By James Matteson
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Additional resources for A collection of diophantine problems with solutions
Without loss of generality we may assume that pja, so a D pr for some r 2 R. rb/. 1 rb/ D 0 implies that 1 rb D 0 and hence rb D 1. Therefore b is a unit and hence p is irreducible. (2) Suppose that p is a prime element. Then p ¤ 0. Consider the ideal pR and suppose that ab 2 pR. Then ab is a multiple of p and hence pjab. Since p is prime it follows that pja or pjb. If pja then a 2 pR while if pjb then b 2 pR. Therefore pR is a prime ideal. Conversely suppose that pR is a prime ideal and suppose that p D ab.
Since p is a prime it follows easily that either n D 1 or n D p. 3 the conclusion of this lemma will be taken as the deﬁnition of a maximal ideal. 3 we make the following general deﬁnition for commutative rings R with identity. 1. Let R be a commutative ring. An ideal P in R with P ¤ R is a prime ideal if whenever ab 2 P with a; b 2 R then either a 2 P or b 2 P . This property of an ideal is precisely what is necessary and sufﬁcient to make the factor ring R=I an integral domain. 2. Let R be a commutative ring with an identity 1 ¤ 0 and let P be a nontrivial ideal in R.
B C ac/ and so d1 jb. Therefore d1 jd so d1 Ä d . Combining these we must have d1 D d . The next result, called the Euclidean algorithm, provides a technique for both ﬁnding the GCD of two integers and expressing the GCD as a linear combination. 9 (Euclidean algorithm). Given integers b and a > 0 with a − b, form the repeated divisions b D q1 a C r1 ; 0 < r1 < a a D q2 r1 C r2 ; 0 < r2 < r1 :: : rn 2 D qn rn rn 1 D qnC1 rn : 1 C rn ; 0 < rn < rn 1 The last nonzero remainder rn is the GCD of a; b.
A collection of diophantine problems with solutions by James Matteson