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Unformatted text preview: Lecture 1 1 Lecture 1 § 1 Three theorems of McCoy R is always a commutative ring with unity 1 R . U ( R ) is the group of units of R . C ( R ) is the set of regular elements of R , i.e. the set of nonzerodivisors. Z ( R ) is the set of zerodivisors, the complement of C ( R ). ( { a i } ) = ( a 1 ,a 1 ,... ) is the ideal generated by the set { a i } ; this is also written ∑ a i R . The principle ideal ( a ) is aR = Ra . The notation I ⊳ R means that I is an ideal in R . Theorem 1.1. Let A = R [ x ] . Let f = a n x n + ··· + a ∈ A . If there is a nonzero polynomial g ∈ A such that fg = 0 , then there exists r ∈ R integerdivide { } such that f · r = 0 . Proof. Choose g to be of minimal degree, with leading coefficient bx d . We may assume that d > 0. Then f · b negationslash = 0, lest we contradict minimality of g . We must have a i g negationslash = 0 for some i . To see this, assume that a i · g = 0, then a i b = 0 for all i and then fb = 0. Now pick j to be the largest integer such that a j g negationslash = 0. Then 0 = fg = ( a + a 1 x + ··· a j x j ) g , and looking at the leading coefficient, we get a j b = 0. So deg( a j g ) < d . But then f · ( a j g ) = 0, contradicting minimality of g . Theorem 1.2 (Prime Avoidance) . Let I 1 ,...,I n ⊳ R . Let A ⊂ R be a subset which is closed under addition and multiplication. Assume that at least n − 2 of the ideals are prime. If A ⊆ I 1 ∪ ···∪ I n , then A ⊆ I j for some j . Proof. Induct on n . If n = 1, the result is trivial. The case n = 2 is an easy argument: if a 1 ∈ A integerdivide I 1 and a 2 ∈ A integerdivide I 2 , then a 1 + a 2 ∈ A integerdivide ( I 1 ∪ I 2 ). Now assume n ≥ 3. We may assume that for each j , A negationslash⊆ I 1 ∪ ···∪ ˆ I j ∪ ··· I n . 1 Fix an element a j ∈ A integerdivide ( I 1 ∪ ··· ∪ ˆ I j ∪ ··· I n ). Then this a j must be contained in I j since A ⊆ uniontext I j . Since n ≥ 3, one of the I j must be prime. We may assume that I 1 is prime. Define x = a 1 + a 2 a 3 ··· a n , which is an element of A . Let’s show that x avoids all of the I j . If x ∈ I 1 , then a 2 a 3 ··· a n ∈ I 1 , which contradicts the fact that a i negationslash∈ I j for i negationslash = j and that I 1 is prime. If x ∈ I j for j ≥ 2. Then a 1 ∈ I j , which contradicts a i negationslash∈ I j for i negationslash = j . Definition 1.3. An ideal I ⊳ R is called dense if rI = 0 implies r = 0. This is denoted I ⊆ d R . This is the same as saying that R I is a faithful module over R . If I is a principal ideal, say Rb , then I is dense exactly when b ∈ C ( R ). The easiest case is when R is a domain, in which case an ideal is dense exactly when it is nonzero....
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This note was uploaded on 10/26/2011 for the course MATH 8020 taught by Professor Clark during the Summer '11 term at UGA.
 Summer '11
 Clark
 Algebra

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