$\DeclareMathOperator{gal}{Gal}$
 Let $Φ_{m}(x) ∈ ℂ[x]$ be the $m$th cyclotomic polynomial, the monic polynomial whose roots are the primitive $m$ th roots of 1 in $ℂ$. Show that
 $Φ_{1}(x)=x1 ; Φ_{2}(x)=x+1 ; Φ_{3}(x)=x^{2}+x+1 ; Φ_{4}(x)=x^{2}+1$.
 $\prod_{dm} Φ_{d}(x)=x^{m}1$.
 $Φ_{m}(x) ∈ ℤ[x]$. [Hint: prove first that $Φ_{m}(x) ∈ ℚ[x]$ by induction on $m$).
 If $p$ is prime then $Φ_{p}(x)=1+x+x^{2}+⋯+x^{p1}$ and $Φ_{p^{n}}(x)=Φ_{p}\left(x^{p^{n1}}\right)$.
 $\deg Φ_{n m}=\deg Φ_{m} \deg Φ_{n}$ if $(m, n)$ are relatively prime.
 Let $n$ be a positive integer and $f=x^{p^{n}}x ∈ 𝔽_{p}[x]$. Let $M$ be the splitting field of $f$ over $𝔽_{p}$. Show that $M$ consists exactly of the set of roots of $f$. Show that $\left[M: 𝔽_{p}\right]=n$. Explain why this fact also shows the existence of an irreducible polynomial of degree $n$ in $𝔽_{p}[x]$.

 Prove that $Φ_{12}(x)=x^{4}x^{2}+1$, and that it is irreducible over $ℚ$. Factorise it into irreducibles over $𝔽_{p}$ when $p=2,3,5,13$.
 If $p$ is any prime with $p>3$ show that $p^{2}1$ is divisible by 12, and deduce that $Φ_{12}$ is reducible over $𝔽_{p}$ for every prime $p$.
 For this exercise recall the definition of a group action on a set. Let $f ∈ K[x]$ be a separable degree $n$ polynomial, let $M$ be its splitting field and $G=Γ(M: K)$ be the Galois group of $M$. Let $A=\left\{α_{1}, … α_{n}\right\} ⊆ M$ be the set of roots of $f$. Let $S(A)$ be the set of permutations of the roots of $f$.
 Show that $G$ acts faithfully on $A$ (this is equivalent to showing that there is an injective group homomorphism between $G$ and $S(A)$).
 Show that if $f$ is irreducible, then $G$ acts transitively on $A$ (this is equivalent to show that for any $α_{i}, α_{j} ∈ A$ there exists $σ ∈ G$ such that $σ\left(α_{i}\right)=α_{j}$).
 Find the Galois groups of the following polynomials and for each subgroup identify the corresponding subfield of the splitting field:
 $x^{2}+1$ over $ℝ$;
 $x^{3}1$ over $ℚ$;
 $x^{3}5$ over $ℚ$;
 $x^{6}3 x^{3}+2$ over $ℚ$;
 $x^{5}1$ over $ℚ$;
 $x^{6}+x^{3}+1$ over $ℚ$.
 Find the Galois group of the polynomial $x^{p^{n}}xt$ over $𝔽_{p^{n}}(t)$ (you can assume that this polynomial is irreducible over $𝔽_{p^{n}}(t)$; you need not determine the subfield subgroup correspondence here).
 Prove that $ℚ(\sqrt{2+\sqrt{2}})$ is Galois over $ℚ$, and find its Galois group.
 This is clear for $Φ_1(x)$ and $Φ_2(x)$. For $Φ_3(x)$, notice that $x^31=(x1)\left(x^2+x+1\right)$ and for $Φ_4(x)$ notice that $(xi)(x+i)=x^2+1$.
 Denoting by $\operatorname{ord}(k, ℤ/m ℤ)$ the order of $k\pmod m$ in $ℤ/m ℤ$, we have
$$
x^m1=\prod_{k=0}^{m1}\left(xe^{2 i k π/m}\right)=\prod_{dm}\prod_{\operatorname{ord}(k, ℤ/m ℤ)=d}\left(xe^{2 i k π/m}\right)=\prod_{dm} Φ_d(x).
$$
 By induction on $m$. The statement is clear for $m=1$. By (b) we have
$$
Φ_m(x)=\frac{x^m1}{\prod_{dm, d ≠ m} Φ_d(x)}
$$
Now for any $dm$ with $d ≠ m$, the content of $Φ_d(x)$ is one by the inductive hypothesis. Hence the content of $Φ_m(x)$ is also one since the content function is multiplicative and the content of $x^m1$ is 1. Thus $Φ_m(x) ∈ ℤ[x]$ (see before Lemma 2.13 in the notes).
