Algebra Notes Nov. 12: FTOGT examples Geoffrey Scott Monday, we stated the fundamental theorem of Galois theory: Fundamental Theorem of Galois Theory Let F be a finite field or a field of characteristic zero. Let E be the splitting field over F of some polynomial in F [x]. Then the following two maps are inverses (and hence define a bijection) Subfields of E containing F ↔ Subgroups of Gal(E/F) K 7→ Gal(E/K) EH ←[ H Furthermore, for any subfield K of E containing F , the following are true 1. [E : K] = |Gal(E/K)| and [K : F ] = |Gal(E/F )|/|Gal(E/K)|. 2. If K is the splitting field over F of some polynomial, then Gal(E/K) is a normal subgroup of Gal(E/F ), and Gal(K/F ) ∼ = Gal(E/F )/Gal(E/K). Today, we do two examples to illustrate the theory, in which we compare the lattice of subfields of an extension with the lattice of subgroups of its Galois group. Example Let E be the splitting field for x7 −1, and suppose we know the (true) fact that x6 +x5 +· · ·+x+1 2πi is irreducible in Q[x]. Let ξ = e 7 . The roots of x7 − 1 are the elements {1, ξ, ξ 2 , . . . , ξ 6 }, so E = Q(ξ, ξ 2 , . . . , ξ 6 ) = Q(ξ) Because the minimal polynomial for ξ is x6 + x5 + · · · + x + 1, it follows that [E : Q] = 6. How many elements are there of Gal(E/Q)? Any Q-automorphism of E is determined by where it sends ξ. Where could a Q-automorphism of E possibly send ξ? Draw the Q-automorphism of E given by ξ 7→ ξ 2 by showing how it permutes the roots of x7 − 1. What is the order of the element of Gal(E/F ) that it represents? Draw the Q-automorphism of E given by ξ 7→ ξ 3 by showing how it permutes the roots of x7 − 1. What is the order of the element of Gal(E/F ) that it represents? Notice that this second Q-automorphism, τ , has order 6, so it must generates the entire group Gal(E/F ) ∼ = Z6 . Draw the lattice of subgroups of Gal(E/F ) and the lattice of subfields of E containing F . Example Let E be the splitting field for x4 − 2 over Q. Let’s study the subfields of E and the subgroups of Gal(E/Q). √ √ √ √ First, we know that the roots of x4 − 2 in C are { 4 2, i 4 2, − 4 2, −i 4 2}, so √ √ √ √ 4 4 4 4 E = Q( 2, i 2, − 2, −i 2) √ 4 = Q( 2, i) What is the degree [E : Q]? What does this tell you about Gal(E/Q)? Describe the elements of Gal(E/Q) in terms of where they send the elements the permutation of the roots of x4 − 2 given by each. √ 4 2 and i. Draw What familiar group is Gal(E/Q) isomorphic to? Draw the lattice of subgroups of Gal(E/Q). Solvability The fundamental theorem of Galois theory gives us the understanding about Galois groups that we need to start proving things about solvability of quintics. Recall that we were able to show that certain geometric constructions are impossible by translating facts about geometric constructions into facts about degrees of field extensions. Similarly, we can also translate facts √ about “having roots that can be expressed just in terms of rational numbers and symbols n ” into facts about Galois groups. To study this problem, we need to define rigorously the concept √ of “having roots that can be expressed in terms of rational numbers and n .” Definition: Let F be a field. A polynomial f (x) ∈ F [x] is solvable by radicals over F if it splits in some extension field of the form F (a1 , . . . , an ), where the elements ai have the property that there are positive integers k1 , . . . , kn such that ak11 ∈ F ak22 ∈ F (a1 ) ak33 ∈ F (a1 , a2 ) .. . aknn ∈ F (a1 , a2 , . . . , an−1 ) Soon, we will prove that whenever a polynomial is solvable by radicals, the Galois group of its splitting field must be a solvable group. Definition: A group G is solvable if there are subgroups H0 , H1 , . . . , Hk such that {e} = H0 ⊆ H1 ⊆ H2 ⊆ · · · ⊆ Hn = G with the property that each Hi is normal in Hi+1 , and Hi+1 /Hi is abelian. Next class, we’ll see examples of solvable groups and study their properties.