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Two families where the Galois group can be named outright: the roots of unity, whose group is the units modulo $n$, and the finite fields, whose group is cyclic and generated by raising to the $p$th power.
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 compute the degree of a cyclotomic extension as $\phi(n)$, identify its Galois group as the units modulo $n$ and say when that group is cyclic, find the real subfield and its degree, use the subgroup lattice to list the subfields, describe the Frobenius map and check it is an automorphism, and prove that it generates the Galois group of a finite field extension.
The fundamental theorem turns a Galois group into the complete structure of an extension. It is only as useful as the groups one can actually compute. This lesson supplies the two families where the group is known outright: the roots of unity, where it is the units modulo $n$, and the finite fields, where it is cyclic and generated by one explicit map.
A root of unity of order $n$ satisfies $x^{n} = 1$; it is primitive when its order is exactly $n$. The $n$th cyclotomic polynomial $\Phi_n$ is the monic polynomial whose roots are the $\phi(n)$ primitive $n$th roots; it has integer coefficients and is irreducible over $\mathbb{Q}$. The Frobenius map of a field of characteristic $p$ is $x \mapsto x^{p}$.
Cyclotomic extensions. Let $\zeta_n = e^{2\pi i / n}$. The roots of $x^{n} - 1$ are the powers of $\zeta_n$, so the splitting field is the simple extension $\mathbb{Q}(\zeta_n)$ — every other root is already a power of the one adjoined.
The minimal polynomial is $\Phi_n$, of degree $\phi(n)$, so
$$[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \phi(n).$$
Irreducibility of $\Phi_n$ is a genuine theorem; for prime $p$ it follows from Eisenstein applied to $\Phi_p(x+1)$.
The group. An automorphism sends $\zeta_n$ to another primitive root, so to $\zeta_n^{a}$ with $\gcd(a, n) = 1$, and composing corresponds to multiplying the exponents. So
$$\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^{\times},$$
which is abelian of order $\phi(n)$. Every subgroup is therefore normal, so every subfield of a cyclotomic field is Galois over $\mathbb{Q}$. The converse — every abelian extension of $\mathbb{Q}$ sits inside a cyclotomic field — is the Kronecker-Weber theorem, and it is deep.
The group is cyclic exactly when $n$ is $1, 2, 4, p^{k}$ or $2p^{k}$ for an odd prime $p$. For $n = 8$ it is the Klein four-group.
The real subfield. Complex conjugation is the automorphism $a = -1$, of order $2$ for $n > 2$. Its fixed field is $\mathbb{Q}(\zeta_n + \zeta_n^{-1}) = \mathbb{Q}(2\cos(2\pi/n))$, of degree $\phi(n)/2$. That degree is what decides whether a regular $n$-gon is constructible.
Finite fields. The Frobenius map $\varphi(x) = x^{p}$ is an automorphism of any field of characteristic $p$: multiplicative trivially, additive because $\binom{p}{k}$ is divisible by $p$ for $0 < k < p$. On $\mathbb{F}_{p^{n}}$ it fixes $\mathbb{F}_p$ pointwise by Fermat's little theorem, and $\varphi^{k}(x) = x^{p^{k}}$ is the identity only for $k \ge n$ — so it has order $n$. Since the degree is $n$,
$$\operatorname{Gal}(\mathbb{F}_{p^{n}}/\mathbb{F}_p) = \langle \varphi \rangle \cong \mathbb{Z}_n.$$
A cyclic group of order $n$ has one subgroup per divisor, so $\mathbb{F}_{p^{n}}$ has one subfield per divisor of $n$ — the theorem proved by hand earlier, now a corollary.
Why these two matter. They are the computable abelian cases, and between them they carry most of the applications: cyclotomic fields for number theory and for constructibility, finite fields for coding theory and cryptography.
