# Structure theorem for abelian groups

1. Jul 3, 2011

### Jim Kata

So say you have a presentation matrix A for a module, and you diagonalise it and you get something like diag[1,5] well you can interpret that as A can be broken down as the direct sum of 1+Z/5Z. That is the trivial group of just the identity plus cyclic 5. What if your module is defined over the ring of polynomials, and after you diagonalise your presentation matrix you get something like
diag[1,(t-1)^2(t-2)]? How do I interpret this? What is the cyclic group something like
(t-1)^2(t-2) would represent?

2. Jul 4, 2011

### micromass

I assume you are working with k[X]. Your presentation matrix of M is

$$A=\left(\begin{array}{cc} 1 & 0\\ 0 & (X-1)^2(X-2)\\ \end{array}\right)$$

Let the map corresponding to this matrix by $\alpha:k[X]^2\rightarrow k[X]^2$. Then we know that

$$M\cong k[X]^2/\alpha k[X]^2$$

Now, let {e,e'} be a basis corresponding to A, then we know that $\alpha k[X]$ is generated by $\{e,(X-1)^2(X-2)e'\}$

Thus

$$M\cong k[X]^2/\alpha k[X]^2=k[X]\times (k[X]/(X-1)^2(X-2))$$