 If $p$ is prime, we have $Φ_p(x)=\left(x^p1\right) /(x1)=1+x+⋯+x^{p1}$ by (b).
For the equality $Φ_{p^n}(x)=Φ_p\left(x^{p^{n1}}\right)$, notice that $z ∈ ℂ$ is a primitive $p^n$th root of unity iff $z^{p^{n1}}$ is a primitive $p$th root of unity (why?). Thus both polynomials have the same roots. Since $Φ_{p^n}(x)$ has no multiple roots by definition we thus have $Φ_{p^n}(x)Φ_p\left(x^{p^{n1}}\right)$. On the other hand we have
$$
\deg\left(Φ_{p^n}(x)\right)=\#\left(ℤ/p^n ℤ\right)^*=p^{n1}(p1)
$$
 It was shown in Q6 sheet 1 that the set of roots of $f(x)$ in $M$ is a subfield of $M$, which must therefore coincide with $M$, since $M$ is generated by the roots of $f(x)$. Since $f(x)$ is separable (see Q6 of sheet 1), there are exactly $p^n$ elements in $M$. On the other hand $p^{[M:𝔽_p]}=p^n$, so $n=[M: 𝔽_p]$. For the second statement let $α ∈ M$ be a generator of the multiplicative group $M^*$ (which is cyclic). We then clearly have $M=𝔽_p(α)$. The minimal polynomial of $α$ is an irreducible polynomial of degree $n$ in $𝔽_p[x]$.
 Let $ρ_3$ be a primitive 3 rd root of 1 and let $ρ_4$ be a primitive 4 th root of 1 (eg $ρ_4=i$). If $\left(ρ_3 ρ_4\right)^k=1$ then $ρ_3^k=ρ_4^{4k}$, so that the order of $ρ_3^k$ in $ℂ^*$ divides 4. Since the order of $ρ_3^k$ also divides 3, we conclude that $ρ_3^k=ρ_4^{4k}=1$. Hence $3 ∣ k$ and $4 ∣ 4k$ (equivalently $4 ∣ k$) and so $12 ∣ k$. In particular, 12 divides the order of $ρ_3 ρ_4$. Since we also have $\left(ρ_3 ρ_4\right)^{12}=1$, we conclude that the order of $ρ_3 ρ_4$ is 12, ie $ρ_3 ρ_4$ is a primitive 12th root of 1. On the other hand,
$$
ℚ\left(ρ_3 ρ_4\right)=ℚ\left(ρ_4, ρ_3\right)=ℚ\left(ρ_4\right)\left(ρ_3\right)
$$
(because $ρ_4=\left(ρ_3 ρ_4\right)^9$ and $ρ_3=\left(ρ_3 ρ_4\right)^4$). Now $ρ_3=\frac12+\frac{\sqrt3}2i∉ℚ\left(ρ_4\right)=ℚ(i)$.
Now the extension $ℚ\left(ρ_3\right)(i) ∣ ℚ(i)$ has degree at most 2 since $ρ_3^2+ρ_3+1=0$. So it has degree 2. On the other hand, we have $[ℚ(i): ℚ]=2$ since $i^2+1=0$. We deduce from the tower law that $\left[ℚ\left(ρ_3 ρ_4\right): ℚ\right]=4$ so that the minimal polynomial of $ρ_3 ρ_4$ has degree 4. Since we also have $Φ_{12}\left(ρ_3 ρ_4\right)=0$ and $\deg\left(Φ_{12}(x)\right)=4$, we conclude that $Φ_{12}(x)$ is the minimal polynomial of $ρ_3 ρ_4$ and is hence irreducible.
Note. In Prop. 5.5 in the notes it is proven by a different argument that all the cyclotomic polynomials over $ℚ$ are irreducible.
We have the following decompositions into irreducibles in small finite fields.
 over $𝔽_2$
$$
Φ_{12}(x)=\left(x^2+x+1\right)^2\pmod2
$$
To find this, notice that $Φ_{12}(x)$ divides $x^{12}1$ in $ℤ[x]$ by Q1 (b) so the irreducible factors of $Φ_{12}(x)\pmod2$ are all irreducible factors of $x^{12}1\pmod2$. Now compute
$$
x^{12}1\pmod2=\left(x^31\right)^4\pmod2
$$
and
$$
\left(x^31, x^4x^2+1\right)=x^2+x+1\pmod2.
$$
One verifies that $x^2+x+1\pmod2$ has no roots in $𝔽_2$ so it must be an irreducible factor. Hence $Φ_{12}(x)=\left(x^2+x+1\right)^2\pmod2$.