Another way: picture
Draw the $n$th roots of unity as $n$ equally spaced points on a circle. An automorphism must permute them while respecting multiplication, so it is multiply every angle by $a$ for some $a$ coprime to $n$ — spin the whole picture up by a whole number of turns. That is why the group is the units modulo $n$: it is the ways of winding the circle onto itself without collapsing it.
Another way: steps
For a cyclotomic field: 1. Compute $\phi(n)$; that is the degree. 2. The group is $(\mathbb{Z}/n\mathbb{Z})^{\times}$; identify it by factoring $n$. 3. Its subgroups give all the subfields, all of them normal over $\mathbb{Q}$. For a finite field: 4. The group is generated by $x \mapsto x^{p}$, cyclic of order $n$. 5. Subfields are one per divisor of $n$.
$\mathbb{Q}(\zeta_5)$ has degree $\phi(5) = 4$ and Galois group $(\mathbb{Z}/5\mathbb{Z})^{\times}$, which is cyclic of order $4$ generated by the residue $2$ — its powers are $2, 4, 3, 1$.
The subgroups. A cyclic group of order $4$ has exactly three: the trivial one, the one of order $2$ generated by $4 \equiv -1$, and the whole group.
So there are exactly three intermediate fields, and the middle one is the fixed field of $\{1, -1\}$ — that is, of complex conjugation. It is $\mathbb{Q}(\zeta_5 + \zeta_5^{-1}) = \mathbb{Q}(2\cos 72^\circ)$, of degree $2$.
Naming it. Let $y = \zeta + \zeta^{-1}$. From $\zeta^{4} + \zeta^{3} + \zeta^{2} + \zeta + 1 = 0$, divide by $\zeta^{2}$:
$$\zeta^{2} + \zeta^{-2} + \zeta + \zeta^{-1} + 1 = 0, \qquad (y^{2} - 2) + y + 1 = 0,$$
so $y^{2} + y - 1 = 0$ and $y = \frac{-1 + \sqrt5}{2}$. The intermediate field is $\mathbb{Q}(\sqrt5)$.
What that settles. $\cos 72^\circ$ has degree $2$ over $\mathbb{Q}$, a power of $2$, so the regular pentagon is constructible — as it has been known to be for two thousand years, now with a reason. The same computation for $n = 7$ gives $\phi(7)/2 = 3$, and the heptagon is not constructible.
And a harder fact worth knowing exists. The Kronecker-Weber theorem says every Galois extension of $\mathbb{Q}$ with abelian group lies inside some $\mathbb{Q}(\zeta_n)$. So the cyclotomic fields are not merely a convenient family — they contain every abelian extension of the rationals there is.
Taking the degree of $\mathbb{Q}(\zeta_n)$ to be $n$. It is $\phi(n)$. The polynomial $x^{n} - 1$ has degree $n$ but is not irreducible — it always has the factor $x - 1$.
Assuming $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is cyclic. For $n = 8, 12, 15, 16$ it is not. It is cyclic exactly for $n = 1, 2, 4, p^{k}, 2p^{k}$ with $p$ an odd prime.
Confusing the additive and multiplicative groups modulo $n$. The Galois group is the units under multiplication, of order $\phi(n)$, not $\mathbb{Z}/n\mathbb{Z}$ of order $n$.
Thinking Frobenius is the identity because $a^{p} = a$. That holds on the prime field only. On $\mathbb{F}_{p^{n}}$ with $n > 1$ it moves most elements, and that is the whole point.
Forgetting Frobenius is additive. $(x+y)^{p} = x^{p} + y^{p}$ is true here and nowhere else; it depends entirely on the characteristic dividing the middle binomial coefficients.
Expecting a cyclotomic extension to need several generators. One root of unity generates them all, because the others are its powers — which is why these splitting fields are simple extensions.
$\phi(8) = 4$, so $\mathbb{Q}(\zeta_8)$ has degree $4$ and the group has order $4$.
Order four.
The units modulo $8$ are $1, 3, 5, 7$, and $3^{2} = 9 = 1$, $5^{2} = 25 = 1$, $7^{2} = 49 = 1$ modulo $8$.
