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

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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 < 0 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; you’re 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

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2 PETE L. CLARK at every prime ideal p of K ; and (iii) a prime p is principal iff its Frobenius ele- ment τ 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 . We’re done.
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