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A formula in radicals is a tower of extensions, each contributing an abelian layer to the Galois group — so a formula exists exactly when the group is solvable, and the general quintic has none.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read a radical formula as a tower of field extensions, say why each step contributes a cyclic layer to the Galois group, state Galois's criterion for solvability by radicals, decide whether a named polynomial satisfies it, show that a particular quintic has Galois group $S_5$, and explain why the impossibility of a quintic formula is a theorem about a group rather than about degree five.
The second unit defined a solvable group and proved that $S_n$ is not one for $n \ge 5$, with the word solvable unexplained. The fifth unit built a correspondence between subgroups and subfields. This lesson joins them, and the join is the theorem both halves were written for.
$L/K$ is a radical extension when $L = K(\alpha)$ with $\alpha^{n} \in K$ for some $n$ — adjoining an $n$th root. A radical tower is a chain of such steps. A polynomial is solvable by radicals over $K$ when its roots lie in some radical tower over $K$. The general polynomial of degree $n$ has indeterminate coefficients, and its Galois group is $S_n$.
What a formula is. The quadratic formula builds its answer from the coefficients using $+, -, \times, \div$ and one square root. Cardano's cubic formula uses square roots and cube roots. To say a polynomial is solvable by radicals is to say its roots can be built that way — which, read from the inside out, is a tower
$$K = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_m, \qquad F_{i+1} = F_i(\alpha_i) \text{ with } \alpha_i^{n_i} \in F_i,$$
with all the roots of the polynomial in $F_m$.
Galois's theorem. Over a field of characteristic zero, a polynomial is solvable by radicals exactly when the Galois group of its splitting field is a solvable group.
Why a radical step gives an abelian layer. Adjoin the $n$th roots of unity first; that step alone has abelian group $(\mathbb{Z}/n\mathbb{Z})^{\times}$. Now over a field containing them, $x^{n} - a$ has roots $\zeta^{k}\alpha$, so an automorphism is determined by which $k$ it picks, and composing adds the $k$s modulo $n$: the group is cyclic. A tower of such steps therefore gives a chain of subgroups with abelian factors, which is what solvable means.
The converse — a solvable group produces a formula — is the harder direction and uses Lagrange resolvents to build the radicals back out of the group.
The quintic. The general polynomial of degree $n$ has Galois group $S_n$. For $n \le 4$ these are solvable:
| $n$ | group | chain | formula |
|---|---|---|---|
| $2$ | $S_2$ | abelian | quadratic formula |
| $3$ | $S_3$ | $1 \lhd A_3 \lhd S_3$ | Cardano |
| $4$ | $S_4$ | $1 \lhd V_4 \lhd A_4 \lhd S_4$ | Ferrari |
| $5$ | $S_5$ | factors $A_5$ and $\mathbb{Z}_2$ | none exists |
$A_5$ is simple and not abelian, so it is its own composition factor and cannot be refined. So $S_5$ is not solvable, and no formula in radicals solves the general quintic.
A specific example. $x^{5} - 6x + 3$ is irreducible by Eisenstein at $3$, and calculus shows it has exactly three real roots. So complex conjugation is a transposition in its Galois group, which also contains a $5$-cycle by Cauchy; a transposition and a $5$-cycle generate $S_5$. That particular equation has no radical solution.
What is not claimed. The roots exist — five of them, by the fundamental theorem of algebra — and can be computed numerically to any accuracy. What fails is a formula of one specific shape. Other expressions do work: elliptic and hypergeometric functions solve the general quintic, as Hermite showed in 1858.
Galois theory replaces an infinite search among fields with a finite list of subgroups. Every question about what lies between $K$ and $L$ — how many fields, which are normal, what their degrees are — is answered by reading a subgroup lattice upside down.
Another way: picture
A formula in radicals is a staircase, each step an extraction of one root. Galois's theorem says the staircase is visible in the group: each step shows up as an abelian layer. A group with a non-abelian simple factor has a slab in it that cannot be cut into layers at all — so no staircase can reach the top, however long it is, and the search for a quintic formula was a search for a staircase through solid rock.
Another way: steps
To decide whether a polynomial is solvable by radicals: 1. Find its splitting field and Galois group. 2. Compute the composition factors of that group. 3. Solvable by radicals exactly when every factor is abelian, hence of prime order. 4. For degree at most $4$ the answer is always yes. 5. For a quintic, showing the group is $S_5$ — irreducible with exactly three real roots is the standard route — shows the answer is no.
The general quintic is unsolvable because its group is $S_5$ by construction. Showing a particular rational quintic has group $S_5$ takes an argument, and the standard one is short and worth knowing.
Take $f(x) = x^{5} - 6x + 3$.
Step one: irreducible. Eisenstein at $p = 3$: the prime divides $-6$ and $3$, does not divide the leading $1$, and $9$ does not divide $3$. So $f$ is irreducible over $\mathbb{Q}$, and its Galois group acts transitively on the five roots. By orbit-stabiliser, $5$ divides the order of the group, so by Cauchy the group contains an element of order $5$ — a $5$-cycle.
