Hilbert’s Basis Theorem

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:

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 R be a Noetherian ring. Then R[X] is Noetherian, and via induction, R[X_1,\dots,X_k] is Noetherian as well. 

Proof: Assume towards contradiction that R[X] was not Noetherian. Using the Axiom of Choice, we may select a sequence of polynomials f_0,f_1,\dots, f_k,\dots such that f_k\in R[X]\setminus (f_0,\dots,f_{k-1}) and is of minimal degree in R[X]\setminus(f_0,\dots,f_{k-1}). For each k, let a_k be the leading coefficient of f_k, and consider the chain of ideals

(a_0)\subset (a_0,a_1)\subset\dots\subset (a_0,a_1,\dots,a_k)\subset\dots\subset R.

Now R is Noetherian, so this chain stabilizes for some integer N. Thus a_{N+1}\in (a_0,\dots,a_N), so that a_{N+1}=\sum_{i=0}^Nr_ia_i for some r_i\in R. Now consider the polynomial

\left(\sum_{i=0}^Nr_if_ix^{\text{deg}(f_{N+1})-\text{deg}(f_i)}\right)-f_{N+1}.

First, this is well defined, since the degree sequence of f_0,f_1,\dots is non-decreasing, so each summand is multiplied by a nonnegative power of x. Next, this polynomial lies outside of the ideal (f_0,\dots,f_N) since f_{N+1} lies outside of this ideal. Finally, note that by choice of the r_i, the degree of this polynomial is strictly less than the degree of f_{N+1}. This contradicts our original choice of f_{N+1}, and hence it must be that R[X] is Noetherian.

\square.

Some concluding remarks. First, using induction and the fact that R[X_1,\dots,X_k]\cong (R[X_1,\dots,X_{k-1}])[X_k], 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!

Advertisements
This entry was posted in Ring and Field Theory. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s