Torsion 2

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December 4th 2011, 08:03 AM
jcir2826

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Torsion 2

If G is a torsion abeliean group
December 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?
December 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
December 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?
December 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.
December 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 $\text{Hom}_\mathbb{Z}(\mathbb{Z}_5,\mathbb{Z}_7)$ ?
December 4th 2011, 02:11 PM
jcir2826

Re: Torsion 2

All the Z-homomorphism from Z_5 to Z_7.
December 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.
December 4th 2011, 02:41 PM
jcir2826

Re: Torsion 2

Thanks I get your point.
December 6th 2011, 05:41 AM
jcir2826

Re: Torsion 2

Hom_z(Z_5,Z_7) is isomorphic to Z_7. I am still not sure why the last three summans can be reduced. Can it be since there is an isomorphism from the from Hom(diret product of M_p) to direct sum (M_p,R^m)
December 6th 2011, 03:07 PM
Drexel28

Re: Torsion 2

Quote:

Originally Posted by jcir2826

Hom_z(Z_5,Z_7) is isomorphic to Z_7. I am still not sure why the last three summans can be reduced. Can it be since there is an isomorphism from the from Hom(diret product of M_p) to direct sum (M_p,R^m)

$\text{Hom}(\mathbb{Z}_5,\mathbb{Z}_7)\cong\{0\}$ . 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?