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Home , Algebraic extension, Galois extension, Galois group, Galois theory

Galois Theory: of Degree 5 and Up

Tokuei Higashino

Union College

ax2 + bx + c = 0 √ −b ± b2 − 4ac x = 2a Cubic Formula ax3 + bx2 + cx + d = 0

b x = − 1 3a s 1 1 q  − 3 2b3 − 9abc + 27a2d + (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 3a 2 s 1 1  q  − 3 2b3 − 9abc + 27a2d − (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 3a 2 b x = − 2 3a √ s 1 + i 3 1  q  + 3 2b3 − 9abc + 27a2d + (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 6a 2 √ s 1 − i 3 1  q  + 3 2b3 − 9abc + 27a2d − (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 6a 2 b x = − 3 3a √ s 1 − i 3 1  q  + 3 2b3 − 9abc + 27a2d + (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 6a 2 √ s 1 + i 3 1  q  + 3 2b3 − 9abc + 27a2d − (2b3 − 9abc + 27a2d)2 − 4 (b2 − 3ac)3 6a 2 Question Are there general solutions by radicals for polynomials of degree 5 and up? Answer No. How do we prove this?

I translate into question about fields

I use to translate into question about groups Definition A field is a closed, associative, and commutative under + and ·, contains 0, 1, negatives, and reciprocals, and satisfies the distributive laws of · over +. Example Q, R, and C are fields. Definition A field extension of a field K is a field L that contains K. Example R is a field extension of Q. Example √ Q( 2) is a field extension of Q. Question Given a p(x) with coefficients in K and of degree 5 or up, is there a sequence of radical extensions K ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kn such that all of the roots of p(x) are in Kn? √ Kn = Kn−1( rn an)

. ⊆ .

⊆ √ K2 = K1( r2 a2)

⊆ √ K1 = K0( r1 a1)

⊆ ri ∈ N K0 = K ai+1 ∈ Ki Definition An L of K is a field extension such that for all a ∈ L, there exists a polynomial p(x) with coefficients in K such that p(a) = 0. Non-example R is not an algebraic extension of Q, since π ∈ R. Example √ √ Q( 3) = {√a + b 3 | a, b ∈ Q} is an algebraic extension of Q, since a + b 3 is a root of the polynomial x2 − 2ax + a2 − 3b2.

All radical extensions are algebraic extensions. Definition A L of K is a field extension such that for every polynomial p(x) with coefficients in K, if L contains one of its roots, then L contains all of its roots. Example C is a normal extension of R, which follows from the Fundamental Theorem of . Non-example √3 √3 √3 Q( 2) = {a + b 2 + c 4 | a, b, c ∈ Q} is not a normal√ extension 3 3 of Q, since the complex roots of x − 2 are not in Q( 2). Theorem L is a normal extension of K iff for some polynomial p(x) with coefficients in K, L contains all of p’s roots. Example √ √ √ 2 Q( 6) contains 6 and − 6, which are roots of x − 6, which is a polynomial with coefficients in Q. Definition A L of K is a field extension such that for all a ∈ L, there exists an m(x) with coefficients in K with distinct roots. Example √ √ Any algebraic extension of Q, such as Q( 2, 3) is a separable extension. Definition A of K is a field extension that is algebraic, normal, and separable over K. Definition The Galois of a field extension L over K is the set of of L that preserve K. It is denoted Gal(L/K). Fundamental Theorem of Galois Theory If L is a finite Galois extension of K, then there is a one-to-one correspondence between the field extensions of K that are contained in L and the subgroups of Gal(L/K). p(x) = x4 − 5x2 + 6 √ √ 2 2 Q( 2, 3) = (x − 2)(x − 3)

√ √ √ Q( 2) Q( 3) Q( 6)

Q

{id, σ, τ, στ}

{id, σ} {id, τ} {id, στ}

√ √ {id} σ = (√3 −√3) √ √ τ = ( 2 − 2) Gal(Q( 2, 3)/Q) = {id, σ, τ, στ} Definition A polynomial p(x) with coefficients in K is solvable by radicals if there exists a sequence of radical extensions K ⊆ K1 ⊆ K2 ⊆ · · · ⊆ Kn such that all the roots of p(x) are in Kn. Definition A group G is solvable if there exists a sequence of subgroups {id} = G1 ⊆ G2 ⊆ · · · ⊆ Gm = G, such that Gj is normal in Gj+1 and |Gj+1|/|Gj | is prime. Theorem p(x) is solvable by radicals iff Gal(Kn/K) is solvable. Abel-Ruffini Theorem There exist polynomials of every degree ≥ 5 which are not solvable by radicals. Lemma If f (x) is an irreducible polynomial over Q, of prime degree p, and if f has exactly p − 2 real roots, then its is Sp. Lemma If n ≥ 5 and Gal(L/K) = Sn, then Gal(L/K) is not solvable. y 10 y = x5 − 4x + 2 5 x −2 −1 1 2 −5

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