{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ahw1 - Math 5126 Wednesday January 23 First Homework...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 rm , and then we let I be the kernel). Note that I = 0 because M is not free, so there exists 0 = x I . Then x ( R / I ) = 0 which shows that M is a torsion module.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}