This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 4400/6400 PROBLEM SET 1 Key: (E) denotes easy. If you honestly feel the problem is too easy, just write okay, but try to solve some harder problems as well. 1.1)(E) Prove the Division Theorem: If a b > 0 are integers, then there ex- ist unique non-negative integers q and r such that a = qb + r and 0 r < b . Hint: It suffices to take q to be the largest non-negative integer such that a- qb 0. 1.2)(E) In the notation of Problem 1.1), show that b | a r = 0. 1.3) Prove the converse of Euclids Lemma: suppose d is a positive integer such that whenever d | ab , d | a or d | b . Then d is prime. Remark: Among other things, this allows us to generalize the notion of primes to not-necessarily principal ideals. 1.4)a)(E) To contain is to divide: for integers a and b , we have a | b ( a ) ( b ). b) Confirm that part a) holds true for elements a and b in any commutative ring. c) For elements a , b in an integral domain R , show that the following are equivalent: (i) There exists a unit u R such that b = ua . (ii) There exist units u, v R such that b = ua , a = vb . (iii) a | b and b | a . (iv) There is an equality of principal ideals ( a ) = ( b ). d)* Find a commutative ring R (not an integral domain) and elements a and b such that in part c) above, (iii) and (iv) hold but (i) and (ii) do not. In other words, in a general commutative ring, being associates is a stronger relation than generating the same principal ideal....
View Full Document