Homework 1 - Aaron Goldsmith Homework 1 Section 3.4 Algebra...

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Aaron Goldsmith Homework 1 Section 3.4 Algebra II 3 a) Let E = { 2 n | n Z + } . If 2 n 1 , 2 n 2 E , then the product (2 n 1 )(2 n 2 ) = 2(2 n 1 n 2 ) E holds, and E is multiplicative. Now, we see that E - 1 Z is a com- mutative ring (Thm 4.3) and the subring of Q trademarked by even denominators. For the reverse inclusion, let p/q Q . Then, p/q = 2 p/ 2 q E - 1 Z . b) In addition to 0 6∈ S and of course multiplicativity, S must in- tersect every nonzero ideal. Equivalently, for every prime p , there must be an s S such that p divides s . This is necessary because if there were a prime p that divided no s S , then 1 /p 6∈ S - 1 Z . It is sufficient because we can take the prime factorization of any de- nominator, replace the primes p with elements of S that p divides, and choose a suitable numerator to cancel all that’s extra. 10 Let S = R - M . By definition, R M = S - 1 R . To see that 1 S , note that R is an integral domain containing 1, and if 1
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Homework 1 - Aaron Goldsmith Homework 1 Section 3.4 Algebra...

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