MA3412S2_Hil2014.pdf

Proof let i be the ideal of r generated by x 1 x 2 x

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Proof Let I be the ideal of R generated by x 1 , x 2 , . . . , x k . Then I = ( d ) for some d R , since R is a principal ideal domain. Then d divides x i for i = 1 , 2 , . . . , k , and therefore d is a unit of R . It follows that I = R . But then 1 I , and therefore 1 = a 1 x 1 + a 2 x 2 + · · · + a k x k for some a 1 , a 2 , . . . , a k R , as required. Lemma 2.8 Let p be an irreducible element of a principal ideal domain R . Then the quotient ring R/ ( p ) is a field. Proof Let x be an element of R that does not belong to ( p ). Then p does not divide x , and therefore any common divisor of x and p must be a unit of R . Therefore there exist elements y and z of R such that xy + pz = 1 (Lemma 2.7). But then y + ( p ) is a multiplicative inverse of x + ( p ) in the quotient ring R/ ( p ), and therefore the set of non-zero elements of R/ ( p ) is an Abelian group with respect to multiplication. Thus R/ ( p ) is a field, as required. Theorem 2.9 An element of a principal ideal domain is prime if and only if it is irreducible. Proof We have already shown that any prime element of an integral domain is irreducible (Lemma 2.4). Let p be an irreducible element of a principal ideal domain R . Then p is neither zero nor a unit of R . Suppose that p | yz for some y, z R . Now any divisor of p is either an associate of p or a unit of R . Thus if p does not divide y then any element of R that divides both p and y must be a unit of R . Therefore there exist elements a and b of R such that ap + by = 1 (Lemma 2.7). But then z = apz + byz , and hence p divides z . Thus p is prime, as required. 2.4 Fermat’s Two Squares Theorem We shall use the fact that the ring of Gaussian integers is a principal ideal domain to prove a theorem, originally claimed by Fermat, that states that 17

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an odd prime number p can be represented in the form p = x 2 + y 2 for some integers x and y if and only if p 1 (mod 4). We make use of the following theorem, claimed and used by Ibn al-Haytham some time around the year 1000, and subsequently stated by Leibniz and by John Wilson, and proved by Lagrange in 1771. Theorem 2.10 (Wilson’s Theorem) ( p - 1)!+1 is divisible by p for all prime numbers p . Proof Let p be a prime number. If x is an integer satisfying x 2 1 (mod p ) then p divides ( x - 1)( x + 1) and hence either p divides either x - 1 or x + 1. Thus if 1 x p - 1 and x 2 1 (mod p ) then either x = 1 or x = p - 1. For each integer x satisfying 2 x p - 2, there exists exactly one integer y satisfying 2 y p - 2 such that xy 1 (mod p ), and moreover y 6 = x . It follows that ( p - 2)! is a product of numbers of the form xy , where x and y are distinct integers between 2 and p - 2 that satisfy xy 1 (mod p ). It follows that ( p - 2)! 1 (mod p ). But then ( p - 1)! p - 1 (mod p ), and hence ( p - 1)! + 1 is divisible by p , as required. Corollary 2.11 Let p be an odd prime number, and let m = 1 2 ( p - 1) . Then ( m !) 2 + ( - 1) m is divisible by p .
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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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