Math 5126 Wednesday, January 23 First Homework Solutions 1. Since R / I is a Noetherian ring, it is Noetherian as an R / I-module. The submodules of R / I considered as an R-module are precisely the same as the submodules of R / I considered as an R / I-module; thus R / I is a Noetherian R-module. Obviously I is a Noetherian R-module, because | I | is ﬁnite. Therefore R is a Noetherian R-module ( N and M / N are Noetherian if and only if M is Noetherian) and the result is proven. 2. If M is free (and nonzero), then M has a submodule isomorphic to R and so certainly M is not a torsion module. Conversely suppose M is not free. Since M is a nonzero cyclic R-module, it is isomorphic to R / I for some ideal I of R (reason: if M = Rm , then we can deﬁne an R-module epimorphism R ± M by r 7→ rm , and then we let I be the kernel). Note that I 6 = 0 because M is not free, so there exists 0 6 = x ∈ I . Then x ( R / I ) = 0 which shows that M is a torsion module.
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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.