 over $𝔽_3$
$$
Φ_{12}(x)=\left(x^2+1\right)^2\pmod3
$$
Similarly, we have $\left(x^{12}1\right)\pmod3=\left(x^41\right)^3\pmod3=\left(x^21\right)^3\left(x^2+1\right)^3\pmod3$ and we try $\left(x^21\right)$ and $\left(x^2+1\right)$.
 over $𝔽_5$
$$
Φ_{12}(x)=\left(x^2+2 x+4\right)\left(x^2+3 x+4\right)\pmod 5
$$
To find this, consider a possible decomposition
$$
x^4x^2+1=\left(x^2+a x+b\right)\left(x^2+c x+d\right)
$$
into irreducibles in $𝔽_5$ (having checked that $Φ_{12}(x)$ has no roots in $𝔽_5$). We compute
$$
\left(x^2+a x+b\right)\left(x^2+c x+d\right)=x^4+(a+c) x^3+(c a+b+d) x^2+(d a+c b) x+d b
$$
and we conclude that $a=c$. We compute again
$$
\left(x^2+a x+b\right)\left(x^2a x+d\right)=x^4+\left(b+da^2\right) x^2+(db) a x+d b
$$
We now check that $a=0$ does not work, because otherwise $b=1d$ and
$$
d b=d(1d)=1
$$
and $d^2d1$ has no solution in $𝔽_5$. Hence $d=b$ and we compute again
$$
\left(x^2+a x+b\right)\left(x^2a x+b\right)=x^4+\left(2 ba^2\right) x^2+b^2
$$
so that $b= ± 1$ and $a$ is a square root of $1+2 b$. We check that 3 has no square root in $𝔽_5$ so it must be $b=1\pmod5=4\pmod5$. Hence $a= ± 2$ and we check that $a=2$ does not work by computing $\left(Φ_{12}(x), x^22 x+4\right)\pmod5=1$. So $a=2$ and $b=4$ and the values of $c$ and $d$ follow from the above computation.
This method can also be applied to the $p=2$ and $p=3$ cases.
 over $𝔽_{13}$
$$
Φ_{12}(x)=(x+2)(x+6)(x+7)(x+11)\pmod{13}
$$
To find this, just look for all the roots of $Φ_{12}(x)$ in $𝔽_{13}$.
 Suppose that $p>2$. Note first that $p^21=(p1)(p+1)$. Note also that one of $p1, p$ or $p+1$ is divisible by 3 (since one of any three consecutive natural numbers is divisible by 3). Hence $p^21$ is divisible by 3. Also, since $p$ is odd (being prime and$>2$), we have that $p+1$ and $p1$ are both even and so $p^21$ is divisible by 4. Hence $p^21$ is divisible by 12.
Now by Q2, the splitting field $𝔽_{p^2}$ of $x^{p^2}x$ over $𝔽_p$ has order $p^2$. Hence the multiplicative group $𝔽_{p^2}^*$ has order $p^21$. Since $12 ∣ p^21$, the cyclic group $𝔽_{p^2}^*$ contains a cyclic group of order 12. Thus the polynomial $x^{12}1$ splits in $𝔽_{p^2}$. On the other hand, $Φ_{12}(x)$ divides $x^{12}1$ in $ℚ[x]$ hence there is a monic polynomial $Q(x) ∈ ℚ[x]$ such that $Φ_{12}(x) Q(x)=x^{12}1$. Looking at contents (and using that the content function is multiplicative), we conclude that the content of $Q(x)$ is 1 and hence $Q(x) ∈ ℤ[x]$. Reducing the equality
$$
Φ_{12}(x) Q(x)=x^{12}1
$$
modulo $p$, we conclude that $Φ_{12}(x)$ divides $x^{12}1$ in $𝔽_p[x]$. Hence $Φ_{12}(x)$ splits in $𝔽_{p^2}$ by the above. Now let $ρ$ be any root of $Φ_{12}(x)$ in $𝔽_{p^2}$ and let $P(x) ∈ 𝔽_p[x]$ be the minimal polynomial of $ρ$. Then $\deg(P) ≤\left[𝔽_{p^2}: 𝔽_p\right]=2$ (see Q2) and $P(x) ∣ Φ_{12}(x)$. Since $\deg\left(Φ_{12}(x)\right)=4$, we see that $Φ_{12}(x)$ is not irreducible over $𝔽_p$.
 If an element of $G$ acts trivially on $A$ then it acts trivially on $M$ since $A$ generated $M$.
 See Lemma 4.6 in the notes.
