Galois Theory

Splitting field of a separable polynomial in $K[x]$.
1. Let $K / F$ be a Galois extension and let $F ⊆ L ⊆ K$ be an intermediate field. Then $$L=K^{\operatorname{Gal}(K / L)}$$
2. Let $K / F$ be a Galois extension and let $H ⩽ \operatorname{Gal}(K / F)$. Then $\operatorname{Gal}(K / K^H)=H$.
3. Let $K / F$ be a Galois extension with $G=\operatorname{Gal}(K / F)$ and let $L$ be an intermediate subfield with $H:=\operatorname{Gal}(K / L)$.
  - $H$ is normal in $G$ if and only if $L$ is Galois over $F$.
  - If $H$ is normal in $G$, the restriction map $\operatorname{Gal}(K / F) → \operatorname{Gal}(L / F)$ induces a group isomorphism $$G / H ≅ \operatorname{Gal}(L / F)$$
  - $[L: F]=[G: H]$ and $[K: L]=|H|$

For example $ℚ ≤ ℚ(\sqrt{2}) ≤ ℚ(\sqrt[4]2)$, $m_{\sqrt[4]2}(x)=x^4-2$ does not split in $ℚ(\sqrt[4]2)$.

Remark: This correpond to the fact in group theory that $\operatorname{Gal}(ℚ(\sqrt[4]2,i):ℚ(\sqrt[4]2))≅C_2$ is not a normal subgroup of $\operatorname{Gal}(ℚ(\sqrt[4]2,i):ℚ)=D_4$.
Error! Click to view log.

Subgroups of $D_4$ seen as symmetry group of a square
A normal subgroup $C_2$	Generated by rotation 180°
Four conjugate subgroups $C_2$	Each generated by a reflection
A normal subgroup $C_4$	Rotations
Two normal subgroups $C_2^2$	One is the stabilizer of a diagonal: Error! Click to view log.
Two normal subgroups $C_2^2$	One is stabilizer of a midline: Error! Click to view log.

$L_1=\operatorname{split}(f_1), L_2=\operatorname{split}(f_2)$. Then $F=\operatorname{split}(f_1 f_2)$.
\begin{aligned} ϕ: G &→ H \\ g&↦ (g|_{L_1},g|_{L_2}) \end{aligned}

$ϕ$ is well-defined:

$L_1/K,L_2/K$ are Galois $⇒g|_{L_1}, g|_{L_2}$ send $L_1 → L_1, L_2 → L_2$ respectively.

$ϕ$ is injective:

$g ∈ \ker(ϕ)$ $⇒ g$ fixes $L_1, L_2$ $⇒ g$ fixes $F$ $⇒ g=\text{id}$
Next, assume $L_1 ∩ L_2=K$, want to show $ϕ:G→H$ is isomorphism.
$G, H$ finite, so enough to show $|G|=|H|$. \begin{aligned} & |H|=\left|Γ\left(L_1: K\right)\right|\left|Γ\left(L_2: K\right)\right| \\ & \frac{|G|}{\left|Γ\left(L_2: K\right)\right|}=\frac{|Γ(F: K)|}{\left|Γ\left(L_2: K\right)\right|}=\left|Γ\left(F: L_2\right)\right| \end{aligned} so enough to show $|Γ\left(F: L_2\right)|=|Γ\left(L_1: K\right)|$
Consider $S:=Γ(F: L_2)$ and the restriction \begin{aligned} i: Γ(F: L_2)&→ Γ(L_1: K) \\ g&↦ g|_{L_1} \end{aligned}

$i$ is well-defined:

$F/L_1$ is Galois $⇒g$ send $L_1 → L_1$.

$i$ is injective:

$g ∈ \ker(i)$ $⇒ g$ fixes $L_1, L_2$ $⇒ g$ fixes $F$ $⇒ g=\text{id}$

$i$ is surjective:

Consider $i(S) ≤ Γ(L_1: K)$, need to show $i(S) = Γ(L_1: K)$.
By Fundamental Theorem of Galois Theory, it suffices to prove $L_1^{i(S)}=K$. Now $\begin{aligned}[t] L_1^{i(S)}&=\{x ∈ L_1: g(x)=x \; ∀ g ∈ Γ(F: L_2)\} \\ & =L_1 ∩ F^{Γ(F: L_2)} \\ & =L_1 ∩ L_2\\ & = K \end{aligned}$
Let $K=ℚ(ξ),L_1=ℚ(ξ,\root3\of2),L_2=ℚ(ξ,\sqrt2),F=ℚ(ξ,\root3\of2,\sqrt2)$. Then $F$ is the splitting field of $f$ over $K$.
$L_1∩L_2=K$ since $[L_1∩L_2:K]$ divides $[L_1:K]=3$ and $[L_1:K]=2$ and $\gcd(3,2)=1$.
$L_1/K$ is Kummer extension$⇒Γ(L_1:K)≅C_3$
By (e) $Γ(F:K)≅Γ(L_1:K)×Γ(L_2:K)≅C_3×C_2$.
Generators: $σ_1,σ_2$ \begin{array}{ll} σ_1:&\sqrt2↦\sqrt2,&\root3\of2↦ξ\root3\of2\\ σ_2:&\sqrt2↦-\sqrt2,&\root3\of2↦\root3\of2 \end{array}

For any field $F$, let us denote by $F(t)$ the field of rational functions in the variable $t$ over the field $F$. Let $𝔽_{125}$ be the finite field of order 125, let $K=𝔽_{125}(t)$, and let $f=x^{124}-t∈K[x]$. In this exercise, you can assume without proof that $f$ is irreducible in $K[x]$.

Define what is meant by the splitting field of a polynomial.
Let $E$ be the splitting field of $f$. Show that $E/K$ is separable.
Prove that the multiplicative group of a finite field is cyclic.
Find $g ∈ 𝔽_5[y]$ such that $𝔽_{125} ≅ 𝔽_5[y] /(g(y))$.
Prove that $E ≅ K[x] /(f)$.
Compute $Γ(E: K)$ and compute the number of subfields of $E$ containing $K$.

splitting field of a polynomial $f$ over $F$ is $K$ iff $f$ splits on $K$ and its roots generate $K$
\begin{aligned} & f'=124 x^{123} \\ & \gcd(f, f')=\gcd(x^{124}-t, 124 x^{123})=1 \\ & ⇒ f \text{ separable.} \end{aligned} Just as an example, if we had used $f=x^{125}-t$, then $f'=0$, so $f$ is inseparable.
Claim
Let $A$ be a non-cyclic finite ablian group. Then $A$ has two distinct cyclic subgroups of the same order.

Proof
Let $a$ have maximal order. Take $b∈A∖⟨a⟩$. I claim that $o(b)$ must divide $o(a)$. Indeed, in an abelian group, the orders of elements are lcm-closed; that is, if you have an element of order $m$ and an element of order $n$, then there must exist an element of order $\mathrm{lcm}(m,n)$. Since $n$ is the largest possible order, we must have $\mathrm{lcm}(m,n)≤n$, and therefore $\mathrm{lcm}(m,n)=n$, proving that $m$ divides $n$, as claimed.
Since $m$ divides $n$, let $c=a^{n/m}$, then $⟨c⟩⊂⟨a⟩$, so $b∉⟨c⟩$, so $⟨b⟩≠⟨c⟩$ but both has order $m$.
Suppose $𝔽^×$ is non-cyclic, then it has two distinct cyclic subgroups of the same order $k$, then $x^k=1$ has more than $k$ roots in $𝔽^×$, contradiction. Another proof
First we find a cubic irreducible polynomial on $𝔽_5$. Let $g(x)=x^3+x-1$. Since $𝔽_5[y]/(g(y))$ has degree 3 of $𝔽_5$, it has cardinality 125, so $𝔽_5[y]/(g(y))≅𝔽_{125}$.
Set $α$ equal the class of $x$ modulo $f$ in the field $K(α)=K[x]/(f)$. The field $K(α)$ is a subfield of $E$ as it contains $α$, a root of $f$. Observe that for any $u ∈ 𝔽_{125}^×$, we have that \[ (uα)^{124}-t=α^{124}-t=0 \] The set of all roots of $f$ is $\{uα\}_{u ∈ 𝔽_{125}^×}⊆ K(α)$, which forces $E=K(α)$.
Let $α=[x]$ be a root of $f$ in $E=K[x]/(f)$. An automorphism of $K(α)$ is determined by its action on $α$, and it must send $α$ to another root $uα$ of $f$. Thus, for $u ∈ 𝔽_{125}^×$ $$Φ_u:E → E, [g(x)] ↦[g(ux)]$$ is an automorphism fixing $𝔽_{125}$ $$⇒ 𝔽_{125}^× ⩽ Γ(E: K)$$ both has size 124 (as $Γ(E:K)=Γ(K(α):K),|Γ(K(α):K)|=[K(α):K]=124$), so they are equal. \[ ⇒ Γ(E:K)=𝔽_{125}^×=C_{124} \] cyclic, one subgroup for each divisor of $124=2^2 ⋅ 31⇒ 6$ subgroups.

Let $n$ be a positive integer and let $G$ be a subgroup of the symmetric group $S_n$, acting on $\{1, …, n\}$ on the right. Let $ℚ$ be the field of rational numbers. Let $L=ℚ(x_1, … x_n)$ be the field of rational functions in the variables $x_1, … x_n$. Consider the action of $G$ on $L$ by relabelling the variables, i.e. for any $g ∈ G$ and any $f(x_1, … x_n) ∈ L, f^g(x_1,…,x_n)=f(x_{1 g}, …, x_{n g})$. Write $L^G=\{f ∈ L: f^g=f\text{ for all }g ∈ G\}$.

- Give the definition of the Galois group of a separable polynomial over a base field $F$.
- Show that a finite extension of a finite field is separable.
- Let $F$ be a field of characteristic zero and $F'$ be a finite extension field of $F$. Show that if $F' / F$ satisfies $\left|\operatorname{Aut}_F(F')\right|=[F': F]$, then $F'$ is Galois over $F$.
- Show that $L$ is a Galois extension of $L^G$. In addition, show that $G ≅ Γ(L:L^G)$ (you may use the primitive element theorem, provided that you state it correctly.) Conclude that any finite group appears as the Galois group of some field extension (Hint: Cayley's Theorem might be useful here).
- Let $M$ be a subfield of $ℂ$ such that $M / ℚ$ is a finite Galois extension that if $[M: ℚ]$ is an odd number, then $M ⊆ ℝ$.
- Let $ε$ be the primitive ninth root of unity. Find a subfield $E$ of $ℚ(ε)$ which is over $ℚ$ and such that $Γ(E: ℚ)=C_3$. (You may use standard facts about cyclic extensions provided that they are clearly stated.)

- Galois group of a separable polynomial $f$ over a field 𝔽 is $\operatorname{Gal}(\operatorname{split}(f):𝔽)$
- Let $L/K$ be a finite extension of finite fields. Then there exist a prime $p$ and a positive integer $n$ such that $L$ is the splitting field of $x^{p^n}-x$. Since $\gcd(x^{p^n}-x,-1)=1$, then $L/𝔽_p$ is separable. Since $K ⊃ 𝔽_p$, then also $L / K$ is separable.
- $G=Γ(F': F)$ acts faithfully on $F'$, by Artin's lemma $F'/(F')^G$ is Galois and ${|Γ(F':(F')^G)|}=[F':(F')^G]$.
  By 3.12 $Γ(F':(F')^G)=Γ(F':F)$. Thus $[F':(F')^G]={|Γ(F':F)|}=[F':F]$ by assumption. \begin{aligned}{} [F':F]&=[F':(F')^G][(F')^G:F]\\ &=[F':F][(F')^G:F]\\ &⇒(F')^G=F⇒F'/F\text{ is Galois} \end{aligned}
- Let $G$ be a given group, and let $n=|G|$.
  There is a field extension $K/E$ with $\mathrm{Aut}_E(K)≅ S_n$ (e.g., $E=ℚ(s_1,…,s_n)$, $K=ℚ(x_1,…,x_n)$, where $s_1,…,s_n$ are the symmetric polynomials in $x_1,… x_n$).
  By Cayley's Theorem, we know that $G$ is isomorphic to a subgroup $H$ of $S_n$, so if we let $L$ be the fixed field of $H$ in $K$, then by the Fundamental Theorem of Galois Theory we know that $K/L$ is Galois and $\mathrm{Aut}_{L}(K) = H≅G$.
- \begin{aligned} M&=\operatorname{Split}(f) \text{ for some } f \\ M(i)&=\operatorname{Split}(f×(x^2+1)) \text{ is Galois over }ℚ \end{aligned} $ℚ ≤ M ≤ M(i)$. Let $ι$ be complex conjugation \[ ι∈ Γ(M(i):ℚ) \] $ι$ restricts to $ι|_M: M → M$.
  Since $M/ℚ$ is Galois, $|Γ(M:ℚ)|=[M:ℚ]$ is odd. If $ι|_M ≠ \text{id}$ it has order 2, contradiction \begin{aligned} & ⇒ ι|_M=\text{id} \\ & ⇒ M ⊆ ℝ . \end{aligned}
- $φ(9)=6$. One of the primitive root modulo 9 is 2 since we have 2,4,-1,-2,-4,1.
  So $C_6$ acts on $ℚ(ε)$ by $ε↦ε^2$.
  So $C_2⩽C_6$ acts on $ℚ(ε)$ by $ε↦ε^{-1}$, want to find $E=ℚ(ε)^{C_2}$.
  Let $E=ℚ(ε)^{C_2}=ℚ(ε+ε^{-1})$ then $E≤ℝ$, so $E⊂ℚ(ε)^{C_2}$.
  $E≠ℚ$ [If $E=ℚ$ then $[ℚ(ε):ℚ]≤2$, a contradiction], so $[E:ℚ]=2$, so $E=ℚ(ε)^{C_2}$.

Appendix

Let $G$ be a finite group acting faithfully on $K$. Then the extension $K/K^G$ is Galois, we have $[K:K^G]=|G|$, and the Galois group of the extension is $G$.

Lemma 9.21.6.

Pick finitely many elements $α _ i ∈ K$, $i = 1, … , n$ such that $σ (α _ i) = α _ i$ for $i = 1, … , n$ implies $σ $ is the neutral element of $G$. Set \[ L = K^ G(\{ σ (α _ i); 1 ≤ i ≤ n, σ ∈ G\} ) ⊂ K \] and observe that the action of $G$ on $K$ induces an action of $G$ on $L$. We will show that $L$ has degree $|G|$ over $K^ G$. This will finish the proof, since if $L ⊂ K$ is proper, then we can add an element $α ∈ K$, $α ∉ L$ to our list of elements $α _1, … , α _ n$ without increasing $L$ which is absurd. This reduces us to the case that $K/K^ G$ is finite which is treated in the next paragraph.

Assume $K/K^G$ is finite. By Lemma 9.19.1 we can find $α ∈ K$ such that $K = K^G(α)$. By the construction in the first paragraph of this proof we see that $α $ has degree at most $|G|$ over $K$. However, the degree cannot be less than $|G|$ as $G$ acts faithfully on $K^G(α) = L$ by construction and the inequality of Lemma 9.15.9.