# Consequences Galois II. Exercises.

Understanding is a kind of ecstasy, Carl Sagan

Exercise. In each of the following extension K/F, find all Gal(K/F) and determine the intermediate fields.

• K be the splitting fields of x3 -2 over $\mathbb{F_2}$: x3 - 2 = [characteristic 2] x3, the only root is zero, the splitting fields is $\mathbb{F_2}$ itself, G = {identity}.

• K be the splitting field of x3 -2 over $\mathbb{F_3}$: x3 - 2 = [If a polynomial has degree 2 or 3 and has no roots in F, then f is irreducible in F[x]. We can check directly each element {0, 1, 2} ], 03 = 0 ≠ 2, 13 = 1, 23 = 2, therefore 2 is the only root, and x3 -2 = (x -2)3 over $\mathbb{F_3}$. It has a unique multiple root, namely 2, that is in $\mathbb{F_3}$, so the splitting field of f is $\mathbb{F_3}$ itself. K = $\mathbb{F_3}$ and G = {identity}.

• K be the splitting fields of x3 -2 over $\mathbb{F_5}$. Notice that 33 = 27 ⇒ 33 -2 = 25 ≡ 0 in $\mathbb{F_5}$. -2 = 3 is a root of x3 -2 = (x +2)(x2 -2x +4) = (x -3)(x2 -2x -1), and g(x) = x2 -2x -1 has not roots in $\mathbb{F_5}$ (-2 = 3 is not a root of g(x)), it is indeed irreducible over $\mathbb{F_5}$ because it has no root, 32 -2·3 -1 = 2 ≠ 0.

Since K is the splitting fields of x3 -2 over $\mathbb{F_5}$, then it is the splitting field of x2 -2x -1 (irreducible) over $\mathbb{F_5}$ and is $\mathbb{F_5}(α)$ for some root α of x2-2x-1. The roots of x3 -2 are then -2, α, and β = 2 -α, and β = 2 -α (since the sum of the roots of x2 -2x -1 is 2 and so α + β -2 = 0). K is the splitting field of x2 -2x -1 (irreducible) ⇒ [K : $\mathbb{F_5}$] = 2.

Recall. Any irreducible polynomial f(x) ∈ $\mathbb{F_q}$[x] of degree n has the Galois group of f over $\mathbb{F_q}$ cyclic of degree n and the Galois group is generated by the Frobenius map, Φ(x) = xq, Gal($\mathbb{F_q}(α)/\mathbb{F_q}$) = ⟨Φ⟩ ≋ Cn where the field extension $\mathbb{F_q}(α)/\mathbb{F_q}$ is of degree n and $\mathbb{F_q}(α)$ is the splitting field for f over $\mathbb{F_q}$. In particular, any finite extension of finite fields has a cyclic Galois group.

The Frobenius map Φ (Φ(x) = x5) then acts by Φ: -2 → -2 (-25 = -32 =F5 -2), α → β, β → α. So numbering the roots 1 (-2), 2 (α), 3 (beta), we have Φ as the permutation (1)(23), Gal(K/ℚ) ≋ C2 ≋ ℤ/2ℤ and there are no non-trivial intermediate fields.

• K be the splitting fields of x3 -2 over $\mathbb{F_7}$. x3 -2 is irreducible over $\mathbb{F_7}$ because [If a polynomial has degree 2 or 3 and has no roots in F, then f is irreducible in F[x]. ] it has no roots (0 -2 = -2; 1 -2 = -1; 23 -2 = 8 -2 = 6, etc.) and has degree 3.

Let α ∈ K be a root of x3 -2 in $\mathbb{F_7}$. There are two possibilities:

1. F(α) = K ⇒ [K : F] = 3 ⇒ Gal(K/F) ≋ ℤ/3ℤ.
2. F(α) ≠ K ⇒ f(x) = x3 -2 = (x-α)g(x) ⇒ deg(g(x)) = 2, and g(x) will be irreducible over F(α) ⇒ [K : F(α)] = 2 and [K : F] = 3·2 = 6, Gal(K/F)≋ ℤ/6ℤ.