 $x^2+1$ over $ℝ$. The splitting field is $ℂ$ and the Galois group is $ℤ/2 ℤ$. There are no non trivial intermediate fields.
 $x^31$ over $ℚ$. We have $x^31=(x1)\left(x^2+x+1\right)$ so the splitting field is of degree 2, the Galois group is $ℤ/2 ℤ$ and there are no non trivial intermediate fields. This is an example of a cyclotomic field.
 $x^35$ over $ℚ$. Let $j=e^{2 i π/3}$. The splitting field of $x^35$ is $ℚ(\sqrt[3]{5}, j)$, which is of degree 6 over $ℚ$, since $j$ is not real, $j^2+j+1=0$ and $x^35$ is irreducible by Eisenstein's criterion (see Q4 (b) in sheet 2 for a completely analogous reasoning). Hence the Galois group must be $S_3$, since $\# S_3=6$ and the Galois group can be realised as a subgroup of $S_3$ (see Q4 (a)). There is a unique subgroup of order 3 in $S_3$ and it is isomorphic to $ℤ_3$ and there are three subgroups of order 2 in $S_3$, which are all conjugate to each other. For this, see eg
groupprops
Note. These groups are Sylow subgroups. See Th. 6.1 in the notes.
We proceed to find these subgroups in $\gal(ℚ(\sqrt[3]{5}, j) ∣ ℚ)$. The subgroup isomorphic to $ℤ/3 ℤ$ is $\gal(ℚ(\sqrt[3]{5}, j) ∣ ℚ(j))$. The corresponding subfield is thus $ℚ(j)$.
One of the subgroups of order two is $\gal(ℚ(\sqrt[3]{5}, j) ∣ ℚ(\sqrt[3]{5}))$ and the two other ones are obtained by conjugation. The corresponding subfields are $ℚ(\sqrt[3]{5}), ℚ(j \sqrt[3]{5})$ and $ℚ\left(j^2 \sqrt[3]{5}\right)$. To see this, note that clearly $ℚ(j \sqrt[3]{5}) ≠ ℚ(\sqrt[3]{5})$ and $ℚ\left(j^2 \sqrt[3]{5}\right) ≠ ℚ(\sqrt[3]{5})$ (because $j$ is not real). Furthermore, $ℚ(j \sqrt[3]{5}) ≠ ℚ\left(j^2 \sqrt[3]{5}\right)$ for otherwise $\left(j \sqrt[3]{5} j^2 \sqrt[3]{5}\right)^2=5\sqrt[3]{5} ∈ ℚ(j \sqrt[3]{5})$, which is not true, since $ℚ(j \sqrt[3]{5}) ≠ ℚ(\sqrt[3]{5})$.
 $x^63 x^3+2$ over $ℚ$. We have $x^63 x^3+2=(x1)\left(x^2+x+1\right)\left(x^32\right)$ so the splitting field is $ℚ(\sqrt[3]2, j)$, with $j$ as in (c). The Galois group and the subfields are now the same as in (c), replacing 5 by 2.
 $x^51$ over $ℚ$. The splitting field is $ℚ\left(e^{2 i π/5}\right)$. The minimal polynomial of $e^{2 i π/5}$ is $P(x)=x^4+x^3+x^2+x+1$. Now $P(x)$ is irreducible by Eisenstein's criterion because
$$
P(x+1)=x^4 + 5x^3 + 10x^2 + 10x + 5
$$
Using Theorem 5.4(ii) in the notes we now conclude that $\gal\left(ℚ\left(e^{2 i π/5}\right) ∣ ℚ\right)$ is naturally isomorphic to $(ℤ/5 ℤ)^* ≃ ℤ/4 ℤ$. This group has only one non trivial subgroup (of index $2)$ and the corresponding intermediate field is clearly $ℚ\left(e^{2 i π/5}+e^{2 i π/5}\right)$.
 $x^6+x^3+1$ over $ℚ$. The roots of $P(x)=x^6+x^3+1$ are cube roots of $j$ and $j^{1}$($j$ as before). Let $ρ ∈ L$ be a root of $P(x)$. Notice now that
$$
P(x+1)=x^6+6 x^5+15 x^4+21 x^3+18 x^2+9 x+3
$$
is irreducible by Eisenstein's criterion and thus $P(x)$ is irreducible. Thus $P(x)=Φ_9(x)$ and thus its Galois groups is $(ℤ/9 ℤ)^* ≃ ℤ/2 ℤ × ℤ/3 ℤ$.
