Theorem 15 for any a b z there exists a greatest

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Theorem 1.5 For any a, b Z , there exists a greatest common divisor d of a and b , and moreover, a Z + b Z = d Z ; in particular, as + bt = d for some s, t Z . Proof. We apply the previous theorem to the ideal I = a Z + b Z . Let d Z with I = d Z , as in that theorem. Note that a, b, d I . Since a I = d Z , we see that d | a ; similarly, d | b . So we see that d is a common divisor of a and b . Since d I = a Z + b Z , there exist s, t Z such that as + bt = d . Now suppose a = a 0 d 0 and b = b 0 d 0 for a 0 , b 0 , d 0 Z . Then the equation as + bt = d implies that d 0 ( a 0 s + b 0 t ) = d , which says that d 0 | d . Thus, d is the greatest common divisor of a and b . 2 For a, b Z , we denote by gcd( a, b ) the greatest common divisor of a and b . We say that a and b are relatively prime if gcd( a, b ) = 1. Notice that a and b are relatively prime if and only if a Z + b Z = Z , i.e., if and only if there exist s, t Z such that as + bt = 1. Theorem 1.6 For a, b, c Z such that c | ab and gcd( a, c ) = 1 , we have c | b . Proof. Suppose that c | ab and gcd( a, c ) = 1. Then since gcd( a, c ) = 1, by Theorem 1.5 we have as + ct = 1 for some s, t Z . Multiplying this equation by b , we obtain abs + cbt = b. (1.2) Since c divides ab by hypothesis, and since c clearly divides cbt , it follows that c divides the left-hand side of (1.2), and hence that c divides b . 2 As a consequence of this theorem, we have: Theorem 1.7 Let p be prime, and let a, b Z . Then p | ab implies that p | a or p | b . Proof. The only divisors of p are ± 1 and ± p . Thus, gcd( p, a ) is either 1 or p . If p | a , we are done; otherwise, if p - a , we must have gcd( p, a ) = 1, and by the previous theorem, we conclude that p | b . 2 3
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1.3 Finishing the Proof of Theorem 1.2 Theorem 1.7 is the key to proving the uniqueness part of Theorem 1.2. Indeed, suppose we have p 1 · · · p r = p 0 1 · · · p 0 s , where the p i and p 0 i are primes (duplicates are allowed among the p i and among the p 0 i ). If r = 0, we must have s = 0 and we are done. Otherwise, as p 1 divides the right-hand side, by inductively applying Theorem 1.7, one sees that p 1 is equal to some p 0 i . We can cancel these terms and proceed inductively (on r ). That proves the uniqueness part of Theorem 1.2. 1.4 Further Observations For non-zero integers a and b , it is easy to see that gcd( a, b ) = Y p p min( ν p ( a ) p ( b )) , where the function ν p ( · ) is as implicitly defined in Theorem 1.2. For a, b Z a common multiple of a and b is an integer m such that a | m and b | m ; moreover, m is a least common multiple of a and b if m is non-negative and m divides all common multiples of a and b . In light of Theorem 1.2, it is clear that the least common multiple exists and is unique; indeed, if we denote the least common multiple of a and b as lcm( a, b ), then for non-zero integers a and b , we have lcm( a, b ) = Y p p max( ν p ( a ) p ( b )) . Moreover, for all a, b Z , we have gcd( a, b ) · lcm( a, b ) = ab. Finally, we recall the basic fact that that there in infinitely many primes. For a proof of this, suppose that there were only finitely many primes, call them p 1 , . . . , p k . Then set x = 1 + Q k i =1 p i , and consider any prime p that divides x . Clearly, p cannot equal any of the p i , since if it did, we would would have p | 1, which is impossible. Therefore, the prime
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