Math 121: Modern Algebra II

Topics: fields, polynomials, Galois theory, solvable groups and simple groups. Prerequisite: math 120.

Lectures: Tuesdays and Thursdays 11-12:15 in 381T.

Course assistant: Dmitriy Ivanov. Office: 380-U1. Email: dmivanov (at math.stanford.edu).

Office hours: My office hours are Tuesdays 1:30-3:00 and Thursdays 2:30-4:00 (or by appointment) in my office 383-EE (third floor of the math building). Dmitriy's office hours are Mondays 2:30 - 4:30 and Thursdays 12:30 - 2:30 (or by appointment).

Text: Dummit and Foote's Absract Algebra, third edition (though the second edition should also be fine).

There will be weekly homework assignments, an in-class midterm, and a final exam. The midterm will be on Thursday, February 21. The final exam will be 7-10pm on Wednesday, March 19. That time is set by the registrar. Be sure you can take the exam at that time before signing up for the course.

Problem sets: Homework is an essential part of the course. In general, problem sets will be posted each Friday and will be due on 4:00pm the following Friday. There will be a large envelope for hw assignments in my mail box. (The faculty mailboxes are opposite the elevator on the first floor.) Do not leave hws under my office door. No late homework will be accepted.

Some of the problems are meant to be quite challenging, so don't be discouraged if you have difficulty. If you get stuck, you can ask for hints. (But it's best to try hard before getting a hint.)

I recommend that you start working on each problem set as soon as possible. If you work on a problem and put it aside, often it will seem easier when you come back to it later. Perhaps part of your brain keeps working on it while you're doing other things...

Be sure to look at the posted solutions even for the problems you succeed in solving. In some cases, you may find the posted solutions to be more elegant than the ones you discover.

  • hw1, due Friday, Jan 18. Solutions.
  • hw2, due Friday, Jan 25. Solutions.
  • hw3, due Fri., February 1. Solutions.
  • hw4, due Fri, February 8. Solutions.
  • hw5, due Fri, February 15. Solutions.

    Midterm solutions.

  • hw6, due Fri, February 29. Solutions.
  • hw7, due Fri, March 7. Solutions.
  • hw8, due Fri, March 14. Solutions.

    The course so far:

  • Class 1 (Tu Jan. 8): Historical remarks (solvability of polynomial equations, impossibility of certain compass and straightedge constructions). Gauss's lemma. The Eisenstein criterion for irreducibilty of polynomials over Q. The polynomial p(x) in F[x] has x-a as a factor if and only if p(a)=0. lecture notes.
  • Class 2 (Th Jan. 10): Fields and extension fields. The degree [E:F] of an extension. Algebraic and transcendental elements of an extension field. Minimal polynomials. Structure of F(a) where a is algebraic over F. If p(x) in F[x] is irreducible, then K=F[x]/(p(x)) is an field containing F and containing a root of p(x).
  • Class 3 (Tu Jan. 15): Characteristic of a field. Prime subfield. An element a of an extension field is algebraic over F if and only if [F(a):F] is finite. If F, K, and L are nested fields, then [L:F]=[L:K][K:F]. The degree [F(a,b):F] is divisible by [F(a):F] and [F(b):F], and it is less than or equal to their product. Quadratic extensions for characteristic not =2.
  • Class 4 (Th Jan. 17): An extension field E of F is finitely generated and algebraic if and only if [E:F] is finite. The elements of an extension of F that are algebraic over F form a subfield. If L is an algebraic extension of K and if K is an algebraic extension of F, then K is an algebraic extension of F. The composite K_1_2 of two subfields K_1 and K_2. If K_1 and K_2 contain F, then [K_1K_2:F] is at most [K_1:F] [K_2:F]. A real number is constructible by compass and straightedge if and only if it lies in a field obtained from Q by repeated quadratic extensions. The cube root of 2 is not constructible.
    • Class 5 (Tu, Jan 22): existence of splitting fields. If p is prime, then (x^p-1)/(x-1) is irreducible over Q and the spitting field has degree p-1 over Q. Examples. The splitting field of a polynomial of degree n has at most n! elements.
  • Class 6 (Th, Jan 24): Uniqueness of splitting field up to isomorphism: If F is isomorphic to F', E is a splitting field of f(x) and E' is a spitting field for the corresponding polynomial f'(x), then the isomorphism from F to F' extends to an isomorphism from E to E'. Splitting field for an infinite set of polynomials. Algebraic closures. Formal derivative of a polynomial. A polynomial f(x) has multiple roots (in the splitting field) if and only if f(x) and Df(x) have a nontrivial common factor. Irreducible polynomials over a field of characteristic 0 are separable (i.e. do not have multiple roots.)
  • Class 7 (Tu, Jan 29): For nonzero characteristic p, an irreducible polynomial f(x) is separable if and on if f(x) = g(xp) for some polynomial g(x). The Frobenius endomorphism a --> ap. For finite fields, the Frobenius endomorphism is an isomorphism. Every irreducible polynomial over a finite field is separable. Perfect fields.
  • Class 8 (Th, Jan 31):
  • Class 9 (Tu, Feb 5):
  • Class 10 (Th, Feb 7): Suppose K/F is Galois and E is an intermediate field. Theorem: E/F is Galois if and only if Gal(K/F) stablizes E, which is if and only if H (the subgroup of Gal(K/F) that fixes elements of E) is a normal subgroup of Gal(K/F). Theorem: If K/F is Galois and r is in K, then the minimal polynomial in F[x] of r is (x-r1)...(x - rk), where r1, ..., rk is the orbit of r under Gal(K/F).
  • Class 11 (Tu, Feb 12): Gal(Fpn/Fp) is a cyclic group of order n generated by the Frobenius automorphism. In any field, the multiplicative group of nth roots of unity is cyclic. Simple extensions. Theorem: Every finite, separable extension is simple.
  • Class 12 (Th, Feb 14): Gal(Q(zn)/Q) is isomorphic to (Z/nZ)x. Abelian extensions. If K/F is an abelian extension and if E is an intermediate field, then K/E and E/F are abelian extensions. Constructiblity of regular n-gons by compass and straightedge.
  • Class 13 (Tu, Feb 19): K/F Galois and F' any extension of F implies that KF'/F' is Galois and Gal(KF'/F') is isomorphic to Gal(K/K intersect F"). If K1 and K2 are Galois extensions of F, then K1 intersect K2 and K1K2 are Galois extensions of F. Description of Gal(K1K2/F). Also: Theorem: (char 0) If F contains the nth roots of unity and if a is an nth root of some element in F, then F(a)/F is Galois and Gal(F(a)/F) is cyclic.
  • Class 14 (Thur, Feb 21): Midterm.
  • Class 15 (Tue, Feb 26): Solvable groups. For H a normal subgroup of G, is solvable if and only if H and G/H are solvable. All groups of order less than 60 are solvable, but A5 (of order 60) is not. Root extensions. A composite of roots extensions is also a root extension. Galois closures. The Galois closure of a root extension is also a root extension. Definition of "f(x) is solvable by radicals over F''.
  • Class 16 (Thur, Feb 28): (For characteristic 0.) Theorem: If f(x) is solvable by radicals over F, then its Galos group is solvable. Example: x6-6x+3 has Galois group S5, therefore is not solvable by radicals. Characters of a group. Distinct characters are linearly independent. Lagrange resolvents. If F contains a primitive root of unity and if K/F is Galois with Galois group a cyclic group of order n, then K=F(r) where rn is in F.
  • Class 17 (Tue, March 4): If the Galois group of a polynomial is solvable, then the splitting field is contained in a root extension.
  • Class 18 (Thur, March 6): The discriminant. The Galois group of a nth degree polynomial in F[x] is contained in An if and only if the discriminant is a square in F. Cardano's formula for the solution of a cubic. The reduction of the quartic to the cubic. Artin's proof by Galois Theory of the fundamental theorem of algebra (C is algebraically closed.)
  • Class 19 (Tue, March 11):
  • Class 20 (Thur, March 13): A subgroup of a finitely generated abelian group is finitely generated. A complex number r is an algebraic integer if and only if Z[r] (under addition) is contained in a finitely generated subgroup of C. The algebraic integers form a ring. An algebraic integer that is rational is an integer. Every algebraic number is the quotient of an algebraic number and an integer. If r is a root of a monic polynomial with algebraic integer coefficients, then r is an algebraic integer.

    Brian White
    Math Dept, Room 383-EE
    Phone: 650-723-0952 (but e-mail is better)
    Fax: 650-725-4066
    Email: white (at math.stanford.edu)

    To my home page.