Key Questions Flashcards
What does it mean for an algebraic number to be expressible by radicals
An algebraic number alpha is expressible by radicals if alpha is an element of an algebraic extension of Q obtained by iterated radical extensions - adjoin nth roots of rational numbers repeatedly
D&F 627
Are all algebraic number expressible by radicals?
No. Key solvable by radicals => solvable Galois extension
1. Add roots of unity –> (Z/nZ)* Abelian so solvable extension
- Add nth root and all conjugates –> ensures Galois extension gives a subgroup of Z/nZ, cyclic
- Repeat. After finite number of steps - solvable
Example: f(x) = x^5 -4x +2
We chose this polynomial since it is irreducible by Eisenstein and contains 3 real roots, 2 complex.
Let K be the splitting field of f(x) over Q. We have a degree 5 extension Q[x] / f(x) => [K:Q] is divisible by 5 since degree multiplicative. Since, [K:Q] = |Gal(K/Q)|, |Gal(K/Q)| is divisible by 5 and hence contains a 5 cycle by Cauchy’s Theorem. Exactly 2 non-real roots so G has a transposition - complex conjugation - transposition plus 5 cycle generate S_5 which is not solvable - A_5 first simple non-abelian group.
Or in general terms - if we perform a chain of cyclic extensions, the Galois group will be solvable. But a generic polynomial has Galois group S^5
Discuss x^3 -2 = 0 over Q
Prove Irreducible: Eisenstein Criterion
Draw picture of roots - triangle in S^1
Construct splitting field - adjoin all 3 roots (-1+i root(3) )/2 etc. see that root 3 is in here. So Q( 3rd root(2), root(3) ) - or recall constructing splitting field, we have a root after first extension - next must be degree 2!
Show Galois group is S_3 - We know it is order 6, we also know it acts on the three roots of x^3 - 2, so it must be a subgroup of S_3 - hence all of S_3 since degree 6. (Or observe there are only 2 groups of order 6 - cyclic and S_3 - can’t be cyclic since a root of cyclotomic polynomial x^2 + x +1 is not root of x^3 -2)
Classify finite fields, discuss Galois groups
Characteristic must be prime - otherwise zero divisors.
So look at any finite field as vector space over prime subfield - must be order p^n.
Existence: Splitting field of x^p^n -x
Uniqueness: In F_p^n any nonzero x must satisfy x^p^n-1 = 1 (mult group has one less element) by Lagrange. So x^p^n -x = 0 - a root of magic polynomial
Galois extension: The extension F_p^n / F_p is Galois since it was constructed as the splitting field of a separable polynomial: x^p^n-x alpha^p. This is injective clearly since no power of nonzero field element can be 0. It surjective since finite.
Galois Group: Galois group must be order n since we have a degree n extension. The Frobenius automorphism generates Gal(F_p^n / F_p). This is easy to see. Let g be Frobenius automorphism, the g^i(alpha) = alpha^p^i. Since alpha^p^n = alpha, we see alpha^n = identity. Any lower power identity would imply alpha^p^i = alpha for all alpha in F_p^n which is impossible since only p^i roots of this equation. Since we have exhibited an element of order n, it follows that Gal(F_p^n / F_p) is Z/nZ.
Subfields: By Fundamental Theorem, every subfield corresponds to a subgroup of Z/nZ - d divisor of n, then F_p^d is subfield corresponding to Z/dZ. Abelian so everything normal - everything Galois
The multiplicative group F_p* is cyclic
Define: analytic function
Here I think he’s talking about Holomorphic function. Complex differentiable
Derive the Cauchy-Riemann equations
Do as in Stein. Consider if we can discuss differential forms, topology - commutes with i
Show analytic function who’s image is real must be constant. Use this to show no analytic function maps onto a line
Seems like multiple options here - try Cauchy Riemann or power series expansion.
Use Mobius transformation to take line onto R
Or immediately by open mapping theorem?
Define: mobius transformation
Discuss Hopf fibration
Look in Conway - na Wikipedia is better - do as in discussion with Chirs - RP^1 = riemann sphere….
Integrate 1/(1+x^4) from -inf to inf
Use contour integration
Compute homology for S^2, Klein bottle, torus, closed surfaces
notes
Discuss various long exact sequences
pair, Mayer Vetoris
Prove fundamental theorem of algebra
topology, complex, algebra
What is relationship of pi_n to h_n?
Hurewicz Thm
Is homology complete invariant for homotopy type? Cohomology? Homotopy groups?
Whitehead Thm
Discuss x^3 -2 over finite field - polynomials over finite fields more generally
Every polynomial over a finite field is separable. If p(x) irreducible of degree d dividing x^p^n - x and alpha is a root of p(x), then F_p(alpha) is a subfield of F_p^n of degree d. Galois since every extension of a finite field is Galois - thus contains all roots of p(x).
In fact x^p^n-x is precisely the product of all distinct irreducible polynomials in F_p[x] of degree d as d runs through the divisors of n. Gives a recursive way to compute irreducibles. Irreducible elements important because we can mod out by any degree d irreducible to get a version of F_p^d - they are all isomorphic so just get different reps of same field
Example: p(x) = x^3 -2
Degenerate over F_2. Consider over F_7 - irreducible. Taking F_7[x] / (x^3 -2) gives a degree 3 extension - isomorphic to F_7^3 - Galois so contains all roots.
Galois Group: Z_3 instead of S_3 cyclic, generated by Frobenius
