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Unformatted text preview: 8430 HANDOUT 1 PETE L. CLARK 1. Statement of the Problem For which primes p do there exist integers x,y such that p = x 2 + ny 2 ? We will abbreviate the clause “there exist integers x,y such that p = x 2 + ny 2 ” to either “ p is of the form x 2 + ny 2 ” or (worse!) to “ p = x 2 + ny 2 .” Unless otherwise noted , the following conventions are in force: p > 2, ( n,p ) = 1, and n > 0. These are in increasing order of seriousness: p > 2 is almost harmless; ( n,p ) = 1 will be an important technical assumption, and if n were negative the theory would have a quite different flavor. At times we will relax one or more of these conventions to see what happens, but this will happen with clear warning. 2. The fundamental congruence The following observation will be our constant companion throughout the course. Proposition 1. Let n be any integer and p any prime. Then p = x 2 + ny 2 implies n is a square modulo p . Proof: Reducing the equation mod p , we get x 2 ≡  ny 2 (mod p ). If y ≡ 0 (mod p ), then x 2 ≡ 0 (mod p ) hence x ≡ 0 (mod p ); then p  x, p  y and thus p 2  x 2 + ny 2 = p , a contradiction. So y is invertible mod p and we may write n ≡ ( x y ) 2 (mod p ). 3. When O n is a PID For any n ∈ Z , define the quadratic ring O n as the quotient ring Z [ t ] / ( t 2 + n ). Exercise 1.3.1. a) Show that in all cases the map the map x + yt + ... + ( t 2 + n ) 7→ ( x,y ) defines an isomorphism of additive groups from O n to Z 2 . b) Observe that if n = 0, the ring O n is nonreduced, i.e., has nilpotent elements. c) If n = m 2 , show that O n is isomorphic, as a ring, to Z × Z . d) Otherwise, show that O n is an integral domain, with quotient field Q ( √ n ). e) Explain why the notation Z [ √ n ] is appropriate for O n under the conditions of part d) but not those of part b) or c). Exercise 1.3.2: Suppose that n = m 2 , so that O n is not an integral domain. Determine exactly which integers N are of the form x 2 + ny 2 . In the sequel we shall assume that n is not of the form m 2 . Again, the ma jority of our interest will be in the case n > 0, but for the rest of this section we also entertain the case of an arbitrary integer n not of the form m 2 . 1 2 PETE L. CLARK Theorem 2. Let n 6 = m 2 be an integer. Suppose that O n is a principal ideal domain (henceforth PID). Then for any prime p (including p = 2 and p  n ), if n is a square mod p then p =  x 2 + ny 2  . Proof: To say that n is a square mod p is to say that there exists x ∈ Z such that p  x 2 + n . In the ring O n the latter factors as ( x + √ n )( x √ n ). Suppose, for the sake of contradiction, that p is irreducible in O n . Then, since O n is a PID, the principal ideal ( p ) is prime, i.e., it satisfies Euclid’s Lemma : if α,β ∈ O n are such that p  αβ , then p  α or p  β . Thus under our assumption we get p  x ± √ n , which would mean that the element x p ± √ n p of the quotient field Q ( √ n ) actually lies in...
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 Fall '11
 Clark
 Geometry, Number Theory, Integers, Algebraic number theory, discriminants, pete l. clark

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