of NN.

So, all together, the relationships between the field extensions and their respective groups of our tower are as follows:

At the moment, it may seem like we're no closer to knowing anything about quintics though...

If we reference the initial target tower:

Additionally, if we simply further adjoin the roots of unity ζ\zeta with Q(23)\mathbb Q(\sqrt[3]{2}) yielding Q(23,ζ)\mathbb Q(\sqrt[3]{2}, \zeta) this would not be a radical extension.3 Instead, we would want to first adjoin the unities with the rationals which (a familiar cyclotomic extension), then adjoin the cube roots of θ=2\theta = 2 which is also a valid Kummer extension:

We know that Cyclotomic and Kummer extensions are both Galois extensions, so now we're kosher again. This tells us that, when we're constructing our tower, we want to first add all the roots via Galois extension, then add just the roots at each stage ensure that these are radical extensions.

We construct it this way so that we can reuse the results of previous constructions, namely that Galois groups of both types of extensions are Abelian.

To construct the target tower, we add all nin_i-th roots of unity: βi=θini\beta_i = \sqrt[n_i]{\theta_i} where nin_i are the orders of radicals βi\beta_i making all these radical extensions either Cyclotomic or Kummer, and thus all Galois.

For the big, direct extension from QQ(β1,β2,...,βr)\mathbb Q \rightarrow \mathbb Q(\beta_1, \beta_2, ..., \beta_r), there is a tedious and recursive construction algorithm:

  1. Let K0=QK_0 = \mathbb Q
  2. Before adjoining βi=θini\beta_i = \sqrt[n_i]{\theta_i} to Ki1K_{i-1}, first adjoin those nin_i-th roots of unity.
  3. After adjoining βi\beta_i, then add all numbers of the form βi(σ)=σ(θi)ni\beta_i(\sigma) = \sqrt[n_i]{\sigma(\theta_i)} where σAut(Ki1/Q)\sigma \in Aut(K_{i-1}/\mathbb Q)

This works via the fact that the only thing fixed by all automorphisms in Aut(Ki1/Q)Aut(K_{i-1}/\mathbb Q) must already be in Q\mathbb Q to begin with.

This big Galois extension is the splitting field of some polynomial ff with rational coefficients in Q\mathbb Q. Since Q\mathbb Q is our base field, the coefficients are necessarily also in all our intermediate extensions, so all extensions from an intermediate field to the biggest target field are also Galois – including the splitting field of the quintic Q(α1,α2,...,αn)\mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n). So we have:

Which has the same form as our pedagogical tower:

We can label all the extension's Galois groups of the target tower G,G1,...,N,HG, G_1, ..., N, H and prove that the target tower is "reducible" to the trivial tower.

An Aside about Group Solubility

We've now introduced enough concepts about groups (namely, normal subgroups and quotient groups) to define group solubility.

There is a very deep theorem in finite group theory which is known as the Feit-Thompson theorem. It states that if G|G| is odd, then GG is a solvable group. The paper giving this proof, entitled ”Solvability of Groups of Odd Order,” was published in 1963 and is more than 250 pages long.4

Suppose that GG is a finite group. We say that GG is soluble if the following statements about it are true:

  • If G>1|G| > 1, then GG has a normal subgroup H1GH_1 \vartriangleleft G such that the quotient group G/H1G/H_1 is abelian
  • If H1>1|H_1| > 1 then H1H_1 also has a normal subgroup H2H1H_2 \vartriangleleft H_1 such that the quotient group H1/H2H_1/H_2 is abelian
  • Etc. till the base case of Hk=1|H_k| = 1 (with the possibility that G=1|G| = 1)

Alternatively, if we denote G=H0G = H_0, GG is soluble if there exists a sequence of subgroups H1,...,HkH_1, ..., H_k s.t. for each j,1jkj, 1 \leq j \leq k:

  • HjH_j is a normal subgroup of its predecessor HjHj1H_j \vartriangleleft H_{j-1}
  • the quotient group Hj1/HjH_{j-1}/H_j is abelian
  • and Hk={ϵG}H_k = \{ \epsilon_G\}, that is, the final subgroup contains merely the identity on H0H_0

If GG is a solvable group, then every subgroup of GG is a solvable group, and every quotient group of GG is also a solvable group. Suppose that GG is a group and that NN is a normal subgroup of G: NGN \vartriangleleft G, then it can be proved that GG is a solvable group iff both G/NG/N and NN are solvable.

