# Can you have arbitrarily long chains of submodules in a noetherian $R$-module?

If we look at a finite dimensional vector space over a field $F$ as a noetherian $F$-module, we can view the dimension of the vector space as the length of the maximal ascending chain of subspaces. A chain being a sequence of subspaces which contain each other, without multiplicity, e.g., $V_1 \subsetneq V_2 \subsetneq\ ...\ \subsetneq V_n$.

Can we extend the same idea to any noetherian $R$-module? Or is there an example of a noetherian $R$-module which for any $N > 0$ has a chain of length larger than $N$?

## 2 Answers

This does not extend to $R$-modules. For example, consider $\mathbb Z$ as a $\mathbb Z$-module, in which we have arbitrarily large chains of submodules of the form $(2^n)\subset (2^{n-1})\subset\cdots \subset (2)\subset \mathbb Z$. More generally, we have arbitrarily large chains for any non-Artinian $R$-module, and so in particular for nontrivial modules over a non-Artinian ring $R$ (which in the case of Noetherian rings is the same as saying $\dim R>0$).

• Thanks! I knew there was a simple example that I was missing. Just a clarification. Are you saying that any $R$-module which is noetherian and artinian has a maximal chain whose length I can define as the dimension? Is there any interesting cases of noetherian and artinian $R$-modules which are not finite dimensional vector spaces? – zrbecker Mar 2 '12 at 9:09
• You should learn about composition series and the Jordan-Hölder theorem for modules. – Curufin Mar 2 '12 at 11:18

Consider the polynomial ring $k[x]$. This ring is clearly noetherian.

For every natural number $n$ there is a chain of ideals $$(x^n)\subset (x^{n-1})\subset\ldots\subset (x)\subset k[x]$$ of length $n+1$.