What is a Noetherian ring?

Noetherianness is about making sure small modules don’t have big parts. But how do you measure how big a module is? Remember that while every vector space has a dimension—the size of a basis—most modules don’t have a basis. We can still talk about a generating set for an R-module M, though, like \{1, 1/2, 1/3, \dots\} for the \mathbb{Z}-module \mathbb{Q}: a subset of M for which every element can be written as an R-linear combination of those generating elements.

Question: For a vector space, its dimension is its minimum number of generators. Can we use the minimum number of generators as a measure of the size of a module in general?

Answer: Not really. You’d expect that for any good notion of size \mu, we’d have \mu(M)\leq\mu(N) whenever a module M is contained in a larger module N (i.e. M is a submodule of N). Dimension of vector spaces certainly has this property, but the minimum size of a generating set for modules does not.

Example. Take the polynomial ring in two variables R = \mathbb{R}[x,y]. As an R-module itself, R can be generated by a single element, 1. But the submodule of R generated by x and y (i.e. the elements of R that are R-linear combinations of x and y) cannot be generated by fewer than two elements. So in this sense we have a smaller module with a bigger “size.”

In general, an R-submodule of a ring R is called an ideal of R, and the ideal generated by a collection of elements \{a_1,a_2,\dots\} is denoted (a_1,a_2,\dots). So denoting the minimal number of generators of an R-module M by \mu_R(M), in the above example we have \mu_R(R) = 1 but \mu_R((x,y)) = 2.

If we modify the example so that R is a polynomial ring in infinitely many variables, R=\mathbb{R}[x_1,x_2,\dots], we get an even more striking example: \mu_R(R) is still 1, but \mu_R((x_1,x_2,\dots)) is now infinite.

Question: Do you need such a big ring to get an example of a finitely generated module with a submodule that isn’t finitely generated?

Answer: Yes! Rings with the property that their finitely generated modules only have finitely generated submodules are called Noetherian—so a polynomial ring in infinitely many variables is not Noetherian—and there are two main theorems that make it easy to prove rings are Noetherian:

  1. It’s sufficient to check only that all ideals of R are finitely generated to prove that R is Noetherian.

So since every ideal of \mathbb{Z} can be generated by a single element—i.e. every ideal of \mathbb{Z} is principal—we automatically know that every submodule of a finitely generated \mathbb{Z}-module is also finitely generated.

The other theorem has to do with finitely generated algebras over a field, of which \mathbb{R}[x,y] is one but \mathbb{R}[x_1,x_2,\dots] isn’t. An algebra over a ring R is just another ring A equipped with a ring homomorphism R\to A; a generating set for an algebra is a subset \{a_i:i\in I\}\subseteq A such that every element of A can be written as a polynomial in the elements of a_i with coefficients in R. (In other words, the ring homomorphism from the polynomial ring R[x_i:i\in I] to A sending each x_i to a_i is surjective.) A finitely generated R-algebra is just an algebra with a finite generating set, i.e. a quotient of the polynomial ring \mathbb{R}[x_1,\dots,x_n] for some n.

  1. Every finitely generated algebra over a field is Noetherian.

The proof goes by first checking that fields are Noetherian—they only have two ideals, \{0\} and the whole field, which are both principal—then showing that if R is a Noetherian ring then R[x] is Noetherian too (this is called the Hilbert Basis Theorem) and so is any quotient of R. By induction, this gives Noetherianness of every finitely generated algebra over a field.

This means that any non-Noetherian algebra over a field must need infinitely many algebra generators. In that context, even though the number of generators of a module isn’t a good notion of size, we can still distinguish “small” modules (the finitely generated ones) from “large” ones (the rest), and a submodule of a small module is small.

Advertisements

Has this post helped you, or left you with further questions? Let me know with a comment!

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