Proof Let J be the product of all the nonzero prime ideals If I is any nonzero

# Proof let j be the product of all the nonzero prime

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Proof. Let J be the product of all the nonzero prime ideals. If I is any nonzero ideal, then by (7.8.2) there is a nonzero ideal I such that II is a principal ideal ( a ), with I relatively prime to J . But then the set of prime factors of I is empty, so that I = R . Thus ( a ) = II = IR = I . The next result shows that a Dedekind domain is not too far away from a principal ideal domain.
7.9. P-ADIC NUMBERS 29 7.8.4 Corollary Let I be a nonzero ideal of the Dedekind domain R , and let a be any nonzero element of I . Then I can be generated by two elements, one of which is a . Proof. Since a I , we have ( a ) I , so I divides ( a ), say ( a ) = IJ . By (7.8.2) there is a nonzero ideal I such that II is a principal ideal ( b ) and I is relatively prime to J . If gcd stands for greatest common divisor, then the ideal generated by a and b is gcd(( a ) , ( b )) = gcd( IJ,II ) = I since gcd( J,I ) = (1). Problems For Section 7.8 1. Let I ( R ) be the group of nonzero fractional ideals of a Dedekind domain R . If P ( R ) is the subset of I ( R ) consisting of all nonzero principal fractional ideals Rx , x K , show that P ( R ) is a subgroup of I ( R ). The quotient group C ( R ) = I ( R ) /P ( R ) is called the ideal class group of R . Since R is commutative, C ( R ) is abelian, and it can be shown that in the number field case, C ( R ) is finite. 2. Continuing Problem 1, show that C ( R ) is trivial iff R is a PID. We will now go through the factorization of an ideal in a number field. The neces- sary background is developed in a course in algebraic number theory, but some of the manipulations are accessible to us now. By (7.2.3), the ring B of algebraic integers of the number field Q ( 5) is Z [ 5]. (Note that 5 3 mod 4.) If we wish to factor the ideal (2) = 2 B in B , the idea is to factor x 2 +5 mod 2, and the result is x 2 +5 ( x +1) 2 mod 2. Identifying x with 5, we form the ideal P 2 = (2 , 1+ 5), which turns out to be prime. The desired factorization is (2) = P 2 2 . This technique works if B = Z [ α ], where the number field L is Q ( α ). 3. Show that 1 5 P 2 , and conclude that 6 P 2 2 . 4. Show that 2 P 2 2 , hence (2) P 2 2 . 5. Expand P 2 2 = (2 , 1 + 5)(2 , 1 + 5), and conclude that P 2 2 (2). 6. Following the technique suggested in the above problems, factor x 2 + 5 mod 3, and conjecture that the prime factorization of (3) in the ring of integers of Q ( 5) is (3) = P 3 P 3 for appropriate P 3 and P 3 . 7. With P 3 and P 3 as found in Problem 6, verify that (3) = P 3 P 3 . 7.9 p-adic Numbers We will give a very informal introduction to this basic area of number theory. ( For further details, see Gouvea, “p-adic Numbers”.) Throughout the discussion, p is a fixed prime.
30 CHAPTER 7. INTRODUCING ALGEBRAIC NUMBER THEORY 7.9.1 Definitions and Comments A p -adic integer can be described in several ways. One representation is via a series x = a 0 + a 1 p + a 2 p 2 + · · · , a i Z . (1) (Let’s ignore the problem of convergence for now.) The partial sums are x n = a 0 + a 1 p + · · · + a n p n , so that x n x n 1 = a n p n . A p -adic integer can also be defined as a sequence of integers x = { x 0 ,x 1 ,..., } satisfying x n x n 1 mod p n , n = 1 , 2 , .... (2) Given a sequence satisfying (2), we can recover the coeﬃcients of the series (1) by a 0 = x 0 , a 1 = x 1 x 0 p , a 2

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• Spring '11
• sdd
• Algebra, Number Theory, Ring, Algebraic number theory, Ring theory, Principal ideal domain

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