When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes
Theorem: Let f(x) be the general nth degree polynomial in F[x] of deg(f) = n ≤ 4, then f is solvable (by radicals). Mod p Irreducible Test. Let p be a prime and suppose that f(x)∈ ℤ[x] with deg f(x) ≥ 1. Let f′(x) be the polynomial in ℤ_{p}[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f′(x) is irreducible over ℤ_{p} and deg f(x)= deg f′(x), then f(x) is irreducible over ℚ.
Resolvent cubic is a cubic polynomial defined from a monic polynomial of degree four, g(x) = (x -β_{1})(x -β_{2})(x -β_{3}) ∈ K[x]. We define β_{1} = α_{1}α_{2} + α_{3}α_{4}, β_{2} = α_{1}α_{3} + α_{2}α_{4}, β_{3} = α_{1}α_{4} + α_{2}α_{3}, β_{i}∈ K.
D square ∈ F | D square ∉ F | |
---|---|---|
g is reducible over F | D_{2} | G = D_{4} or G = C_{4} |
g is irreducible over F | G = A_{4} | G = S_{4} |
Corollary. If g splits completely in F ↭ G = D_{2}. If g has exactly one root in F ↭ G = D_{4} or C_{4}. g is irreducible ↭ G = S_{4} or A_{4}
Disc(x^{4} +cx +d) = 256d^{3} -27c^{4}. Resolvent cubic of x^{4} +cx + d = x^{3} -4dx -c^{2}
K = $ℚ(\sqrt[4]{2}, \sqrt[4]{2}i) = ℚ(\sqrt[4]{2}, i)$, and this can be easily seen because i = $\frac{i\sqrt[4]{2}}{\sqrt[4]{2}}$ so $ℚ(\sqrt[4]{2}, i) ⊆ ℚ(\sqrt[4]{2}, \sqrt[4]{2}i)$, but obviously it is also the case that $\sqrt[4]{2}i = \sqrt[4]{2}·i$ so $ℚ(\sqrt[4]{2}, \sqrt[4]{2}i) ⊆ ℚ(\sqrt[4]{2}, i)$.
$β_1 = 2i\sqrt{2}, β_2=0, β_3=-2i\sqrt{2}, g = (x -2i\sqrt{2})(x + 2i\sqrt{2})x = x^3+8x$, g is reducible, and has exactly one root in ℚ ⇒ Gal(f) = D_{2} (it is not possible because it is not completely reducible -Corollary-) D_{4} or C_{4}. To discriminate between these two choices, further analysis is required.
Therefore, [K: ℚ] = [K : ℚ(i, ∜2)]·[ℚ(∜2) : ℚ] = 8 ⇒ Gal(f) = D_{4}.
The dihedral group D_{4} is the symmetry group of the square. D_{4} = ⟨r, s : r^{4} = s^{2} = id, rs = sr^{-1}⟩.
Since [K : ℚ] = 8, and Δ is not a square in F (G = D_{4}, C_{4} -|C_{4}| = 4, or S_{4} -|S_{4}| = 24), the only option with order 8 is D_{4} = ⟨a, b: a^{4} = b^{2} = e, ab = ba^{-1}⟩.
It is obviously solvable by radicals: $\mathbb{Q}⊆\mathbb{ℚ}(\sqrt{2})⊆{ℚ}(\sqrt{2},\sqrt{3+\sqrt{2}})⊆{ℚ}(\sqrt{2},\sqrt{3+\sqrt{2}},\sqrt{3-\sqrt{2}})$
f(x) = x^{4} +1, roots: ±1, ±i, g = x^{3} -4x. g is completely reducible in ℚ, its roots are -2, 0, and 2 ⇒ Gal(f) = D_{2}. Another way of obtaining the same result is seeing that K = ℚ(i), [K : ℚ] = 2, K/ℚ ≋ D_{2}.
f(x) = x^{4} -16x^{2} + 4. It is irreducible modulo 3 (or over ℤ_{3}, Mod p Irreducible Test) ⇒ f is irreducible in ℚ[x]. Its discriminant (Disc(f) = Disc(g) = ) Δ = 1920^{2}, so $\sqrt{Δ} = 1920 ∈ \mathbb{Q}$. Futhermore, the resolvent is g = x^{3} -128x^{2} +3840x, so it is reducible over ℚ ⇒ Gal(K/Q) ≋ D_{2}.
