# Torsion 2

• Dec 4th 2011, 08:03 AM
jcir2826
Torsion 2
If G is a torsion abeliean group
• Dec 4th 2011, 09:39 AM
Drexel28
Re: Torsion 2
Quote:

Originally Posted by jcir2826
If G is a torsion abeliean group

I'll prod you in the right direction with 1., and then we'll work through 2. TOGETHER, ok? You know that $M\cong R^m\oplus M_{p_1}\times\cdots M_{p_n}$ and so by the way Hom reacts with products

$\displaystyle \text{Hom}_R(M,M)\cong \prod_p \text{Hom}_R(M_p,M_p)\oplus\prod_{p\ne q}\text{Hom}_R(M_p,M_q)\oplus \prod_p \text{Hom}_R(M_p,R^m)\oplus\prod_p\text{Hom}_R(R^m ,M_p)$

So, now, tell me, why do the last three direct summands (groups of direct summands) dissapear?
• Dec 4th 2011, 10:31 AM
jcir2826
Re: Torsion 2
Is it because the coproducts and summands can be switched? Like on Rotman advanced algebra pg 431
• Dec 4th 2011, 12:05 PM
Drexel28
Re: Torsion 2
Quote:

Originally Posted by jcir2826
Is it because the coproducts and summands can be switched? Like on Rotman advanced algebra pg 431

Well, that is true in some sense, by why does that help us? What we really want to show is that each of the three Homs $\text{Hom}_R(M_p,M_q), p\ne q$, $\text{Hom}_R(M_p,R^m)$, and $\text{Hom}_R(R^m,M_p)$ can be reduced. How can we reduce each of these?
• Dec 4th 2011, 01:36 PM
jcir2826
Re: Torsion 2
Hey drexel I feel bad cause i know you are trying to help me but I just cant see the answer.
• Dec 4th 2011, 01:38 PM
Drexel28
Re: Torsion 2
Quote:

Originally Posted by jcir2826
Hey drexel I feel bad cause i know you are trying to help me but I just cant see the answer.

Ok man, so let's see if we can make things easier. What is $\text{Hom}_\mathbb{Z}(\mathbb{Z}_5,\mathbb{Z}_7)$?
• Dec 4th 2011, 02:11 PM
jcir2826
Re: Torsion 2
All the Z-homomorphism from Z_5 to Z_7.
• Dec 4th 2011, 02:30 PM
Drexel28
Re: Torsion 2
Quote:

Originally Posted by jcir2826
All the Z-homomorphism from Z_5 to Z_7.

Right. But what is it isomorphic to? I think you need to play around with some examples before you tackle this problem man, or your just going to be taking what I say as true--you need to see it for yourself.
• Dec 4th 2011, 02:41 PM
jcir2826
Re: Torsion 2
$\text{Hom}(\mathbb{Z}_5,\mathbb{Z}_7)\cong\{0\}$. Do you see why this is now true? Think about torsion, etc. etc. See if you can then use this to see why our problem is reuced?