If G is a torsion abeliean group

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- Dec 4th 2011, 08:03 AMjcir2826Torsion 2
If G is a torsion abeliean group

- Dec 4th 2011, 09:39 AMDrexel28Re: Torsion 2
I'll prod you in the right direction with 1., and then we'll work through 2. TOGETHER, ok? You know that $\displaystyle M\cong R^m\oplus M_{p_1}\times\cdots M_{p_n}$ and so by the way Hom reacts with products

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

So, now, tell me, why do the last three direct summands (groups of direct summands) dissapear? - Dec 4th 2011, 10:31 AMjcir2826Re: Torsion 2
Is it because the coproducts and summands can be switched? Like on Rotman advanced algebra pg 431

- Dec 4th 2011, 12:05 PMDrexel28Re: Torsion 2
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 $\displaystyle \text{Hom}_R(M_p,M_q), p\ne q$, $\displaystyle \text{Hom}_R(M_p,R^m)$, and $\displaystyle \text{Hom}_R(R^m,M_p)$ can be reduced. How can we reduce each of these?

- Dec 4th 2011, 01:36 PMjcir2826Re: 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 PMDrexel28Re: Torsion 2
- Dec 4th 2011, 02:11 PMjcir2826Re: Torsion 2
All the Z-homomorphism from Z_5 to Z_7.

- Dec 4th 2011, 02:30 PMDrexel28Re: Torsion 2
- Dec 4th 2011, 02:41 PMjcir2826Re: Torsion 2
Thanks I get your point.

- Dec 6th 2011, 05:41 AMjcir2826Re: 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)

- Dec 6th 2011, 03:07 PMDrexel28Re: Torsion 2