However, K = F(α), the second option is not possible. We know that α is a root of f(x). α3 = 2 ⇒ (2α)3 = 23α3 = 8α3 [8≡1 in $\mathbb{F_7}$] = α3 = 2 ⇒ 2α is a root of x3 -2. Similarly, (4α)3 = 64α3 = α3 = 2 ⇒ 4α is a root of x3 -2, too. The roots of x3 -2 in F(α) are α, 2α, and 4α ⇒ K = F(α).

Futhermore, since f is irreducible, G acts transitively on the roots and its roots lie in $\mathbb{F_{7^3}}$. If α is a root of x3 - 2 in $\mathbb{F_{7^3}}$ (α3 = 2), then the roots are generated by the Frobenius map, Φ(x) = x7, and are Φ0(α) = $α^{7^0}=α$, Φ1(α) = $α^{7^1}=α^7=α^{3^2}α=2^2α=4α$, Φ2(α) = $α^{7^2}=α^{49}=α^{3^{16}}α=2^{16}α=65536α=2α$, Φ3(α) = $α^{7^3}=α^{7^2}α^{7^1}=4α2α=α$, the Frobenius map takes us back to α, and thus acts by cyclically permuting the roots Φ: α → 4α → 2α → α, α ∈ $\mathbb{F_{7^3}}$ is a primitive element for the finite extension $\mathbb{F_{7^3}}/\mathbb{F_3}, \mathbb{F_{7^3}}=\mathbb{F_7}(α)$, Gal($\mathbb{F_7}(α)/\mathbb{F_7}$) = ⟨Φ⟩ ≋ C3. So numbering the roots 1, 2, and 3 whe have Φ as the permutation (123) and there are no non-trivial intermediate groups.

C3 is the unique group of order 3. It is both Abelian and cyclic.

Exercise. Let K/F be a Galois extension such that Gal(K/F) ≋ ℤ/2ℤ x ℤ/12ℤ. Find the number of intermediate fields L such that (a)[L : F] = 4, (b)[L : F] = 9, (c) Gal(K/L) ≋ ℤ/4ℤ.

Solution.

(a) [L:F]

[K:F] = because K/F is Galois |Gal(K/F)| = 24 ⇒ [K : L] = 6.

By the Main Theorem, any intermediate field L such that [K : L] = 6 must be the fixed field of a subgroup H of Gal(K/F) such that |H| = 6. Therefore, #number of intermediate fields L such that [L : F] = 4 = #number of intermediate fields L such that [K : L] = 6 = #the number of subgroups H ≤ Gal(K/F) such that |H| = 6.

|H| = 6 ⇒ H ≋ S3 or H ≋ ℤ/6ℤ, but H ≤ ℤ/2ℤ x ℤ/12ℤ (Abelian) ⇒ H ≋ ℤ/6ℤ and cyclic.

Recall.

1. If G is a finite cyclic group of order n, then G has a subgroup of order m if and only if m | n.
2. G = G1xG2, G1 and G2 Abelian, a ∈ G1, b ∈ G2, (a, b) ∈ G, ord(a, b) = lcm(ord(a), ord(b)).

What are the order 6 elements?

• 6 = lcm(1, 6): {(0, 0), (0, 2), (0, 4),(0, 6), (0, 8), (0, 10)}
• 6 = lcm(2, 6) → (1, 2): {(0, 0), (1, 2), (0, 4), (1, 6), (0, 8), (1, 10)}
• 6 = lcm(2, 3) → (1, 4): {(0, 0), (1, 4), (0, 8), (1, 0), (0, 4), (1, 8)}

#the number of subgroups H ≤ Gal(K/F) such that |H| = 6, H (ℤ/ℤ6) is cyclic = #the number of order 6 elements of ℤ/2ℤ x ℤ/12ℤ = 3. (a) These 3 subgroups are all the subgroups of order 6.

(b) Answer: 0, it is not possible.

(c) #number of intermediate fields such that Gal(K/L) ≋ ℤ/4ℤ = #number of subgroups of ℤ/2ℤ x ℤ/12ℤ that are isomorphic to ℤ/4ℤ = #the number of order 4 elements of ℤ/2ℤ x ℤ/12ℤ = 2, namely, ⟨(0, 3)⟩ = {(0, 3), (0, 6), (0, 9), (0, 0)} -2 = lcm(1, 4)- and ⟨(1, 3)⟩ = {(1, 3), (0, 6), (1, 9), (0, 0)} -2 = lcm(2, 4)-

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