Finitely generated abelian groups

[ehrenfest]

[SOLVED] finitely generated abelian groups

Homework Statement

My book states that
(\mathbb{Z} \times\mathbb{Z} \times\cdots \times \mathbb{Z})/(d_1\mathbb{Z} \times d_2\mathbb{Z} \times \cdots d_s\mathbb{Z} \times {0} \times \cdots \times {0})
is isomorphic to
\mathbb{Z}_{d_1} \times\cdots \times \mathbb{Z}_{d_s} \times \mathbb{Z} \times\cdots \times \mathbb{Z}

with absolutely no explanation of any sort. I don't know why this is so obvious to everyone because it is NOT TRUE that you can just mod out things by there components as you can clearly see from the fact that: (Z_4 \times Z_6)/<(2,3)> is not isomorphic to Z_2 \times Z_3.
I have absolutely no idea what thought process went into that statement above and why they think they can just mod things out by there components when that is just wrong.

Homework Equations

The Attempt at a Solution

 

[morphism]

Modding out by <(2,3)> is not quite like what they're doing.

Can you think of a surjective homomorphism from RxS onto (R/I) x (S/J)? What's its kernel?

 

[ehrenfest]

Define a homomorphism \phi: (\mathbb{Z} \times\mathbb{Z} \times\cdots \times \mathbb{Z}) \to \mathbb{Z}_{d_1} \times\cdots \times \mathbb{Z}_{d_s} \times \mathbb{Z} \times\cdots \times \mathbb{Z}

as follows. For each of the components 1 \leq i \leq s, take that coordinate to itself mod d_i. For the rest of the cosets take that coordinate to itself. Then the kernel is obviously what we are modding out by. Then apply the first isomorphism theorem. Wow.