Class Field Theory. Steven Charlton. 29th February 2012

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1 Class Theory 29th February 2012

2 Introduction Motivating examples Definition of a binary quadratic form Fermat and the sum of two squares The Hilbert class field form x y 2

3 Motivating Examples p = x 2 + y 2 p = 2, or p 1 (mod 4) p = x 2 + 2y 2 p = 2, or p 1, 3 (mod 8) p = x 2 + 5y 2 p = 5, or p 1, 9 (mod 20)

4 Motivating Examples p = x 2 + y 2 p = 2, or p 1 (mod 4) p = x 2 + 2y 2 p = 2, or p 1, 3 (mod 8) p = x 2 + 5y 2 p = 5, or p 1, 9 (mod 20) For p 2, 17: p = x y 2 {( 17 p ) = 1, and (2x 2 1) 2 17 (mod p) has a solution

5 Binary quadratic form: ax 2 + bxy + cy 2 Discriminant: D = b 2 4ac Notion of equivalence Ideals in a quadratic field correspond to quadratic forms

6 Binary quadratic form: ax 2 + bxy + cy 2 Discriminant: D = b 2 4ac Notion of equivalence Ideals in a quadratic field correspond to quadratic forms Theorem If p n is an odd prime, then ( ) n p = 1 if and only if p is represented by some binary quadratic form of discriminant D = 4n.

7 Fermat and the When is p = x 2 + y 2?

8 Fermat and the When is p = x 2 + y 2? For p 1 an odd prime: p is represented by a form with D = 4 if and only if ( ) 1 p = 1 There is only one quadratic form with D = 4 p is represented by x 2 + y 2 if and only if ( ) 1 p = 1

9 Fermat and the When is p = x 2 + y 2? For p 1 an odd prime: p is represented by a form with D = 4 if and only if ( ) 1 p = 1 There is only one quadratic form with D = 4 p is represented by x 2 + y 2 if and only if ( ) 1 p = 1 Finding the condition: ) = 1 if and only if p 1 (mod 4) by quadratic reciprocity ( 1 p 2 = So p = x 2 + y 2 p = 2, or p 1 (mod 4).

10 Definition The Hilbert class field L is the maximal unramified abelian extension of a number field K.

11 Definition The Hilbert class field L is the maximal unramified abelian extension of a number field K. Theorem A prime ideal p of K is principal if and only if p splits completely in L.

12 Setting Up Which primes are of the form? First look at K = Q( 23) Q 0 = x 2 + xy + 6y 2 corresponds to principal ideals in K Q 0 represents p if and only if p splits into principal ideals in K

13 Hilbert Class of K = Q( 23) The Hilbert class field of K is L = K(α), where α 3 α 1 = 0

14 Hilbert Class of K = Q( 23) The Hilbert class field of K is L = K(α), where α 3 α 1 = 0 L = K(α) F = Q(α) K = Q( 23) Q

15 Hilbert Class of K = Q( 23) The Hilbert class field of K is L = K(α), where α 3 α 1 = 0 L = K(α) F = Q(α) K = Q( 23) Q p = x 2 + xy + 6y 2 if and only if p splits in F and in K

16 Hilbert Class of K = Q( 23) The Hilbert class field of K is L = K(α), where α 3 α 1 = 0 L = K(α) F = Q(α) K = Q( 23) Q p = x 2 + xy + 6y 2 if and only if p splits in F and in K For p 23: p = x 2 + xy + 6y 2 {( 23 p ) = 1, and x 3 x 1 0 (mod p) has a solution

17 Form x y 2 Identity: x 2 + xy + 6y 2 = ( x + y ) 2 ( y 2 If p 2, then p = x 2 + xy + 6y 2 means y is even: 1 x 2 + xy x(x + y) (mod 2) ) 2

18 Form x y 2 Identity: x 2 + xy + 6y 2 = ( x + y ) 2 ( y 2 If p 2, then p = x 2 + xy + 6y 2 means y is even: 1 x 2 + xy x(x + y) (mod 2) p = x 2 + xy + 6y 2 if and only if ) 2 For p 23: {( 23 p ) = 1, and x 3 x 1 0 (mod p) has a solution

19 In summary: Defined binary quadratic forms Proved Fermat s two-squares theorem Used the Hilbert class field to find when

20 In summary: Defined binary quadratic forms Proved Fermat s two-squares theorem Used the Hilbert class field to find when What else could we look at? Ring class fields to deal with all x 2 + ny 2, n > 0 Narrow class fields to study indefinite forms