ahw2 - Math 5126 Monday, January 28 Second Homework...

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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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