I am practicing for myself how to find all possible elementary divisors and the corresponding invariant factors for an $R$-module of order $(x-1)^3(x +1)^2$ where $R = k[x]$ and $k$ is a field.
But it is not very much clear in my mind what are the steps for doing that. I know that at some step I would find the characteristic polynomial of degree 1,2,3 or 1,2 (or maybe I am wrong in that). I think I understand how to find elementary divisors for abelian groups, but how that leads me to the way of finding it for modules?
I have read the question but still I do not understand what I can do in my case as the givens in my problem are different.
Any help will be appreciated!
EDIT:
The elementary divisors are:
Since the elementary divisors are all combinations of the prime ideals we have, then they are:
1-$\{ (x-1), (x-1), (x-1), (x+1), (x+1)\}$
2-$\{ (x-1),(x-1)^2, (x+1), (x+1)\}$
3-$\{(x-1)^3, (x+1), (x+1)\}$
4-$\{ (x-1), (x-1), (x-1), (x+1)^2\}$
5-$\{ (x-1), (x-1)^2, (x+1)^2\}$
6-$\{(x-1)^3,(x +1)^2\}$
Sorry I wrote them before seeing the answer below, thanks @user26857
And the corresponding invariant factors are:
1-$\{(x-1), (x-1)(x+1), (x-1)(x+1),\}$
2-$\{(x-1)(x+1), (x-1)^2(x+1) \}$
3-$\{(x+1), (x-1)^3(x+1)\}$
4-$\{(x-1), (x-1), (x-1)(x+1)^2\}$
5-$\{(x-1)^2, (x-1)(x+1)^2, (x-1)^2\}$
6-$\{ (x-1), (x-1)^2(x+1)^2\}$
7-$\{(x-1)^3(x+1)^2\}$
Best Answer
I would start with the elementary divisors. There are six possibilities corresponding to the partitions of the exponents, three for $3$ and two for $2$, that is,
$x-1,x-1,x-1,x+1,x+1$
$x-1,x-1,x-1,(x+1)^2$
$x-1,(x-1)^2,x+1,x+1$
$x-1,(x-1)^2,(x+1)^2$
$(x-1)^3,x+1,x+1$
$(x-1)^3,(x+1)^2$
I am confident that you can find the invariant factors in each case.