Galois Theory

$\DeclareMathOperator{\gal}{Gal}$

Let $K/F$ be a splitting field of a separable irreducible polynomial $f∈F[t]$.
Show that $\gal(K/F)$ acts transitively on the set of roots $V(f)$ of $f$ in $K$.
$∀α,β∈V(f)$, by 4.12 ∃ $F$-linear automorphism $ϕ\colon F(α)→F(β),α↦β$
$K$ is splitting field of $f$ over $F(α)$.
$K$ is splitting field of $ϕ(f)=f$ over $F(β)$.
By 4.13 extend $ϕ$ to an automorphism $\barϕ\colon K→K$, then $\barϕ(α)=ϕ(α)=β$ and $\barϕ$ fixes $F$ as $ϕ$ does \begin{CD} K@>\barϕ>>K \\ @AAA@AAA\\ F(α)@>ϕ>>F(β)\\ @AAA@AAA\\ F@>\text{id}>>F \end{CD} Another proof: $∀α∈V(f)$ we need to prove orbit of $α$ is all of $V(f)$. The orbit polynomial $g(x)=\prod_{β∈G⋅α}(x-β)∈K^G[x]=F[x]$ annihilates $α$, so $f|g$, so$$\#V(f)=\deg f≤\deg g=\#(G⋅α)$$but $G⋅α⊆V(f)$, so $G⋅α=V(f)$.
- Let $σ ∈ S_n$ be any $n$-cycle and let $τ ∈ S_n$ be any transposition. Prove that $S_n=⟨σ, τ⟩$ whenever $n$ is a prime number. Is this true if $n$ is composite?
- Is it true that $S_5$ is generated by any 5-cycle together with any double transposition?
- Let\[(a \ b … c \ d) \text{ and } (e \ f)\] be any $p$-cycle and any transposition. We can rename $e=1$ such that the permutations are now: \[(a' \ b' … 1 … c' \ d')=(1 … c' \ d' \ a' \ b' …) \text{ and } (1 \ f')\] Now, for some $1 ≤ k ≤ p-1$: \[(1 … c' \ d' \ a' \ b' …)^k=(1 \ f' …)\] and as $p$ is supposed to be prime, $(1 \ f' …)$ is a $p$-cycle. We can again rename the elements of $\{1, …, p\}$ such that $f' = 2$ and the rest accordingly to get: \[(1 \ 2 … p-1 \ p) \text{ and } (1 \ 2)\]According to theorem 3 they generate $S_n$.
  
