Math 5126Wednesday, January 23First Homework Solutions1. SinceR/Iis a Noetherian ring, it is Noetherian as anR/I-module. The submodulesofR/Iconsidered as anR-module are precisely the same as the submodules ofR/Iconsidered as anR/I-module; thusR/Iis a NoetherianR-module. ObviouslyIis aNoetherianR-module, because|I|is finite. ThereforeRis a NoetherianR-module (NandM/Nare Noetherian if and only ifMis Noetherian) and the result is proven.2. IfMis free (and nonzero), thenMhas a submodule isomorphic toRand so certainlyMis not a torsion module. Conversely supposeMis not free. SinceMis a nonzerocyclicR-module, it is isomorphic toR/Ifor some idealIofR(reason: ifM=Rm,then we can define anR-module epimorphismRMbyr→rm, and then we letIbe the kernel). Note thatI=0 becauseMis not free, so there exists 0=x∈I. Thenx(R/I) =0 which shows thatMis a torsion module.
This is the end of the preview.
access the rest of the document.