f(x) = x^{4} -4x^{2} +2, is irreducible by Eisenstein, p = 2 | -4, 2 |2, but 2 ɫ a_{n} = 1, p^{2} = 2^{2} = 4 ɫ a_{0} = a_{2}
x^{4} -4x^{2} +2 = (x^{4} -4x +4) -2 = (x^{2}-2)^{2} -2 = 0 ⇒ (x^{2}-2)^{2} = 2 ⇒ $(x^2-2)=±(\sqrt{2}) ⇒ x^2 = 2±(\sqrt{2})$ ⇒ Its roots are $±α = ±\sqrt{2+\sqrt{2}}, ±β = ±\sqrt{2-\sqrt{2}}.$ Notice that β = $\frac{α^2-2}{α}∈ℚ(α)$, so K = [🚀] $ℚ(α)=ℚ(\sqrt{2+\sqrt{2}})$. An automorphism Φ: ℚ(α) → ℚ(α) is defined by permuting the four roots, e.g., σ: α → β, β → -α, Gal(K/ℚ) ≋ [Since [K : ℚ] = 4, the options with order 4 are Gal(K/ℚ) ≋ D_{2} or C_{4}] D_{2} or C_{4}
🚀 ℚ(α) = $ℚ(\sqrt{2+\sqrt{2}}), β = \sqrt{2-\sqrt{2}} = \frac{(\sqrt{2-\sqrt{2}})·(\sqrt{2+\sqrt{2}})}{(\sqrt{2+\sqrt{2}})}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$. It is obvious that α ∈ ℚ(α) = $ℚ(\sqrt{2+\sqrt{2}}) ⇒ α^2 = 2 + \sqrt{2} ∈ ℚ(α) ⇒ \sqrt{2} ∈ ℚ(α) ⇒ β = \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} ∈ ℚ(α)$ ⇒ K = ℚ(±α, ±β) = ℚ(α).
Let σ be defined by σ: α → β, β → -α. σ^{2}(α) = σ(β) = -α, σ^{3}(α) = σ(-α) = -β, σ^{4}(α) = σ(-β) = α. σ^{2}(β) = σ(-α) = -β, σ^{3}(β) = σ(-β) = α, σ^{4}(β) = σ(α) = β ⇒ σ^{4} = id ⇒ σ has order 4 ⇒ [D_{2} has all elements of order 2 except the identity, but C_{4} has an element of order 4] Gal(K/Q) ≋ C_{4}.
x^{4} +4x^{2} +2 = 0 ↭ $x^4 +4x^2 +4 -2 = 0 ↭ (x+2)^2 = 2 ↭ x = ±\sqrt{-2 ±\sqrt{ 2 }}, α_1 = \sqrt{-2 +\sqrt{ 2 }}, α_2 = -\sqrt{-2 +\sqrt{ 2 }}, α_3 = \sqrt{-2 -\sqrt{ 2 }}, α_4 = -\sqrt{-2 -\sqrt{ 2 }}$
By a similar reasoning than before K = ℚ(α_{1}) because $α_1^2 = -2 + \sqrt{2} ∈ K ⇒ \sqrt{2} ∈ K$. Futhermore, α_{1}α_{3} = $(\sqrt{-2 +\sqrt{ 2 }})(\sqrt{-2 -\sqrt{ 2 }}) = i(\sqrt{2 -\sqrt{ 2 }})i(\sqrt{2 +\sqrt{ 2 }}) = -\sqrt{2} ⇒ α_3 = \frac{-\sqrt{2}}{α_1} ∈ K, α_4 = -α_3 ∈ K$, so K is the splitting field of the irreducible polynomial f over ℚ ⇒ [K : ℚ] = deg(f) = 4 = |Gal(f)|. Since [K : ℚ] = 4, the options with order 4 are Gal(K/ℚ) ≋ D_{2} or C_{4}
Next, σ ∈ Gal(f), σ(α_{1}) = α_{3} is an element of order 4 ⇒ [D_{2} has all elements of order 2 except the identity, but C_{4} has an element of order 4] Gal(K/Q) ≋ C_{4}.
Besides, the discriminant is Δ = 2048, so $\sqrt{Δ}∉\mathbb{Q}$, and the resolvent is g = x^{3} +32x^{2} + 128x which is reducible over ℚ ⇒ Gal(K/ℚ) ≋ D_{2} or C_{4}.
f(x) = x^{4} -x -1 ∈ ℚ[x], f(x) = x^{4} -x -1 is irreducible modulo 2 (Mod p Irreducible Test, 0-0-1=-1≠0, 1-1-1=-1≠0) ⇒ f is irreducible in ℚ[x]. [Disc(x^{4}+ cx +d) = 256d^{3} -27d^{4}. Resolvent cubic(x^{4}+ cx +d) = x^{3} -4dx -d^{2}] Disc(f) = -283 is not a square in ℚ, g = x^{3} +4x -1 ∈ ℚ[x] is irreducible [By rational root test, any root must be ±1, but none of them are roots] ⇒ g is irreducible, Disc(f) is not a square ⇒ Gal(f) ≋ S_{4}.