Everything squares to one.
So the group is the Klein four-group, not cyclic. Its three subgroups of order $2$ give three quadratic subfields: $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt{-2})$ — since $\zeta_8 = \frac{\sqrt2}{2}(1 + i)$.
Three quadratic subfields.
$\mathbb{F}_8 = \mathbb{F}_2(\alpha)$ with $\alpha^{3} = \alpha + 1$. Frobenius is $x \mapsto x^{2}$.
The generator.
It sends $\alpha \mapsto \alpha^{2} \mapsto \alpha^{4} = \alpha^{2} + \alpha \mapsto \alpha^{8} = \alpha$, a cycle of length $3$.
Order three.
The degree is $3$, so the group is cyclic of order $3$. It has no proper non-trivial subgroups, so $\mathbb{F}_8$ has no intermediate field — consistent with $3$ being prime.
No intermediate field.
$\phi(7) = 6$, so the degree is $6$ and the group is $(\mathbb{Z}/7\mathbb{Z})^{\times}$ of order $6$.
Order six.
For a prime modulus the unit group is cyclic, so this is cyclic of order $6$, and a cyclic group has one subgroup per divisor: $1, 2, 3, 6$.
Four subgroups.
So four subfields: $\mathbb{Q}$, a quadratic one ($\mathbb{Q}(\sqrt{-7})$), a cubic one (the real subfield $\mathbb{Q}(\cos(2\pi/7))$), and the whole field. All four are normal over $\mathbb{Q}$, because the group is abelian.
For each $n$, fill in the degree of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$, and then the degree of its real subfield $\mathbb{Q}(\cos(2\pi/n))$.
| The value of $n$ | Degree of the cyclotomic field | Degree of the real subfield | |
|---|---|---|---|
| The first case | 3 | ||
| The second case | 4 | ||
| The third case | 5 | ||
| The fourth case | 7 | ||
| The fifth case | 8 | ||
| The sixth case | 12 |
What is $[\mathbb{Q}(\zeta_{3}) : \mathbb{Q}]$, where $\zeta_{3}$ is a primitive $3$th root of unity?
Answer:
Select every statement that is true of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$.
This task has no paper form; do it on a device.
Match each cyclotomic field to its Galois group over $\mathbb{Q}$.
| Cyclic of order $4$ | Cyclic of order $6$ | The Klein four-group | Cyclic of order $2$ | Cyclic of order $8$ | |
|---|---|---|---|---|---|
| $\mathbb{Q}(\zeta_5)$ | |||||
| $\mathbb{Q}(\zeta_7)$ | |||||
| $\mathbb{Q}(\zeta_8)$ | |||||
| $\mathbb{Q}(\zeta_3)$ |
Put in order the steps of showing that the Galois group of $\mathbb{F}_{p^{n}}$ over $\mathbb{F}_p$ is cyclic of order $n$.
Number the steps in order (write the number in the box):
Build the proof that $\operatorname{Gal}(\mathbb{F}_{p^{n}}/\mathbb{F}_p)$ is cyclic of order $n$, generated by the Frobenius map.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For each $n$, fill in the degree of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$, and then the degree of its real subfield $\mathbb{Q}(\cos(2\pi/n))$.
| The value of $n$ | Degree of the cyclotomic field | Degree of the real subfield | |
|---|---|---|---|
| The first case | 3 | ||
| The second case | 4 | ||
| The third case | 5 | ||
| The fourth case | 7 | ||
| The fifth case | 8 | ||
| The sixth case | 12 |
You can name the Galois group of a cyclotomic field or a finite field extension and list its subfields. Say in your own words why the degree of $\mathbb{Q}(\zeta_n)$ is $\phi(n)$ rather than $n$. Next: what a solvable Galois group has to do with a formula in radicals.
9. Your turn: how many subfields has $\mathbb{Q}(\zeta_7)$?, step 3