$\DeclareMathOperator{\c}{c}\DeclareMathOperator{\gal}{Gal}$
  1. For each $α$ in the list below, find an explicit non-zero $f ∈ ℚ[x]$ of such that $f(α)=0$.
  2. Let $f=x^n+a_{n-1} x^{n-1}+⋯+a_1 x+a_0$ be a polynomial, where $a_0, a_1, ⋯, a_{n-1} ∈ ℤ$.

  3. Let $p$ be a prime, and let $𝔽_p$ the finite field of size $p$. Let $F≔𝔽_p(t)$ be the field of fractions of the polynomial ring $𝔽_p[t]$, and let $K$ be a field extension of $F$ containing a root $α$ of $f≔x^p-t$.
  4. Let $F ⊆ K$ be finite fields. Show that $\operatorname{Stab}_{\gal(K/F)}(z)=\{1\}$ for at least one $z ∈ K$.
    $K$ is a finite field, so $K^×$ is cyclic, let $z$ be a generator. If some $f∈\gal(K/F)$ fixes $z$, then it fixes all of $K$. Therefore $\operatorname{Stab}_{\gal(K/F)}(z)=\{1\}$.
  5. [Primitive element theorem] Let $F$ be a field of characteristic zero and let $E=F(α, β)$ where $α$ and $β$ are algebraic over $F$. Show that $E=F(α+λ β)$ for some $λ ∈ F$. Deduce that for any finite extension $E$ of $F$, there exists an element $θ ∈ E$ such that $E=F(θ)$.
    [Hint: Consider the minimal polynomial of $β$ over $F(α+λ β)$ for a certain $λ ∈ F$.]
    $E=F(α, β)$. $α, β$ algebraic.

    Goal: $∃ λ ∈ 𝔽$ such that $E=F(α + λ β)$.

    To show $β∈F(α + λ β)$, let $g_λ=m_{F(α+λ β),β}$, we want to show $\deg(g_λ)=1$. It is enough to show $g_λ$ has only one root, as $g_λ$ is separable.

    Consider $m_{F, α}(α+λ β-λ x)∈F(α+λ β)[x]$. Since $β$ is a root of $m_{F, α}(α+λ β-λ x)$, $$g_λ∣m_{F, α}(α+λ β-λ x)$$ Suppose for a contradiction: $g_λ$ has a second root $β' ≠ β$, then $∃α'$ such that $m_{F, α}(α')=0$ \begin{aligned} α+λ(β-β') & =α' \\ ⇒ λ & =\frac{α'-α}{β-β'} . \end{aligned} The number of $α'$ is $≤\deg m_{F,α}$ so there are finitely many $λ$ of this form.

    Using ${|F|}=∞$ choose $λ$ not of this form to get a contradiction.

    Use induction to get that $[K: F]<∞ ⇒ K=F(θ)$ for some $θ ∈ K$.
    [Fraleigh page 442] Let $\operatorname{irr}(β, F)$ have distinct zeros $β=β_1, ⋯, β_n$, and let $\operatorname{irr}(γ, F)$ have distinct zeros $γ=γ_1, ⋯, γ_m$ in $\bar{F}$, where all zeros are of multiplicity 1, since $E$ is a separable extension of $F$. Since $F$ is infinite, we can find $a ∈ F$ such that $$ a ≠{β_i-β\overγ-γ_j} $$ for all $i$ and $j$, with $j ≠ 1$. That is, $a(γ-γ_j) ≠ β_i-β$. Letting $α=β+a γ$, we have $α=β+a γ ≠ β_i+a γ_j$, so $$ α-a γ_j ≠ β_i $$ for all $i$ and all $j ≠ 1$. Let $f(x)=\operatorname{irr}(β, F)$, and consider $$ h(x)=f(α-a x) ∈(F(α))[x] . $$ Now $h(γ)=f(β)=0$. However, $h(γ_j) ≠ 0$ for $j ≠ 1$ by construction, since the $β_i$ were the only zeros of $f(x)$. Hence $h(x)$ and $g(x)=\operatorname{irr}(γ, F)$ have a common factor in $(F(α))[x]$, namely $\operatorname{irr}(γ, F(α))$, which must be linear, since $γ$ is the only common zero of $g(x)$ and $h(x)$. Thus $γ ∈ F(α)$, and therefore $β=α-a γ$ is in $F(α)$. Hence $F(β, γ)=F(α)$.
    This doesn't hold for inseparable extension. Example: $𝔽_p(x, y) / 𝔽_p(x^p, y^p)$
    degree of extension$=p^2$
    but every element satisfies a polynomial of degree $p$. So this extension is NOT simple.