In particular $\gal(L ∣ ℚ)$ has only two non trivial subgroups, one of order 2 and the other one of order 3. The group of order 3 corresponds to the field $ℚ(j)$, since the Galois group of $ℚ(j)$ is of order 2. The other group corresponds to the subfield $ℚ\left(ρ+ρ^{1}\right)$. To see this, notice that $\bar{ρ}=ρ^{1}$ and thus $ℚ\left(ρ+ρ^{1}\right) ⊆ ℝ$. Hence $ℚ(ρ+ρ^{1}) ≠ ℚ(j)$ and it must be the field we are looking for by unicity.
 $x^{p^n}xt$ over $𝔽_{p^n}(t)$. Let $P(x)=x^{p^n}xt$. Note that if $P(α)=P(β)=0$ then $(αβ)^{p^n}=αβ$. So that $αβ$ is a root of $x^{p^n}x$. In particular, $αβ ∈ 𝔽_{p^n} ⊆𝔽_{p^n}(t)$. We deduce that the splitting field $L$ of $P(x)$ is generated by any root of $P(x)$ and is thus of degree $p^n$, since $P(x)$ is irreducible. Now choose a root $ρ$ of $P(x)$ in $L$. Let $τ ∈ \gal\left(L ∣ 𝔽_{p^n}(t)\right)$. We see that $τ(ρ)=ρ+μ$, where $μ ∈ 𝔽_{p^n}$. If $ρ'$ is another root then $ρ'=ρ+μ'$ and thus
$$
τ(ρ')=ρ+μ+μ'=ρ'+μ.
$$
In particular, the element $μ$ does not depend on $ρ$. Let us write $μ=μ(τ)$ to emphasize the dependence on $τ$. Notice that if $σ ∈ \gal\left(L ∣ 𝔽_{p^n}(t)\right)$ then
$$
σ τ(ρ)=ρ+μ(τ)+μ(σ)
$$
We have thus constructed an injective group homomorphism $\gal\left(L ∣ 𝔽_{p^n}(t)\right) ↪ 𝔽_{p^n}$. Finally, since $\# \gal\left(L ∣ 𝔽_{p^n}(t)\right)=p^n=\deg(P(x))$ and $\# 𝔽_{p^n}=p^n$, this homomorphism must be an isomorphism.
Note. These kinds of extensions are called ArtinSchreier extensions. They are "additive" analogues of Kummer extensions (see section 5.2 of the notes). The reasoning made here is very similar to the reasoning made in the proof of Lemma 5.6 in the notes (with multiplication replaced by addition).
 The polynomial $P(x)=\left(x^22\right)^22=x^44 x^2+2$ annihilates $\sqrt{2+\sqrt2}$ and it is irreducible by Eisenstein's criterion. Its roots are $± \sqrt{2 ± \sqrt{2}}$. We compute
$$
\sqrt{2\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}
$$
so that all the roots of $P(x)$ are in $L≔ℚ(\sqrt{2+\sqrt{2}})$. In particular, $L$ is Galois over $ℚ$ and it is the splitting field of $P(x)$. Since $\deg(P)=4$ and any group of order 4 is abelian, we see that $\gal(L ∣ ℚ)$ is isomorphic to either $ℤ/2 ℤ × ℤ/2 ℤ$ or $ℤ/4 ℤ$ (by the structure theorem for finite abelian groups—see Rings and Modules). We now analyse the Galois group in more detail. Let $M=ℚ(\sqrt{2}) ⊆ L$. Let $λ$ be a generator of the group $\gal(L ∣ M)$. We then have $λ( ± \sqrt{2 ± \sqrt{2}})=∓ \sqrt{2 ± \sqrt{2}}$. Let $τ ∈ \gal(L ∣ ℚ)∖\gal(L ∣ M)$. After perhaps replacing $τ$ by $λ τ$, we may assume that $τ(±\sqrt{2+\sqrt{2}})= ± \sqrt{2\sqrt{2}}$ (why?). Finally, we know that $τ^{1}$ is equal to either $τ$ or $λ τ$ (since $τ^{1}$ does not fix $M$). If $τ^{1}=τ$, then $τ(\sqrt{2\sqrt{2}})=\sqrt{2+\sqrt{2}}$ and thus
$$
\sqrt{2}=\sqrt{2\sqrt{2}} ⋅ \sqrt{2+\sqrt{2}}=τ(\sqrt{2\sqrt{2}} ⋅ \sqrt{2+\sqrt{2}})=τ(\sqrt{2})
$$
which is not true, since $\left.τ\right_M ≠ \mathrm{Id}$ by construction. So $τ^{1}=λ τ$. In particular, $τ$ is not of order 2 and must therefore be of order 4. Thus $\gal(L ∣ ℚ) ≃ ℤ/4 ℤ$.