$f-cyclic\Rightarrow f-indecomposable$? I am looking for a proofverification, in the book that I am reading there is no proof for this statement. Bosch Linear Algebra page 223 f.
Let $V$ be a vectrospace over the field $F$ and $U$ a subspace, and $f$ an endomorphism $f:V\rightarrow V$. 
$U$ is called $f$-invariant iff $f(U)\subseteq U$. 
$U$ is called $f$ - cyclic if there exists a $u\in U$ such that $(f^n(u))_{n\in\mathbb{N_0}}$ is a generating system, i.e. for all $x\in U$ there exists a $m_1,...,m_r\in \mathbb{N}$ and $\lambda_{1,...,r}\in F$ such that $\sum_{i=1}^r\lambda_i 
 f^{m_i}(u)=x$
$U$ is called $f$ indecomposable iff $U$ is $f$-invariant and there exist no two real subspaces i.e. $A,B\neq 0$ which are $f$ invariant and $A\oplus B=U$
I now assume $U$ to be $f$ cyclic, now I suppose that $U$ is not indecompsable then there exists two subspaces which are both $\neq 0$ and $f-$invariant such that $A\oplus B= U$. Then either $A$ or $B$ must contain $u$ the element which induces a generating system in $U$. Withous loss let $u\in A$, because $A$ is $f$ invariant all $f^n(u)\in A$ therefore $A=U$ and $B=0$ which is a contradiction therefore $U$ is $f$-indecomposable.
 A: Short answer: no, cyclic does not imply indecomposable.
The flaw in your reasoning is that you assume $u$ must be in either $A$ or $B$. There is no reason this should be the case: $u$ must be a sum $a+b$ with $a\in A$ and $b\in B$, but there is no reason why $a$ or $b$ should be $0$.
Let me give you an example: suppose $V$ has dimension 2, choose a basis $(e_1,e_2)$, and write $A=Fe_1$ and $B=Fe_2$. Now define $f:V\to V$ such that $f(e_1)=e_1$ and $f(e_2)=-e_2$ (so $f$ is the reflection through $A$ along $B$). Clearly $V$ is not $f$-indecomposable since $A$ and $B$ are stable by $f$ (they are its eigenspaces). But if you choose any $u\in V$ which is not in $A$ or in $B$, then $f(u)$ is not colinear with $u$ (since that would mean $u$ is an eigenvector, so it would be in $A$ or $B$), so $(u,f(u))$ is a basis for $V$, and $V$ is $f$-cyclic.
If you want to understand the situation better, think about cyclic groups: $\mathbb{Z}/n\mathbb{Z}$ is cyclic (there is one element that generates the whole group), but if $n$ is not prime (say $n=pq$), then you can decompose 
 $$\mathbb{Z}/n\mathbb{Z} \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/q\mathbb{Z}.$$
If you want $\mathbb{Z}/n\mathbb{Z}$ to be indecomposable as a group, you need $n$ to be prime. Likewise, if you want $V$ to be $f$-indecomposable, you need the minimal polynomial of $f$ to be irreducible. In the example I gave above, the minimal polynomial of $f$ was $X^2-1=(X-1)(X+1)$, and this decomposition in two irreducible factors gave the decomposition of $V$ in two subspaces (for exactly the same reason; the Chinese remainder theorem works just as well with polynomials as it does for integers).
