200a-f11-hw6

200a-f11-hw6 - First read the denition of the product of...

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Math 200a Fall 2011 Homework 6 Reading assignment: Section 7.1-7.6. We will cover more general localizations than the book does in section 7.5. This is done later in the text, see pages 706-708. Exercises to be handed in: (all exercise numbers refer to Dummit and Foote, 3rd edition.) Section 7.1: 14, 26, 27 Section 7.2: 3(b, c) (think through part (a), but don’t write it up). Section 7.3: 29 (Think through problem #25 also in this section, which you need to use, but don’t write it up.) Section 7.4: 11, 35, 36, 39 (see exercise 37 for the definition of local ring) Hint: For 7.4 #36 you need to show some poset has a minimal element. The trick for doing this is to apply Zorn’s lemma to the poset which is the opposite of the obvious one, i.e. for prime ideals you define P 1 P 2 if P 2 P 1 . Then a maximal element in this ordering is a prime minimal under inclusion. Exercise not from the text: (to be handed in):
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Unformatted text preview: First read the denition of the product of two ideals given on page 247 in the text, since we did not cover this in class yet. We say that an ideal I is nilpotent if I n = 0 for some n 1. (Note that this is stronger than saying that x n = 0 for all x I ; it means b 1 b 2 ...b n = 0 for all b i I .) 1 . (a). Prove that if N = ( a 1 ,...,a m ) = a 1 R + + a m R is the ideal generated by nitely many elements a i R , and each a i is nilpotent in R , then N is a nilpotent ideal. (b). Prove that a polynomial p ( x ) R [ x ], say p ( x ) = a + a 1 x + + a n x n , is nilpotent in R [ x ] if and only if each a i is nilpotent in R for all 0 i n . (c). Prove that a polynomial p ( x ) R [ x ], say p ( x ) = a + a 1 x + + a n x n , is a unit in R [ x ] if an only if a is a unit in R , and a 1 ,...,a n are nilpotent in R . 1...
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