$\DeclareMathOperator{\gal}{Gal}$

1. • Let $F=ℚ(α)$ where $α^3=2$. Prove that $x^2+α x+α^2$ is irreducible over $F$.
   • Prove that $x^{p-1}+⋯+x+1$ is irreducible over $ℚ$ for any prime $p$.
   • Let $F=ℚ(α)$ where $α^5=2$. Prove that $x^4+α x^3+α^2 x^2+α^3 x+α^4$ is irreducible over $F$.

   ---

   • $x^2+α x+α^2=(x-ωα)(x-ω^2α)$
     If not irreducible, $ω∈F$
     $2=[ℚ(ω):ℚ]∣[F:ℚ]=3$, contradiction.
   • $x^{p-1}+⋯+x+1=\frac{x^p-1}{x-1}$ shift by 1: $$\frac{(1+x)^p-1}x=x^{p-1}+px^{p-2}+⋯+p$$ is irreducible by Eisenstein.
   • Roots of $f$ are $αζ^j,j∈\{1,…,4\}$ where $ζ=e^{2πi\over5}$
     splitting field of $f$ over $ℚ(α)$ is $ℚ(α,ζ)$, I want to show $[ℚ(α,ζ):ℚ(α)]=4=\deg f$, so $f$ is irreducible.
     $[ℚ(α,ζ):ℚ]=5[ℚ(α,ζ):ℚ(α)]≤20$ because ζ has degree ≤4 over ℚ(α), $$\left.\begin{array}r 4=[ℚ(ζ):ℚ]∣[ℚ(α,ζ):ℚ]\\5=[ℚ(α):ℚ]∣[ℚ(α,ζ):ℚ]\\4\text{ and }5\text{ are coprime}\end{array}\right\}⇒20∣[ℚ(α,ζ):ℚ]$$ so $20=[ℚ(α,ζ):ℚ(α)][ℚ(α):ℚ]⇒[ℚ(α,ζ):ℚ(α)]=4$
2. Exhibit a Galois extension $K / F$ and an explicit intermediate subfield $F ⊂ L ⊂ K$ such that $L / F$ fails to be Galois.

   ---

   $ℚ(\root3\of2,ω)/ℚ(\root3\of2)/ℚ$. See Q4(b)
3. Let $p$ be a prime, let $F$ be a field of characteristic $p$, and let $f ∈ F[t]$.
   • Show that $D(f)=0$ if and only if there exists $g ∈ F[t]$ such that $f(t)=g\left(t^p\right)$.
   • We say that $F$ is perfect if the $p$-power map is bijective on $F$. Deduce that if $F$ is perfect, then $f$ is separable.

   ---

   • $Df=0$, $Dx^n=nx^{n-1}$, $p∣$every exponent.
4. For each of the following subfields of ℂ:
   • $K=ℚ(\sqrt{2}, \sqrt{3})$
   • $K=ℚ(\sqrt[3]{2}, e^{\frac{2 π i}{3}})$
   • $K=ℚ(e^{\frac{2 π i}{7}})$
   • prove that $K$ is a Galois extension of ℚ,
   • compute the Galois group $G=\gal(K/ℚ)$,
   • exhibit a generator for each proper subfield of $K$ strictly containing ℚ,
   • calculate the minimal polynomial over ℚ of this generator.

   ---

   • • $K=ℚ(\sqrt{2}, \sqrt{3})=ℚ(±\sqrt{2},±\sqrt{3})$ is the splitting field of $(x^2-2)(x^2-3)$.
       Since char ℚ = 0, $K/ℚ$ is separable and hence Galois.
     • By property of Galois extension, ${|\gal(ℚ(\sqrt2,\sqrt3),ℚ)|}=[ℚ(\sqrt2,\sqrt3):ℚ]=4$
       Groups of order 4 are $C_2^2$ and $C_4$
       $\sqrt2↦±\sqrt2,\sqrt3↦±\sqrt3$ so at least two index-2 subgroups, so must be $C_2^2$.
     • $C_2^2$ has 3 proper subgroups, corresponding to 3 proper subfield of $K$ strictly containing ℚ: $$ℚ(\sqrt2),ℚ(\sqrt3),ℚ(\sqrt6).$$
     • $x^2-2,x^2-3,x^2-6$.
   • • $K=ℚ(\sqrt[3]{2}, e^{\frac{2 π i}{3}})=ℚ(\sqrt[3]{2},\sqrt[3]{2}e^{\frac{2 π i}{3}},\sqrt[3]{2}e^{\frac{4π i}{3}})$ is the splitting field of $x^3-2$.
       Since char ℚ = 0, $K/ℚ$ is separable and hence Galois.
     • ${|\gal(ℚ(\sqrt[3]2,e^{2π i\over3}),ℚ)|}=[ℚ(\sqrt[3]2,e^{2π i\over3}):ℚ]=6$
       Groups of order 6 are $S_3≅D_6$ and $C_6$
       $\gal(ℚ(\sqrt[3]2,e^{2π i\over3}),ℚ)$ permutes 3 roots, so is a subgroup of $S_3$, ${|S_3|}=6$, so must be all of $S_3$.
       Alternatively, use \begin{array}{ll} σ: α→ ω α, &ω → ω\\ τ: α→α,&ω →ω^2 \end{array} generate group and $σ^3=τ^2=τστσ=1$.

