Let be a Noetherian ring. Prove that , as an -module in the usual way, is also Noetherian.
Recall that a ring is Noetherian if every ideal is finitely generated, and a module is Noetherian if every submodule is finitely generated. (That is, a ring is Noetherian (as a ring) if it is Noetherian (as a module) over itself.)
We will proceed by induction on .
For the base case, , is certainly Noetherian as a module over .
For the inductive step, suppose is Noetherian. Now let be a submodule; our goal is to show that must be finitely generated. To that end, let . That is, is the collection (in ) of all zeroth coordinates of elements in .
We claim that is an ideal. If , then there exist elements and in such that and . Now since is a submodule, we have for all , so that . We clearly have , so that by the submodule criterion, is an ideal of .
In particular, since is Noetherian, the ideal is finitely generated – say by . We will let be an element of whose zeroth coordinate is . Now let . Now for some , and so is an element of whose first coordinate is 0. In particular, we have , where . (We showed the direction, and the inclusion is clear.) Now is an ideal of , which is Noetherian by the induction hypothesis. So is finitely generated as an -module, and thus is finitely generated over .
Since was an arbitrary submodule of , is Noetherian as a module.