  Counterexample when $n$ is composite: $n=4,σ=(1234),τ=(13)$. Consider the action on 4 vertices of a square where $σ$ is rotation and $τ$ is a reflection, then $D_4=⟨σ,τ⟩$ is a proper subgroup of $S_4$, for example $(12)∈S_4$ cannot be generated by $σ,τ$.
- Any 5-cycle and any double transposition are even permutations so they cannot generate whole $S_5$.
  For example $σ=(12345)$ and $τ=(14)(23)$. Consider the action on 5 vertices of a pentagon where $σ$ is rotation and $τ$ is a reflection, then $D_5=⟨σ,τ⟩$ is a proper subgroup of $S_5$, for example $(12)∈S_5$ cannot be generated by $σ,τ$.
Let $f=t^3-s_1 t^2+s_2 t-s_3 ∈ F[t]$ have roots $α, β, γ$. Write $$ σ_i≔α^i+β^i+γ^i \text{ for } i ≥ 0 . $$
- Show that $σ_0=3, σ_1=s_1$ and $σ_2=s_1^2-2 s_2$.
- Show that $σ_r=s_1 σ_{r-1}-s_2 σ_{r-2}+s_3 σ_{r-3}$ for all $r ≥ 3$.
- Let $V=\begin{pmatrix}1 & 1 & 1 \\ α & β & γ \\ α^2 & β^2 & γ^2\end{pmatrix}$ be the Van der Monde matrix.
  Show that $V ⋅ V^T=\begin{pmatrix}σ_0 & σ_1 & σ_2 \\ σ_1 & σ_2 & σ_3 \\ σ_2 & σ_3 & σ_4\end{pmatrix}$.
- Let $δ=(α-β)(β-γ)(γ-α)$. Deduce that $$ Δ=δ^2=-4 s_1^3 s_3+s_1^2 s_2^2+18 s_1 s_2 s_3-4 s_2^3-27 s_3^2 . $$
- Equating coefficients in $t^3-s_1 t^2+s_2 t-s_3=(t-α)(t-β)(t-γ)$ gives \[ s_1=α+β+γ, s_2=α β+β γ+γ α, s_3=α β γ . \] Then $σ_0=1+1+1=3$, whereas $σ_1=α+β+γ=s_1$, whereas \[ s_1^2-2 s_2=(α+β+γ)^2-2(α β+β γ+γ α)=α^2+β^2+γ^2=σ_2 . \]
- For any $x ∈\{α, β, γ\}$ we have $x^3=s_1 x^2-s_2 x+s_3$ and hence $x^r=s_1 x^{r-1}-$ $s_2 x^{r-2}+s_3 x^{r-3}$ and all $r ≥ 3$. Summing over all $x ∈\{α, β, γ\}$ shows that $σ_r=$ $s_1 σ_{r-1}-s_2 σ_{r-2}+s_3 σ_{r-3}$.
- $V ⋅ V^T=\left(\begin{array}{ccc}1 & 1 & 1 \\ α & β & γ \\ α^2 & β^2 & γ^2\end{array}\right) ⋅\left(\begin{array}{ccc}1 & α & α^2 \\ 1 & β & β^2 \\ 1 & γ & γ^2\end{array}\right)=\left(\begin{array}{ccc}σ_0 & σ_1 & σ_2 \\ σ_1 & σ_2 & σ_3 \\ σ_2 & σ_3 & σ_4\end{array}\right)$.
- Using (b) we have $σ_3=s_1 σ_2-s_2 σ_1+s_3 σ_0=s_1(s_1^2-2 s_2)-s_2 s_1+3 s_3=s_1^3-3 s_1 s_2+3 s_3$. It turns out we do not need to express $σ_4$ in terms of $s_1, s_2, s_3$. The trick is to perform the row operation \[ R_3:=R_3-s_1 R_2+s_2 R_1 \] to the matrix appearing in part (c): \[ Δ=\left|\left(\begin{array}{ccc} σ_0 & σ_1 & σ_2 \\ σ_1 & σ_2 & σ_3 \\ σ_2 & σ_3 & σ_4 \end{array}\right)\right|=\left|\left(\begin{array}{ccc} σ_0 & σ_1 & σ_2 \\ σ_1 & σ_2 & σ_3 \\ σ_2-s_1 σ_1+s_2 σ_0 & σ_3-s_1 σ_2+s_2 σ_1 & σ_4-s_1 σ_3+s_2 σ_2 \end{array}\right)\right| . \] Using (b) again, we compute Newton-Girard formulas: \begin{aligned} & σ_2-s_1 σ_1+s_2 σ_0=s_1^2-2 s_2-s_1^2+3 s_2=s_2 \\ & σ_3-s_1 σ_2+s_2 σ_1=s_3 σ_0=3 s_3, \text { and } \\ & σ_4-s_1 σ_3+s_2 σ_2=s_3 s_1 \end{aligned} Expanding down the first column, we get \begin{aligned} Δ & =\left|\left(\begin{array}{ccc} 3 & s_1 & s_1^2-2 s_2 \\ s_1 & s_1^2-2 s_2 & s_1^3-3 s_1 s_2+3 s_3 \\ s_2 & 3 s_3 & s_3 s_1 \end{array}\right)\right| \\ = & 3 s_3\left(s_1^3-2 s_1 s_2-3\left(s_1^3-3 s_1 s_2+3 s_3\right)\right)-s_1\left(s_1^2 s_3-3 s_3\left(s_1^2-2 s_2\right)\right) \\ & +s_2\left(s_1^4-3 s_1^2 s_2+3 s_1 s_3-s_1^4+4 s_1^2 s_2-4 s_2^2\right) \\ = & 3 s_3\left(-2 s_1^3+7 s_1 s_2-9 s_3\right)-s_1 s_3\left(-2 s_1^2+6 s_2\right)+s_2\left(3 s_1 s_3-4 s_2^2+s_1^2 s_2\right) \\ = & -6 s_1^3 s_3+21 s_1 s_2 s_3-27 s_3^2+2 s_1^3 s_3-6 s_1 s_2 s_3+3 s_1 s_2 s_3-4 s_2^3+s_1^2 s_2^2 \\ = & -4 s_1^3 s_3+s_1^2 s_2^2+18 s_1 s_2 s_3-4 s_2^3-27 s_3^2 . \end{aligned}
Galois group of a biquadratic quartic
Prove that $f≔t^4-4 t^2+2$ is irreducible over ℚ. Let $α ∈ ℂ$ be a root of $f$. Show that $ℚ(α)$ is the splitting field of $f$. Prove that $\gal(ℚ(α)/ℚ)$ is cyclic of order 4.
By Eisenstein with $p=2$, $f$ is irreducible over ℚ. $$t^4-4t^2 + 2 = 0 ⇔ t^2 =2 ± \sqrt{2}$$ so the roots of $f$ are $±α,±β$ where $α=\sqrt{2+\sqrt2},β=\sqrt{2-\sqrt2}$ $$αβ=\sqrt2=α^2-2⇒β∈ℚ(α)$$ so ℚ(α) is the splitting field of $f$ over ℚ, so $[ℚ(α):ℚ] = 4$, so $\gal(ℚ(α)/ℚ)=C_2^2$ or $C_4$.
Consider $σ∈\gal(ℚ(α)/ℚ);σ(α)=β$ \begin{array}l σ^2(α)=σ(β)=σ\left(\frac{α^2-2}α\right)=\frac{β^2-2}β=\frac{-\sqrt2}β=-α\\σ^3(α)=σ(-α)=-σ(α)=-β\\σ^4(α)=σ(-β)=-σ(β)=α\end{array} so σ has order 4, but no element of $C_2^2$ has order $4$. So $\gal(f) ≅ C_4$.
Remark: $α=2\cos(π/8)$, so $ℚ(α)$ is a subextension of cyclotomic extension $ℚ(ζ_8)$ with Galois group $ℤ^×_8≅C_4×C_2$.
Let $K/F$ be a finite extension. Recall that $K/F$ is said to be separable if $m_{F, α}$ is a separable polynomial for all $α ∈ K$. Prove that $K/F$ is Galois if and only if it is normal and separable.