To reduce the target tower to an instance of the trivial tower, first we focus on G,G1G, G_1, and the Cyclotomic extension.

We know that these are related as subgroups where G1GG_1 \vartriangleleft G, and with the Cyclotomic extension having the Galois quotient group G/G1G/G_1 which means that it's abelian, so any two automorphisms will commute.

Similarly, G2G1G_2 \vartriangleleft G_1 and G1/G2G_1/G_2 is abelian. We can repeat this relation till we reach the last radical extension with Galois group GrG_r which itself also needs to be abelian, giving us a big ol' chian:

GG1G2Gr G \vartriangleright G_1 \vartriangleright G_2 \vartriangleright \cdots \vartriangleright G_r

where the constituent quotient groups Gi/Gi+1G_i/G_{i+1}. This chain of groups is said to be soluble, and the Galois theory states that if there exists abelian subgroups GiG_i to build this chain, then the biggest, direct group GG is solvable. If the tower QQ(β1,β2,...βr)\mathbb Q \rightarrow \cdots \rightarrow \mathbb Q(\beta_1, \beta_2,...\beta_r) is solvable, what about the right side of the tower from QQ(α1,α2,...,αn)Q(β1,β2,...βr)\mathbb Q \rightarrow \mathbb Q(\alpha_1, \alpha_2, ..., \alpha_n) \rightarrow \mathbb Q(\beta_1, \beta_2,...\beta_r)?

We automatically have H=G/NH = G/N as solvable since it's determined by our quintic ff and this setup we've constructed implies that if ff is soluble by radicals, then the Galois group generated by our radical extensions will be soluble. So all that's left to do is find some ff whose Galois group is not soluble.

Again, the comprehensive proof strategy is a bit tedious but I'll outline it below before taking a shortcuit to intuition which leverages the in-solubility of Symmetric Groups shared by most polynomials of degree five and above.

Rigorous Proof Strategy & S5S_5 and Above

  • Claim 1: AnA_n is generated by 3-cycles
  • Claim 2: Any normal subgroup of BnAnB_n \vartriangleleft A_n contains a 3-cycle
    • Case 1: BnB_n contains an nn-cycle for n4n \geq 4
    • Case 2: BnB_n contains at least 2 disjoint 3-cycles
    • Case 3: BnB_n contains one 3-cycle with the remaining permutations being transpositions
    • Case 4: BnB_n consists solely of transpositions

Intuitively, though, what we've done is layer so many constraints that it becomes impossibles to satisfy all of the for polynomials of degree five or higher in the general case.

Recall that a subgroup NGN \vartriangleleft G means that gng1Ngng^{-1} \in N where NN is a group s.t. n1n2,n1,ϵ=nn1Nn_1n_2, n^{-1}, \epsilon = nn^{-1} \in N and. These conditions on NN end up being so restrictive that there's really only a few possible groups which satisfy this criteria e.g. A5S5A_5 \vartriangleleft S_5.

This group we consider now contains all possible permutations of the five roots of a quintic polynomial and it's called the symmetric group of order five, denoted S5S_5. Consider the conjugal5 property of a subgroup NN and how gng1Ngng^{-1} \in N acts on an element:

And consider also the permutation gg on a 5-cycle in S5S_5:

s1s2s3s4s5s1s_1 \rightarrow s_2 \rightarrow s_3 \rightarrow s_4 \rightarrow s_5 \rightarrow s_1

If we fix gg to permute the first two elements of the set that S5S_5 is defined over, gng1gng^{-1} is still a 5-cycle:

In general, any 5-cycle can be obtained by gg in our Galois group. So, if any 5-cycle is in NN, all 5-cycles must be in NN, per the conjugation property. And also, the group property of composition of automorphisms ν1ν2N\nu_1 \circ \nu_2 \in N. So, if the composition of two 5-cycles gives a 3 cycle, then all 3-cycles are in NN as well...

The point being that the existence of even a single 5-cycle quickly expands/reveals NN to be the whole group it's supposed to be normal to. And if we recall that for a group GG to be solvable, every quotient subgroup G/Gi+1G/G_{i+1} as well as the final group GrG_r must all be abelian, so it should be relatively rare that XYZ, considering all these constraints. It's no surprise then that S5S_5 is not solvable in general.

