If you have had some experience with rings and commutative algebra in the past, you may have encountered the idea of a Noetherian ring. This post will not deal with explaining Noetherian rings or their importance; instead I would recommend several resources which undoubtedly handle it much more comprehensively than I can:
- Why are Noetherian rings so natural?
- Why are finiteness conditions important?
- The theory of Noetherian rings (if you have access to SpringerLink)
Today we will prove a fundamental result regarding Noetherian rings, which can be traced back to an 1890 result of Hilbert. In what follows, we will use commutative rings with identity, although analogous results exist in other cases.
Theorem: Let be a Noetherian ring. Then is Noetherian, and via induction, is Noetherian as well.
Proof: Assume towards contradiction that was not Noetherian. Using the Axiom of Choice, we may select a sequence of polynomials such that and is of minimal degree in . For each , let be the leading coefficient of , and consider the chain of ideals
Now is Noetherian, so this chain stabilizes for some integer . Thus , so that for some . Now consider the polynomial
First, this is well defined, since the degree sequence of is non-decreasing, so each summand is multiplied by a nonnegative power of . Next, this polynomial lies outside of the ideal since lies outside of this ideal. Finally, note that by choice of the , the degree of this polynomial is strictly less than the degree of . This contradicts our original choice of , and hence it must be that is Noetherian.
Some concluding remarks. First, using induction and the fact that , one can show that a polynomial ring in finitely many variables over a Noetherian ring is also Noetherian. Second, note that this proof is entirely non-constructive, and tells us nothing about how to get a generating set for an ideal. To do this we must turn to the idea of Grobner bases, which are interesting in their own right but will not be discussed here. As always, thanks for reading!