       ---

       Let $α=\sqrt[3]{2},β=\sqrt[3]{2}e^{\frac{2 π i}{3}},γ=\sqrt[3]{2}e^{\frac{4π i}{3}}$. For each $δ ∈\{α, β, γ\}$, by Lemma 4.12 we have an extension $φ_δ: ℚ(α) → ℚ(δ)$ of $\mbox{id}_ℚ$ that sends $α$ to $δ$. Error! Click to view log. For $δ=α$, $φ_α$ has two extensions $φ_α^{(1)}, φ_α^{(2)}$ to an automorphism of $K$, characterised by $φ_α^{(1)}(β)=β$ and $φ_α^{(2)}(β)=γ$.
       For $δ=β$, $φ_β$ sends $m_{ℚ(α), β}=t^2+α t+α^2$ to $t^2+β t+β^2 ∈ ℚ(β)[t]$, which factorises as $(t-α)(t-γ)$ over $K$. Hence there are two extensions $φ_β^{(1)}, φ_β^{(2)}$ of $φ_β$, characterised by $φ_β^{(1)}(β)=α$ and $φ_β^{(2)}(β)=γ$. Similarly, there are two extensions $φ_γ^{(1)}, φ_γ^{(2)}$ of $φ_γ: ℚ(α) → ℚ(γ)$, characterised by $φ_γ^{(1)}(β)=α$ and $φ_γ^{(2)}(β)=β$. $$ \gal(K/ℚ)=\left\{φ_α^{(1)}, φ_α^{(2)}, φ_β^{(1)}, φ_β^{(2)}, φ_γ^{(1)}, φ_γ^{(2)}\right\} $$ where the effect of these automorphisms on $V(f)=\{α, β, γ\}$ is given as follows: \begin{array}{c|cccccc} & φ_α^{(1)} & φ_α^{(2)} & φ_β^{(1)} & φ_β^{(2)} & φ_γ^{(1)} & φ_γ^{(2)} \\ \hlineα & α & α & β & β & γ & γ \\ β & β & γ & α & γ & α & β \\ γ & γ & β & γ & α & β & α \end{array} Because all possible permutations of $\{α, β, γ\}$ occur, this Galois group has to be isomorphic to $S_3$.
     • There are four proper subgroups of $S_3$; they are all cyclic. The three of order 2 generated by (1 2), (1 3) and (2 3), and the one of order 3 generated by (1 2 3). corresponding to proper subfields of $K$ strictly containing ℚ:$$ℚ(α),ℚ(β),ℚ(γ),ℚ(ω)$$ Error! Click to view log.
     • $m_{α,ℚ}(x)=m_{β,ℚ}(x)=m_{γ,ℚ}(x)=x^3-2,m_{ω,ℚ}(x)=x^2+x+1$.
   • • Let $α=e^{\frac{2 π i}{7}}$. $K=ℚ(α,α^2,…,α^6)$ is the splitting field of $x^7-1$.
       Since char ℚ = 0, $K/ℚ$ is separable and hence Galois.
     • Each $φ∈G$ is determined by $φ(α)$ and $φ(α)≠1$ is a root of $x^7-1$. Since $G=[K:ℚ]=6$ and $φ_k:α↦α^k(k=1,…,6)$ are in $G$, so these are all of $G$.
       $φ_k↦k(k=1,…,6)$ is an isomorphism $G→ℤ_7^×$, so $G≅C_6≅C_2×C_3$.
     • proper subgroups of $C_6$ are $C_2$ and $C_3$.
       3 is a generator of $ℤ_7^×$, so proper subgroups of $ℤ_7^×$ are $⟨3^2⟩=⟨2⟩,⟨3^3⟩=⟨6⟩$, corresponding to proper subfields of $K$ strictly containing ℚ: $K^{φ_2},K^{φ_6}$. $$\small φ_2(q_1α+q_2α^2+q_3α^3+q_4α^4+q_5α^5+q_6α^6)=q_4α+q_1α^2+q_5α^3+q_2α^4+q_6α^5+q_3α^6$$ $$∈K^{φ_2}⟺q_1=q_2=q_4,q_3=q_5=q_6$$ so $K^{φ_2}=ℚ(α+α^2+α^4,α^3+α^5+α^6)$. \begin{array}l α+α^2+α^4+α^3+α^5+α^6=-1\\ (α+α^2+α^4)(α^3+α^5+α^6)=2 \end{array} so $K^{φ_2}=ℚ(α+α^2+α^4)$.