f(x) = x^{4} + x^{3} + 1. It is irreducible modulo 2 or over ℤ_{2} (Mod p Irreducible Test) ⇒ f is irreducible over ℚ. Disc(f) = 229 is not a square in ℚ, g = x^{3} -3x^{2} -61x -1 ∈ ℚ[x] is irreducible [By rational root test, any root must be ±1, but none of them are roots] ⇒ g is irreducible, Disc(f) is not a square ⇒ Gal(f) ≋ S_{4}.
f(x) = x^{4} +8x + 12 ∈ ℚ[x] is irreducible (not easy). Disc(f) = 576, $\sqrt{576}=24∈ ℚ$. The resolvent is g = x^{3} -48x -64, g’= [g modulo 5] x^{3} -3x -4 = x^{3} + 2x +1 is irreducible modulo 5: 0+0+1=1≠0, 1+2+1=4≠0, 8+2+1=1≠0, 27+6+1=34=4≠0, 64+8+1=73=3≠0 ⇒ g is irreducible in ℚ and Disc(f) ∈ ℚ ⇒ Gal(f) ≋ A_{4}.
f(x) = x^{4} +24x +36 ∈ ℚ[x] is irreducible modulo 17 or over ℤ_{17} (Mod p Irreducible Test, tedious 😞). Disc(f) = 1728^{2}, $\sqrt{Δ}=1728 ∈ ℚ$. The resolvent g is g = x^{3} -2304x -36864 is irreducible modulo 3 or over ℤ_{3} ⇒ g is irreducible in ℚ and Disc(f) ∈ ℚ ⇒ Gal(f) ≋ A_{4}.
f(x) = x^{6} -2x^{5} -x^{4} +4x^{2} +2x -4 = (x^{3}-2)(x -1)(x +1)(x -2) so it has three roots in ℚ (namely 1, -1, and 2) and two non-real-roots ⇒ [Insolvability of quintics II Let f ∈ ℚ[x] be an irreducible polynomial of prime degree p with exactly one pair of non-real complex conjugate roots. Then, f is not solvable by radicals, G ≋ ℤ_{p}] Gal(f) = S_{3}
f(x) = x^{5} -7 is irreducible over ℚ (Eisenstein criteria). This is based on an exercise on Fields and Galois Theory, John. M. Howie, Springer. The primitive root w = $e^{\frac{2πi}{5}}$ has minimum polynomial x^{4} +x^{3} + x^{2} +x +1 ⇒ |ℚ(w) : ℚ| = 4.
Futhermore, x^{5} -7 is irreducible over ℚ(w). Suppose for the sake of contradiction, x^{5} -7 is reducible over ℚ(w) ⇒ [Abel’s Theorem. Let F be a field of characteristic 0, let p be a prime number, and a ∈ F. If x^{p} -a is reducible over F, then it has a linear factor x -c in F[x].] ∃b∈ ℚ(w): b = $\sqrt[5]{7}$. However, x^{5} -7 is irreducible over ℚ ⇒ [ℚ($\sqrt[5]{7})$: ℚ] ≥ 5, but [ℚ($\sqrt[5]{7})$: ℚ] ≤_{Since b ∈ ℚ(w)} [ℚ(w) : ℚ] = 4⊥
The roots of x^{5} -7 in ℂ are α, αw, αw^{2}, αw^{3}, αw^{4}, where α = $\sqrt[5]{7}$ and w = $e^{\frac{2πi}{5}}$. The Galois group consist of σ_{p, q}: α → αw^{p}, w → w^{q}, 0 ≤ p, q ≤ 4, σ_{0,1} = id.
σ_{p,q}∘σ_{r,s}(α) = σ_{p,q}(αw^{r}) = σ_{p,q}(α)(σ_{p,q}(w))^{r} = αw^{p}w^{qr} = αw^{p+qr}. σ_{p,q}∘σ_{r,s}(w) = σ_{p,q}(w^{s}) = w^{qs}. Therefore, σ_{p,q}∘σ_{r,s} = σ_{p+qr,qs} where the addition and multiplication p+qr,qs is in mod 5 ⇒ σ_{1,1}^{2} = σ_{2, 1}. More generally, σ_{1, 1}^{p} = σ_{p, 1}. Similarly, σ_{0, 2}^{2} = σ_{0, 22}, and in general, σ_{0, 2}^{q} = σ_{0, 2q}, and σ_{p, 1}σ_{0, 2q} = σ_{p, 2q}.
The Galois group is generated by β = σ_{1, 1} and γ = σ_{0, 2} with presentation ⟨β, γ| β^{5} = γ^{4} = β^{2}γβ^{-1}γ^{-1} = id⟩