S2,S3,S4S_2, S_3, S_4 on the other hand are very solvable, completely solvable in general. Because gng1Ngng^{-1} \in N, and NGN \vartriangleleft G requires fewer conjugations to admit all permutations, so there's fewer constraints.

E.g. for S4S_4, ff GG is abelian, then GG is a solvable group understood as:

  • H0=S4H_0 = S_4
  • H1=A4H_1 = A_4
  • H2={ϵ,(1  2)(3  4),(1  3)(2  4),(1  4)(2  3)}H_2 = \{\epsilon, (1 \; 2)(3 \; 4), (1 \; 3)(2 \; 4), (1 \; 4)(2 \; 3) \}
  • H3={ϵ}H_3 = \{ \epsilon \}

Note that the quotient group H0/H1H_0/H_1 is of order 2, H1/H2H_1/H_2 of order 3, and H2/H3H2H_2/H_3 ≅ H_2 is a group of order 4, and all of these quotient groups are abelian. Every subgroup HkGH_k \subset G is a solvable group, and every quotient group of GG is also a solvable group, therefore S4S_4 is solvable.

More on Insolvency of S5S_5

S5S_5 is not solvable because it has a composition series {ϵ,A5,S5}\{\epsilon, A_5, S_5 \}, giving factor groups isomorphic to A5A_5 and C2C_2, and A5A_5 is not abelian! A5A_5 is a simple, non-trivial (has more than one element), proper (not equal to the whole group) normal subgroup and is non-abelian.

A normal subgroup S5S_5 must be a union of conjugacy classes, that is – a set of element that are connected by conjugation: a,ba, b are conjugates of each other if gG\exists g \in G s.t. a=gbg1a = gbg^{-1}.

Conjugacy classes partition the elements of a group into disjoint subsets, which are the orbits of the group acting on itself by conjugation. Conjugacy classes of a group can be used to classify groups; they can be used to show that two groups are not isomorphic, or to discover properties of an isomorphism between groups if it exists. In general, the sizes of conjugacy classes in a group give information about its structure.

The orbit of an element xXx \in X is the set of elements {yyX:gx=y,gG}\{y | y\in X: g \cdot x = y, g \in G\}, the elements that xx can become by combining it with yXy \in X via an operation GG. Orbits are to groups as characteristics are to fields.

The conjugacy classes in S5S_5 consist of permutations having the same cycle-decompositions. S5S_5 has seven such decompositions shapes with the following cardinalities (number of possible distinct permutations in S5S_5 with that shape):

(a  b)(a \; b)10
(a  b  c)(a \; b \; c)20
(a  b  c  d)(a \; b \; c \; d)30
(a  b  c  d  e)(a \; b \; c \; d \; e)24
(a  b)(c  d)(a \; b)(c \; d)15
(a  b  c)(d  e)(a \; b \; c)(d \; e)20

Note that the sum total of these cardinalities, denoted S5S_5 = 120 = 5!$ which checks out since that's the number of ways to permute 5 distinct elements

Additionally, any subgroup of S5S_5 must contain the identity element and also have order dividing 120. Note that all of these conjugacy classes are disjoint. These criteria indicate that the only possible normal subgroups of S5S_5 will have orders: 1, 40, 60, 120. S5S_5 only has one subgroup of order 60, which is A5A_5 formed by the union of conjugacy classes with the following shapes: (a  b  c)(a  b)(c  d)(a  b  c  d  e)(a \; b \; c) \cup (a \; b)(c \; d) \cup (a \; b \; c \; d \; e) as well as the identity permutation ϵ\epsilon of course.

For order 40, there is one choice of conjugacy classes with total cardinality equal to 40 consisting of the unique conjugacy classes of sizes 1, 24 and 15:

ϵ(a  b  c  d  e)(a  b)(c  d)\epsilon \cup (a \; b \; c \; d \; e) \cup (a \; b) (c \; d)

but the union is not a subgroup because (a  b)(c  d)(a \; b)(c \; d) and (a  b)(c  e)(a \; b)(c \; e) are in the conjugacy class of size 15 and their product (a  b)(c  d)(a  b)(c  e)=(c  d  e)(a \; b)(c \; d)(a \; b)(c \; e) = (c \; d \; e), which is one of the conjugacy classes of size 20. Similarly, the only possibility for n = 60 is the union of the unique conjugacy classes of size 1, 24 and 15 together with the conjugacy class of size 20 consisting of the 3-cycles.

