# An abelian group $A$ equal to $\ker\phi\oplus A'$ where $A'=\Bbb{Z}a_1\oplus\cdots\oplus\Bbb{Z}a_r$.

Let $$A$$ be abelian group and $$L$$ a free abelian group of finite rank, and $$\phi:A\to L$$ a morphism.

Choosing a base $$(l_1,\ldots,l_r)$$ of $$L$$ and $$(a_1,\ldots,a_r)$$ such that $$\forall i: \phi(a_i)=l_i.$$

Show that $$\ker\phi\oplus A'=A$$ where $$A'=\Bbb{Z}a_1\oplus\cdots\oplus\Bbb{Z}a_r$$, not sure how can I prove that, I can prove that the set $$(a_1,\ldots,a_r)$$ is linearly independent.

If I 'play' with $$x\in A$$ such that $$x=y+a'$$ where $$y\in \ker\phi$$ and $$a'\in A'$$, I get $$\phi(x)=\phi(a')$$

As $$x\in A$$ we have $$\phi(x)\in L$$ wich means that $$\phi(x)=\sum_{i=1}^r \alpha_il_i$$ and $$a'\in A'$$ means that $$a$$ can be written as $$a'=\beta_1a_1+\cdots+\beta_ra_r$$ and then $$\phi(a')=\beta_1l_1+\beta_2l_2+\cdots+\beta_rl_r$$, not sure that helps either.

In fact, I am pretty sure that I am not understanding correctly the 'role' of $$(a_1,\ldots,a_r)$$ being linearly independent.

First of all, $\{a_1,\ldots,a_r\}$ must be linearly independent; otherwise $\phi(a_i) = l_i$ would be impossible. (Any linear relation between the $a_i$ would give one on the $l_i$ via $\phi$.)

Now given an element $a \in A$, you get $\phi(a) = \sum_i c_i l_i$ for a unique tuple of integers $(c_i)$. Then $a - \sum_i c_i a_i$ is in $\ker \phi$. Can you show that $a = \left( a - \sum_i c_i a_i \right) + \left(\sum_i c_i a_i\right) \in \ker \phi + A'$ gives you a direct sum decomposition of $A$?

• It was trivial but I didn't see it :( ... Thanks – JeSuis Sep 24 '16 at 19:17

Since $\psi=\phi|_{A'}: A' \to L$ is a bijection, it is pretty straightforward that any element not in the kernel has at least one non-zero coefficient in the $a_i$s expression. That is, if $x \not\in \ker(\phi)$ then $0 \neq \phi(x) = b_1 l_1 + \dots + b_n l_n$, where not all $b_i$ are zero. But then applying $\psi^{-1}$ you get what I said. This means that every element can be decomposed as $$x = \sum b_i a_i + (x-\sum b_i a_i)$$ which is what gives you the decomposition. It remains to prove that it's direct but, since the trivial intersection part is trivial, you get that $A=\ker(\phi) \oplus A'$