Solutions to (Selected) HW Problems

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چکیده

We claim that the prime/irreducible elements of R are the associates of primes in Z different from p. Indeed, suppose that a/p, with a ∈ Z is an irreducible. Since powers of p are units, we may assume k = 0 and p a. Since −1 is also unit, we may also assume a > 0. If a is not prime, it factors in Z as a = bc, with b, c 6= ±1. Since p b, c [as p a and p is prime], they are not units of R and hence a is reducible in R. Now, if a is irreducible in Z, then it is a prime of Z different from p [as a > 0]. Suppose a = (b/p)(c/p). Then, ap = bc in Z, and since a is prime [in Z], we have a | b or a | c. But that being the case, by unique factorization [in Z], we have that either b/p or c/p is a power of p, i.e., a unit. Thus, a is irreducible.

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Solutions to (Selected) HW Problems

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تاریخ انتشار 2014