Hence, the only possible choice of the proper and normal subgroup H1H_1 is A5A_5 since if H1={ϵ}H_1 = \{ \epsilon \}, then G/H1G/H_1 is nonabelian. A5A_5's conjugacy classes and their cardinalities are:

(a  b  c)(a \; b \; c)20
(a  b  c  d  e)(a \; b \; c \; d \; e)2 * 12 = 24
(a  b)(c  d)(a \; b)(c \; d)15

Considering all possible unions of these sets, the total cardinality cannot divide 60 excepts for {ϵ}\{ \epsilon \} which is trivial and A5A_5 itself which is not proper. Therefore, there is no possible choice of a proper, normal subgroup H2H_2 of H1=A5H_1 = A_5 if we require (which we do) that H1/H2H_1/H_2 be abelian. Since A5A_5 is not solvable, S5S_5 is not solvable.

Which ff have Gal(f)=S5Gal(f) = S_5

So, we've found a Galois group that's insoluble: S5S_5, and if we can now find a quintic polynomial with Galois Group S5S_5 that is still soluble by radicals, then we've proved by contradiction the claim.

Turns out, most quintics chosen at random will have Galois group S5S_5 which we can verify with some Mathematica incantations6

(* The group p: *)
p = PermutationGroup[
	{Cycles[{{2, 3, 5, 4}}],
	Cycles[{{1, 2, 3, 4, 5}}]}

(* has elements g: *)
g = GroupElements[p]

g =	[   {Cycles[{}], Cycles[{{2, 3, 5, 4}}],
		Cycles[{{2, 4, 5, 3}}],  Cycles[{{2, 5}, {3, 4}}],
		Cycles[{{1, 2}, {3, 5}}], Cycles[{{1, 2, 3, 4, 5}}],
		Cycles[{{1, 2, 4, 3}}], Cycles[{{1, 2, 5, 4}}],
		Cycles[{{1, 3, 4, 2}}], Cycles[{{1, 3}, {4, 5}}],
		Cycles[{{1, 3, 5, 2, 4}}], Cycles[{{1, 3, 2, 5}}],
		Cycles[{{1, 4, 5, 2}}],  Cycles[{{1, 4, 3, 5}}],
		Cycles[{{1, 4}, {2, 3}}], Cycles[{{1, 4, 2, 5, 3}}],
		Cycles[{{1, 5, 4, 3, 2}}], Cycles[{{1, 5, 3, 4}}],
		Cycles[{{1, 5, 2, 3}}], Cycles[{{1, 5}, {2, 4}}]}

where the order of g given by Length[g] = 20

(* We can see that C_5 is a subgroup of p: *)
c5 = GroupElements[CyclicGroup[5]];
in = Intersection[g, c5]

(* and furthermore, that C_5 is a normal subgroup of p: *)
conj[x_] := PermutationProduct[#, x, InversePermutation[#]] & /@ g
Flatten[conj /@ c5] // Union

(* which yields:
		Cycles[{{1, 2, 3, 4, 5}}], Cycles[{{1, 3, 5, 2, 4}}],
		Cycles[{{1, 4, 2, 5, 3}}], Cycles[{{1, 5, 4, 3, 2}}]}

establishing that gHg' = H, where H = C_5.

And furthermore, we can show that the extension p/C_5 is abelian
since the quotient group elements commute:

cos = {a, b, c, d} =
	Union[RightCosetRepresentative[CyclicGroup[5], #] & /@ g]
su = Subsets[cos, {2}]
sur = Reverse /@ su
PermutationProduct @@@ su
PermutationProduct @@@ sur

(* yielding:
		{Cycles[{{2, 3, 5, 4}}], Cycles[{{2, 4, 5, 3}}],
		Cycles[{{2, 5}, {3, 4}}], Cycles[{}],
		Cycles[{{2, 4, 5, 3}}], Cycles[{{2, 3, 5, 4}}]}

		{Cycles[{{2, 3, 5, 4}}], Cycles[{{2, 4, 5, 3}}],
		Cycles[{{2, 5}, {3, 4}}], Cycles[{}],
		Cycles[{{2, 4, 5, 3}}], Cycles[{{2, 3, 5, 4}}]}

As C_5 <| p, and p/C_5, C_5/{e} are abelian, p is solvable

Example problems

Finally, a few example problems to further illustrate how to take all this lofty theory and apply it to some easy inquiries about splitting fields and Galois groups of various polynomials

Splitting Field of a Quartic

Find the splitting field of f(x)=x4+x2+1f(x) = x^4 + x^2 + 1 over Q\mathbb Q.

