4
$\begingroup$

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 2

3

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).

  • 0
    You should learn about composition series and the Jordan-Hölder theorem for modules.2012-03-02
2

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 $n+1$.