Computing the invariant factors and elementary divisors of finitely generated module over a PID

Question

I have just encountered this question in my abstract algebra class dealing with finitely generated modules over PIDs stating the following:

Let $ D = \mathbb{R}[x] $ be the ring of polynomials over the reals in variable x, and let M be a D module, with elements $ v_1,v_2,v_3,v_4 \in M $ such that $ M = Dv_1 \oplus Dv_2 \oplus Dv_3 \oplus Dv_4 $, satisfying: $ ann(v_1) = (x-1)^3 $ ; $ ann(v_2) = (x^2+1)^2 $ ; $ ann(v_3) = (x-1)(x^2+1)^4 $ ; $ ann(v_4) = (x+2)(x^2+1)^2 $

We are now asked to find the invariant factors and elementary divisors of M

Now all I know is that M is a finitely generated cyclic module over a PID as the direct sum of cyclic modules and I know the fundamental theorems on finitely generated modules over PIDs but I do not know how to extract the invariant factors and elementary divisors from the statement above, so I am hoping for some help on this with a bit of explanation, thanks all helpers

Answer 1

you are given a cyclic decomposition (one generator for each summand). the elementary divisors arise from a cyclic, prime power, decomposition. hence you only have to factor all your anihilators into prime powers. That gives $x+2$, $(x-1)$, $(x-1)^3$, $(x^2+1)^2$, $(x^2+1)^2$, $(x^2+1)^4$. Now to get the invariant factors arrange these as relatively prime products in the only possible way so they successively divide each other. That is, $(x^2+1)^2$, $(x-1)(x^2+1)^2$, $(x+2)(x-1)^3(x^2+1)^4$. voila'.