Without doing any actual number crunching, we can recognize that ff will factor into:

f(x)=(x2+ax+b)(x2+cx+d)=(a+c)x3+(ac+2)x2\begin{aligned} f(x) &= (x^2 + ax + b)(x^2 + cx + d) \\ &= (a + c)x^3 + (ac + 2)x^2 \end{aligned}

for some values of a,b,c,da,b,c,d. We can deduce that:

  • b=d=1b = d = 1 because the constant term in ff is 1,
  • (ac+2)=1(ac + 2) = 1 corresponds to the coefficient of the middle x2x^2 term in our non-factored polynomial,
  • 00 is the coefficient of our x3x^3 term.

These observations are sufficient to tell us what aa and cc are:

a+c=0    c=aac+2=1    c2=1    c=±1\begin{aligned} a + c = 0 &\implies c = - a \\ ac + 2 = 1 &\implies -c^2 = - 1 \\ &\implies c = \pm 1 \\ \end{aligned}

So ff will factor to:

f(x)=(x2+x+1)Φ3(x)(x2x+1)6th roots of unityf(x) = \underbrace{(x^2 + x + 1)}_{\Phi_3(x)}\underbrace{(x^2 - x + 1)}_{6th \text{ roots of unity}}

From here, we can make two claims / observations.

The first being that the first factor with a positive linear xx term is the third cyclic polynomial in xx whose roots are e2πi/3,e2πi/3e^{2 \pi i/3}, e^{-2 \pi i/3}.

The second claim is that the roots of the second factor with a negative linear xx term are eπi/3,eπi/3e^{ \pi i/3}, e^{-\pi i/3}.

We can verify these are roots by converting the roots of unity into rectangular coordinates and plugging them into their respective polynomial factors e.g.:

eπi/3=12±32ie^{\pi i/3} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i

and showing that the expression evaluates to zero. This gives us four roots of our quartic polynomial, all of which are powers of the third roots of unity eπi/3e^{ \pi i/3}, so we can obtain our splitting field by adjoining this third root of unity to the radicals: Q(eπi/3)\mathbb Q(e^{ \pi i/3}). We might equivalently express this field extension in terms of the rectangular coordinates as Q(3i)\mathbb Q(\sqrt{3} i).

And finally, we might inquire about the degree of our extension, which is equal to the degree of the minimal polynomial of 3i\sqrt{3}i. We can simply square this term (3i)2(\sqrt{3}i)^2 to get 3-3 which is a root of x2+3x^2+3 which is irreducible,7 so the degree of our extension is two.


Finding the Galois Group of a Quartic

For some polynomial f(x)=x420x+80f(x) = x^4 -20x + 80, we might inquire about the Galois group. To get started, we find the roots of ff which can be thought of as a quadratic equation in x2x^2. That is, we can solve for the roots by using x2x^2 as our variable, and then plug it into ye ole quadratic equation:

x2=20±4003202    x2=10±25    x=±10±25\begin{aligned} x^2 &= \frac{20\pm \sqrt{400-320}}{2} \\ &\implies x^2 = 10 \pm 2\sqrt{5} \\ &\implies x = \pm \sqrt{10 \pm 2\sqrt{5}} \\ \end{aligned}

We know that 10±2510 \pm 2\sqrt{5} is always non-negative, so we get all four roots as the various permutations of plus and minus signs of within our xx term. We'll alias α,β\alpha, \beta to be partitions of the roots divided by the sign of the inner root:

α=±10+25β=±1025\begin{aligned} \alpha = \pm \sqrt{10 + 2\sqrt{5}} \\ \beta = \pm \sqrt{10 - 2\sqrt{5}} \\ \end{aligned}

And we know that our splitting field, then, is K=Q(α,β)K = \mathbb Q(\alpha, \beta).