       ---

       $$\small φ_6(q_1α+q_2α^2+q_3α^3+q_4α^4+q_5α^5+q_6α^6)=q_6α+q_5α^2+q_4α^3+q_3α^4+q_2α^5+q_1α^6$$ $$∈K^{φ_6}⟺q_1=q_6,q_2=q_5,q_3=q_4$$ so $K^{φ_6}=ℚ(α+α^6,α^2+α^5,α^3+α^4)$. \begin{array}l (α+α^6)+(α^2+α^5)+(α^3+α^4)=-1 \\(α+α^6)(α^2+α^5)+(α+α^6)(α^3+α^4)+(α^2+α^5)(α^3+α^4)=-2 \\(α+α^6)(α^2+α^5)(α^3+α^4)=1 \end{array} so they are three distinct roots of a cubic polynomial $x^3+x^2-2 x-1$, which is irreducible cubic polynomial on ℚ with discriminant $49∈ℚ^2$, so it splits in $ℚ(α+α^6)$, so $K^{φ_6}=ℚ(α+α^6)$. Error! Click to view log.
     • $m_{α+α^2+α^4,ℚ}(x)=x^2+x+2,m_{α+α^6,ℚ}(x)=x^3+x^2-2 x-1$.

   ---

   Equivalent definitions of the Galois conjugates of $α$:
   • $g(α)$ for $g∈G$
   • the roots of $m_{α,F}$
   Proof: $\prod_{β∈ Orb(α)}(x-β)$ is minimal polynomial of $α$ over $F$.
5. Let $f ∈ F[t]$ be a separable polynomial and let $L$ be any field extension of $F$. Prove that $f$ remains separable when viewed as an element of $L[t]$.

   ---

   $f$ is irreducible and separable on $F[t]$, by 4.10 $∃p,q∈F[t]$ such that $pf+qD(f)=1$.
   Suppose $f$ factors into irreducibles $f=f_1⋯f_n$ on $L[t]$, $pf_1⋯f_n+qD(f_1⋯f_n)=1$.
   If $f$ is inseparable on $L(t)$, wlog $D(f_1)=0$, by product rule $pf_1⋯f_n+qf_1D(f_2⋯f_n)=1$, then $f_1|1$, contradiction.

   ---

   Remark: This proof contains the converse of 4.10: If $∃p,q∈L[t]$ st. $pf+qD(f)=1$ then $f$ is separable on $L[t]$.
6. A finite field extension $K/F$ is said to be separable if $m_{F, α}$ is a separable polynomial for all $α ∈ K$. Let $L$ be an intermediate field. Show that if $K/F$ is separable, then so are $K/L$ and $L/F$.

   ---

   To show $L/F$ is separable, for all $α ∈ L$, since $K/F$ is separable, $m_{F, α}$ is separable.
   To show $K/L$ is separable, for all $α ∈ K$, $m_{L,α}(x)|m_{F,α}(x)$. By assumption, $m_{F,α}(x)$ is separable, so $m_{L,α}(x)$ is separable.
7. Let $L=ℚ(2^{\frac{1}{4}}, 3^{\frac{1}{4}})$. Compute the degree $[L: ℚ]$.

   ---

   It suffices to prove $x^4-2=(x-i2^{\frac14})(x+i2^{\frac14})(x-2^{\frac14})(x+2^{\frac14})$ is irreducible on $ℚ(3^{\frac14})$.
   $ℚ(3^{\frac14})/ℚ$ is not Galois extension, but $ℚ(3^{\frac14},i)/ℚ$ is Galois extension.
   If $x^4-2$ has a linear factor, then $2^{\frac14}∈ℚ(3^{\frac14}),\sqrt2∈ℚ(3^{\frac14})$, contradiction
   If $x^4-2$ has a quadratic factor, then $\sqrt2∈ℚ(3^{\frac14})$, contradiction $$\gal(ℚ(3^{\frac{1}{4}},i)/ℚ)=D_8$$ Error! Click to view log. This can be seen as three $V_4$ sticking together:
   Two subgroups $V_4$: $$\gal(ℚ(3^{\frac{1}{4}},i)/ℚ(\sqrt3))=V_4$$ $$\gal(ℚ(3^{\frac{1}{4}},i)/ℚ(\sqrt3i))=V_4$$ This is because $⟨σ,τ|σ^4=τ^2=1,στ=τσ^{-1}⟩=D_8$ has two subgroups $⟨σ^2,τ|σ^4=τ^2=1,στ=τσ^{-1}⟩=V_4$ and $⟨σ^2,τσ|σ^4=τ^2=1,στ=τσ^{-1}⟩=V_4$.
   One quotient group $V_4$: $$\gal(ℚ(\sqrt3,i)/ℚ)=V_4$$ This is because $⟨σ,τ|σ^4=τ^2=1,στ=τσ^{-1}⟩=D_8$ quotiented by $σ^2=1$ will be $⟨σ,τ|σ^2=τ^2=1,στ=τσ⟩=V_4$.