8430notes6 - 8430 HANDOUT 6: PROOF OF THE MAIN THEOREM PETE...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8430 HANDOUT 6: PROOF OF THE MAIN THEOREM PETE L. CLARK 1. Proof of the main theorem for maximal orders We are now going to take a decisive step forward by proving the Main Theorem on which primes p are represented by the principal quadratic form q D ( x,y ) of discriminant D < 0, when D is a fundamental discriminant, i.e., the discriminant of the full ring of integers of Q ( D ). Of course, to solve our original problem of representation of primes by x 2 + ny 2 for arbitrary n , we need to look also at quadratic orders of not necessarily maximal discriminant. But notice that all our efforts so far have yielded solutions for only finitely many values of n , whereas here we can take n to be any squarefree negative number which is not 3 (mod 4). Here is the result: Theorem 1. Let D < be a fundamental quadratic discriminant, the discriminant of (the maximal order of) the imaginary quadratic field K = Q ( D ) . Let h ( K ) = #Pic( O ( D )) be the class number of K . Then there exists a monic polynomial P D ( t ) Z [ t ] such that: (a) P D ( t ) is irreducible of degree h ( K ) , and has at least one real root. (b) A prime number p which is prime to D and to disc( P D ) is of the form q D ( x,y ) iff ( D p ) = 1 and the mod p reduction P D ( t ) F p [ t ] has at least one rational root. The proof occurs in two steps: first we bring our heavy artillery to bear and almost immediately deduce the following result: Theorem 2. With notation as above, a prime p prime to D is of the form q D ( x,y ) iff p splits completely in the Hilbert class field K 1 of K = Q ( D ) . Then we need to use a bit of comparatively elementary (but slightly tricky) reason- ing to get from Theorem 2 to Theorem 1. Well, second things first: Proof of Theorem 2: Suppose p is prime to D . We already know that a necessary condition for p to be represented by q D is the fundamental congruence ( D p ) = 1. (Note that this is to be interpreted as the Kronecker symbol when p = 2; or just assume p is odd; youre not missing out on much.) Further, we saw that the fun- damental congruence buys us precisely a Galois conjugate pair p , p of prime ideals of O ( D ) = O K lying over p . And we also saw that p is of the form q D ( x,y ) iff one of these ideals (equivalently, both) are principal . But now we know more: there is an abelian field extension K 1 /K , called the Hilbert class field, which has the following three remarkable properties: (i) its Ga- lois group is isomorphic to Pic( O K ), the ideal class group of K ; (ii) it is unramified 1 2 PETE L. CLARK at every prime ideal p of K ; and (iii) a prime p is principal iff its Frobenius ele- ment p Pic( O K ) is trivial. As we saw, it immediately follows from Chebotarev Density that the set of principal primes p has density 1 h ( K ) . Compiling these two conditions, we see that indeed what we want is for the rational prime p to split completely from Q to K 1 , the Hilbert class field of K . Were done....
View Full Document

This note was uploaded on 12/03/2011 for the course MATH 8430 taught by Professor Clark during the Fall '11 term at University of Georgia Athens.

Page1 / 13

8430notes6 - 8430 HANDOUT 6: PROOF OF THE MAIN THEOREM PETE...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online