Next, to determine the contents of the Galois group of ff, we might consider what kinds of permutations exist between these roots. We can observe that ff is irreducible per Eisenstein's Criterion which states that a polynomial of the form

f(x)=anxn+an1xn1++a1x1+a0f(x) = a*nx^n + a*{n-1}x^{n-1} + \cdots + a_1x^{1} + a_0

is irreducible iff there is a prime pp s.t.:

  • pp does not divide the coefficient of the highest order term: p∤  anp \not{|} \; a_n
  • pp does divide the coefficients of the rest of the term: p    an1,...,p    a0p \; | \; a_{n-1}, ... , p \; | \; a_{0}
  • p2p^2 does not divide the constant term: p2∤  a0p^2 \not{|}\; a_0

So, our polynomial f(x)=x420x+80f(x) = x^4 -20x + 80 is irreducible for p=5p = 5 since:

  • 55 does not divide 1
  • 5 does divide 20 and 80
  • 25 does not divide 80

This tells us that the minimal polynomial for α\alpha over Q\mathbb Q, denoted irr(α;  Q)irr(\alpha; \; \mathbb Q) is actually ff since it has α\alpha as a root, and is irreducible. This tells us that the degree of this extension [K:Q][K : \mathbb Q] is at least 4 since the degree of the minimal polynomial of Q(α)K\mathbb Q(\alpha) \subset K is 4. But, KK also contains β\beta, which could possibly push the lower bound higher.

We can use this to determine possible elements of Gal(f)Gal(f). Consider the automorphisms of KK:

ψ:KKψ(α)=ψ(α)ψ(β)=ψ(β)\begin{aligned} \psi : K &\rightarrow K \\ \psi(-\alpha) &= -\psi(\alpha) \\ \psi(-\beta) &= -\psi(\beta) \\ \end{aligned}

And while this would be true of any automorphism because it's linear w.r.t. multiplication, but it's particularly special because our roots are ±α,±β\pm \alpha, \pm \beta, so any permutation of the roots our Galois group is singularly determined by where it sends α\alpha and where it sends β\beta.

Using that lil tidbit, it's straightforward to tabulate all possible elements of our Galois group in terms of where they send each positive root respectively:

α\alpha \mapstoα\alphaβ\betaα-\alphaβ-\betaβ\betaα-\alphaβ-\betaα\alpha
β\beta \mapstoβ\betaα-\alphaβ-\betaα\alphaα\alphaβ\betaα-\alphaβ-\beta

Note that these are the possible combinations of places we can send the roots, which do not necessarily fix automorphisms. And though this ordering seems entirely arbitrary, of course I'm cheating since I already know what Gal(f)Gal(f) is, so I structured it accordingly.

α\alpha \mapstoα\alphaβ\betaα-\alphaβ-\betaβ\betaα-\alphaβ-\betaα\alpha
β\beta \mapstoβ\betaα-\alphaβ-\betaα\alphaα\alphaβ\betaα-\alphaβ-\beta

We can retcon an intuitive derivation of these labels starting with σ\sigma which sends αβ\alpha \mapsto \beta. We can convince ourselves that σ2\sigma^2 would send αα\alpha \mapsto -\alpha per:

ασβσα\alpha \underset{\sigma}\longmapsto \beta \underset{\sigma}\longmapsto -\alpha

So these labels are consistent with the laws of composition, what a coincidence! Students of Group theory will recognize this table to be the elements of the dihedral group with eight elements D4D_4 which has one element of order 4 which is σ\sigma and another element with order 2 which is τ\tau, the product of which gives this dihedral group its name.

Because of the way the roots of this particular polynomial work, the possible automorphisms are contrived to be the elements of D4D_4. However, we don't know a priori that each of these elements is an automorphism, so we do a little bit more massaging to narrow down D4D_4 to the elements of our Galois group. We do this by considering a special element qQKq \in Q \subset K and apply each permutation ψD4\psi \in D_4 to see if it fixes qq since we know that Q\mathbb Q-automorphisms of KK leave elements of Q\mathbb Q unchanged. We set qq to some expression evaluates to a rational number, but will also be sufficiently greebled by ψ\psi such as:

q=αβ(α2β2)=(10+25)(1025)(10+25210252)=8080=80Q\begin{aligned} q &= \alpha\beta(\alpha^2 - \beta^2) \\ &=\Big(\sqrt{10 + 2\sqrt{5}}\Big)\Big(\sqrt{10 - 2\sqrt{5}}\Big)\Bigg(\sqrt{10 + 2\sqrt{5}}^2 - \sqrt{10 - 2\sqrt{5}}^2\Bigg) \\ &= \sqrt{80}\sqrt{80} \\ &= 80 \in \mathbb Q \\ \end{aligned}