(⇒) Given a finite Galois extension $K/F$. By Theorem 6.6, $K/F$ is normal.

By Corollary 6.7a, $K/F$ is separable.

(⇐) Given a finite, normal, separable extension $K/F$. Let $α_1,α_2,…,α_n$ be a $F$-basis for $K$.

Since $m_{F,α_i}$ has a root $α_i$ in $K$, by normality, $m_{F,α_i}$ splits over $K$.

Irreducible factors of $f≔\prod_{i=1}^nm_{F,α_i}$ are $m_{F,α_i}$ which are separable since $K/F$ is separable. Thus, $f$ is separable.
Since $α_1,α_2,…,α_n$ generate $K$, $K$ is the splitting field of $f$, then $K/F$ is Galois.
Suppose that $\operatorname{char} F ≠ 2$. Let $f=t^4+p x^2+q t+r$ have roots $α_1, α_2, α_3, α_4$ and let $$ v_1≔α_1 α_2+α_3 α_4, v_2≔α_1 α_3+α_2 α_4, v_3≔α_1 α_4+α_2 α_3 $$ Show that $(t-v_1)(t-v_2)(t-v_3)=t^3-p t^2-4 r t+(4 p r-q^2)$.
By Vieta’s formulas \begin{gather*} 0=α_1+α_2+α_3+α_4 \\p=α_1 α_2+α_1 α_3+α_1 α_4+α_2 α_3+α_2 α_4+α_3 α_4 \\-q=α_1 α_2α_3+α_1α_2 α_4+α_1α_2 α_4+α_2α_3 α_4 \\r=α_1 α_2α_3α_4 \end{gather*} we calculate symmetric polynomials of $v_1,v_2,v_3$ \begin{gather*} v_1 + v_2 + v_3 = p,\\ \begin{split} v_1 v_2 + v_1 v_3 + v_2 v_3 &= (α_1 + α_2 + α_3 + α_4) (α_1 α_2 α_3 + ⋯ + α_2 α_3 α_4) - 4α_1 α_2 α_3 α_4 \\ &= 0(-q) - 4r, \end{split} \\ \begin{split} v_1 v_2 v_3 &= (α_1 α_2 α_3 + ⋯ )^2 + α_1 α_2 α_3 α_4 \bigl\{(α_1 + ⋯ )^2 - 4( α_1 α_2 + ⋯) \bigr\} \\ &= (-q)^2 + r(0^2 - 4p). \end{split} \end{gather*} Hence $v_1,v_2,v_3$ are the roots of $t^3-p t^2-4 r t+(4 p r-q^2)$.
Another method: Let $v=α_1α_2+α_3α_4$, $w=α_1α_2-α_3α_4$, $τ=α_1+α_2$.
Since $α_1+α_2+α_3+α_4=0$ then $-τ=α_3+α_4$.
Also, $\frac{v+w}2=α_1α_2,\frac{v-w}2=α_3α_4$. \begin{align*} t^4+pt^2+qt+r&=(t-α_1)(t-α_2)(t-α_3)(t-α_4)\\&=(t^2-(α_1+α_2)t+α_1α_2)(t^2-(α_3+α_4)t+α_3α_4)\\&=(t^2-τ t+\frac{v+w}2)(t^2+τ t+\frac{v-w}2)\\&=t^4+(v-τ^2)t^2+τ wt+\frac{v^2-w^2}4 \end{align*} Equating coefficients,\[p=v-τ^2,q=τ w,4r=v^2-w^2\] Expressing $w^2,τ^2$ in $v,r,p$ $$4r=v^2-w^2⇒w^2=v^2-4r$$ $$p=v-τ^2⇒τ^2=v-p$$ then$$q^2=τ^2w^2=(v-p)(v^2-4r)=v^3-pv^2-4rv+4rp$$ Hence $v$ is a root of $t^3-pt^2-4rt+(4pr-q^2)$.
Let $K/L$ and $L/F$ be finite separable extensions.
- Suppose that $L/F$ is Galois and that $α ∈ K$ is a root of an irreducible separable polynomial $f ∈ L[t]$. By considering the $\gal(L/F)$-orbit of $f$, show that $α$ is also separable over $F$.
- Show that $K/F$ is also separable.
- Let $f=m_{L,α}(x)∈ L[x]$ and $f_1,…,f_n$ be the orbit of $f$ by $\gal(L/F)$, they are all separable, remains to show they have different roots from each other.
  Since $f$ is a minimal polynomial, $f$ is irreducible, then each $f_i$ is irreducible.
  Suppose $f_i$ and $f_j$ share a root $β$.
  $(f_i,f_j)$ is a proper ideal of $L[x]$ since $h(β)=0∀h∈(f_i,f_j)$
  $(f_i)$ is maximal$⇒(f_i)=(f_i,f_j)=(f_j)⇒f_i=f_j$.
  Let $g=\prod_{i=1}^nf_i$. Then $g$ has no repeated roots.
  The Galois group $G=\gal(L/F)$ fixes $g⇒g∈ L^G[x]=F[x]$.
  Since $g(α)=0⇒m_{F,α}|g$
  Since $g$ has no repeated roots, $m_{F,α}$ has no repeated roots, then $α$ is separable.
- For any $α∈K$, need to show $m_{F,α}$ is separable.
  Since $K/L$ is separable, $m_{L,α}$ is separable.
  Need to form a Galois closure $L'/F$ of $L/F$ in order to apply (i).
  Let $z_1,…,z_n$ be a $F$-basis for $L$, and $L'$ be splitting field of $\prod_{i=1}^nm_{F,z_i}$ then $L'/F$ is Galois.
  (We can't apply (i) directly since $L'⊂K$ is not true.)
  Let $β$ be a root of $m_{L,α}$ in $L'$, then $m_{L,α}=m_{L,β}$.
  $m_{L',β}|m_{L,β}$ and $m_{L,β}$ is separable, so $m_{L',β}$ is separable, so $L'(β)/L'$ and $L'/F$ are separable and $L'/F$ is Galois, so $L'(β)/F$ is separable by (i), then $m_{F,α}=m_{F,β}$ is separable.