0
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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 $A$?

$\endgroup$
1
  • $\begingroup$ It was trivial but I didn't see it :( ... Thanks $\endgroup$ – JeSuis Sep 24 '16 at 19:17
0
$\begingroup$

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 $A=\ker(\phi) \oplus A'$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.