Step two: exactly three real roots. $f'(x) = 5x^{4} - 6$ has two real zeros, so $f$ has two turning points and at most three real roots; checking signs at a few points finds three. So there are exactly two non-real roots, and they are conjugates of each other.
Step three: complex conjugation is a transposition. It fixes the three real roots and swaps the two others — so as a permutation of the roots it is a transposition, and it lies in the Galois group because it is an automorphism of $\mathbb{C}$ fixing $\mathbb{Q}$.
Step four: they generate. A $5$-cycle and a transposition generate $S_5$ whenever the degree is prime. So the group is $S_5$, which is not solvable, and $f$ has no radical solution.
What each unit of the course contributed. Orbit-stabiliser and Cauchy in step one; the class-equation argument for the simplicity of $A_5$ behind the final verdict; the correspondence to convert the group statement into a field statement; and radical towers for the last translation. The proof is a tour of the entire subject, which is why it is the last item.
Thinking an unsolvable quintic has no roots. It has five, in $\mathbb{C}$. What it lacks is a radical expression for them.
Thinking no quintic is solvable. $x^{5} - 2$ and $x^{5} - 1$ both are. The claim is about the general quintic and about particular ones with group $S_5$.
Thinking numerical methods are blocked. They are not. Newton's method converges happily on $x^{5} - 6x + 3$.
Thinking a bigger degree is the obstruction. It is the group. $x^{100} - 2$ has degree $100$ and a solvable group, and is solvable by radicals.
*Reading solvable as has a solution.* It is a technical property of a group, defined two units ago, and the coincidence of names is the whole joke of the terminology.
Assuming no expression of any kind solves the quintic. Hermite solved it with elliptic functions in 1858. The theorem restricts radicals specifically.
The general cubic has Galois group $S_3$, of order $6$.
The group.
$1 \lhd A_3 \lhd S_3$ has factors of orders $3$ and $2$, both cyclic and so abelian: $S_3$ is solvable.
A chain with abelian factors.
So a radical formula exists — and reading the chain backwards is essentially how Cardano's formula is derived, with the cube root corresponding to the factor of order $3$ and the square root to the factor of order $2$.
The chain is the formula.
$x^{5} - 2$ has splitting field $\mathbb{Q}(\sqrt[5]{2}, \zeta_5)$, of degree $5 \times 4 = 20$.
Degree twenty.
The subgroup fixing $\zeta_5$ is normal of order $5$, with quotient of order $4$; both are abelian.
A chain with abelian factors.
So the group is solvable and the equation is solvable by radicals — visibly so, since $\sqrt[5]{2}$ is a radical. Degree five is not itself the obstruction.
Solvable after all.
Its splitting field is $\mathbb{Q}(\sqrt[6]{2}, \zeta_6)$, and $\zeta_6$ generates $\mathbb{Q}(\sqrt{-3})$ of degree $2$.
A tower of two steps.
The degree is $6 \times 2 = 12$, and the subgroup fixing $\zeta_6$ is normal of order $6$ with abelian quotient of order $2$.
A normal subgroup with abelian quotient.
That subgroup is cyclic of order $6$, hence abelian, so the whole group is solvable and the equation is solvable by radicals — as it obviously is, since $\sqrt[6]{2}$ is written with a radical. The theorem agrees with the obvious answer, which is a good check on it.
Put in order the steps of the argument that a formula in radicals forces the Galois group to be solvable.
Number the steps in order (write the number in the box):
Is this solvable by radicals over $\mathbb{Q}$: the general polynomial of degree $5$?
The general polynomial of degree $n$ has Galois group $S_n$. For each degree, fill in the order of $S_n$ and then $1$ if that group is solvable and $0$ if it is not.
| Degree | Order of the group | Solvable: one or zero | |
|---|---|---|---|
| The general quadratic | 2 | ||
| The general cubic | 3 | ||
| The general quartic | 4 | ||
| The general quintic | 5 |
Select every statement that is true about solvability by radicals.
This task has no paper form; do it on a device.
Match each ingredient of Galois's theorem to what it contributes.
| An abelian group, the units modulo $n$ | A cyclic group of order dividing $n$ | A chain of subgroups with abelian factors | A group that is a quotient of a solvable one, hence solvable | A simple non-abelian group | |
|---|---|---|---|---|---|
| Adjoining an $n$th root of unity | |||||
| Adjoining $\sqrt[n]{a}$ when the roots of unity are present | |||||
| A whole tower of such steps | |||||
| The splitting field sitting inside the tower |
$A_{8}$ is simple and not abelian. In a composition series of $S_{8}$, how many of the factors are abelian and come from breaking $A_{8}$ down further?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put in order the steps of the argument that a formula in radicals forces the Galois group to be solvable.
Number the steps in order (write the number in the box):
You can decide whether a polynomial is solvable by radicals from its Galois group, and say what the negative answer does and does not claim. Say in your own words why a group with a non-abelian simple composition factor blocks every possible formula.
9. Your turn: is $x^{6} - 2$ solvable by radicals over $\mathbb{Q}$?, step 3