1
$\begingroup$

My knowledge of algebra is undergraduate linear algebra, so I apologize for my complete ignorance.

Thinking about Jordan normal forms, I unintensionally came to an idea that turned out to be called a cyclic decomposition of a PID module. More precisely, if I have got it correctly, any finitely generated module of a PID can be written as a direct sum of a finite number of cyclic modules and a simple module. I wonder what is wrong with infinitely generated modules: why can't they be written as infinite directs sums of cyclic modules plus a simple module? Intuitively, it seems to me that this has to work in arbitrary dimensions. What I have in mind is in fact a Euclidean domain, to be honest, because in my arguments I use the Euclidean algorithm. I guess that if smart people have chosen to content themselves to the finite case, then there should be something wrong with infinite. I would like to understand in simple terms (undergraduate level) what the matter is there.

To be more precise, here are two questions.

  1. Whether and why does cyclic decomposition fail for infinitely generated PID modules.

  2. The same question for modules of a Euclidean domain.

Thank you.

Edit Prof. Steinberg's comment below made me understand that I had formulated my hypothesis incorrectly. What I actually had in mind is the following:

Every module of a PID (or maybe Euclidean domain) can be written as a direct sum of cyclic submodules plus a module of (i.e., vector space over) the field of fractions of the domain.

This seems to be pretty easy, but is it actually true in this generality? Thank you.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Q (the rationals) cannot be written as a direct sum of cyclic Z-modules. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 3, 2016 at 22:41
  • 1
    $\begingroup$ Thanks for the comment. A good one, actually. This is the situation where the module is non cyclic yet indecomposable. For some reasons I haven't met this in the examples I have in mind. I have to think about it. In view of this my question is probably either trivial or needs modification. $\endgroup$
    – Bedovlat
    Commented Apr 3, 2016 at 23:03
  • 1
    $\begingroup$ I think this is a reasonable question, but even the behaviour of abelian groups (i.e. $\mathbb{Z}$-modules) shows the situation is far more complicated. Some examples to bear in mind are the divisible Prufer groups $\lim \mathbb{Z}/p^n \mathbb{Z}$ (direct limit), the $p$-adic integers $\mathbb{Z}_p$, subgroups of $\mathbb{Q}$ such as $\{ n/2^i3^j : i,j \in \mathbb{N}_0, n \in \mathbb{Z} \}$ and the Baer--Specker group. None of these has a vector space direct summand. $\endgroup$
    – Mark Wildon
    Commented Apr 4, 2016 at 13:55

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .