Math 5126 Monday, January 28 Second Homework Solutions 1. Section 12.1, Exercise 21 on page 472. Prove that a ﬁnitely generated module over a PID is projective if and only if it is free. Free modules are projective, even if the ring is not a PID. Conversely suppose P is a ﬁnitely generated projective module over a PID. Since P is a submodule of a free module, it is torsion free and so by the structure theorem for ﬁnitely generated modules over a PID, we see that P is free. 2. (a) If M ∩ N = 0, then there exist 0 6 = m ∈ M and 0 6 = n ∈ N . We now have nm ∈ M because M is an R-submodule, and mn ∈ N because N is an R-submodule. Since M ∩ N = 0, we deduce that mn = 0, which contradicts the fact that R is an integral domain. (b) Suppose I is cyclic, say I = R f . Then x = r f and y = s f for some r , s ∈ R . Note that f is not a unit, because R f 6 = R . Furthermore x and y are irreducible in the UFD R , hence r and s are units. We deduce that x and y
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This note was uploaded on 01/29/2009 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.