1. 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)
  2. Let $p$ be a prime, let $F$ be a field of characteristic $p$, and let $f ∈ F[t]$.
  3. For each of the following subfields of ℂ:

    Equivalent definitions of the Galois conjugates of $α$: Proof: $\prod_{β∈ Orb(α)}(x-β)$ is minimal polynomial of $α$ over $F$.
  4. 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 $L[t]$.
  5. 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.
  6. 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 $⟨σ,τ|σ^2=τ^2=1,στ=τσ⟩=V_4$.