I am trying to understand a theorem of the book Lineare Algebra by Bosch. It is on page 228 and the corollary 7.
Let $V$ be a vector space 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 $\lambda_1,…,\lambda_r\in F$ 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 proper subspaces i.e. $A,B\neq 0$ which are $f$-invariant and $A\oplus B=U$
$p_f$ is the minimal polynomial, i.e. the uniquely determined polynomial for which $p_f\neq 0$ and for all $r\in K[T]:r(f)=0\Rightarrow p_f|r$
The book first argues that $V\simeq K[T]/(p_f)$. I have understood this part but then he argues that, because of the uniqueness of the decomposition in theorem 6, $V$ is $f$-indecomposable.
My first question is: how does the author conclude from theorem 6 that $V$ is indecomposable?
The theorem $6$ says that
If $V$ is a finite-dimensional $F$-vector space with an endomorphism $f:V\rightarrow V$ then there exist pairwise disjoint monic (and thus not associated) prime polynomials $p_1,…,p_r\in K[T]$ and natural numbers $1\leq n(i,1)\leq…. \leq n(i,s_i)$ and $f$-cyclic subspaces $V_{ij}\subset V,i=1,…,r,j=1,…,s_i$ with
$$V=\bigoplus_{i=1}^{s}\bigoplus_{j=1}^{s_i}V_{ij},\quad V_ij\simeq K[T]/(p_i^{n(i,j)})$$,
in the sense of $K[T]$-modules. Furthermore $p_i^{n(i,j)}$ is the minimal i.e characteristic polynomial of $f|_{V_{ij}}$.
In this decomposition the prime polynomials $p_1,…,p_r$, the numbers $n(i,j)$ and the vector spaces $V_i=\bigoplus_{j=1}^{s_i}V_{ij}$ for $i=1,..,r$ are uniquely determined, but not necessarily the subspaces $V_{ij}$ themselves; we have
$$V_i=\bigcup_{n\in \mathbb{N_0}}\ker p_i^n(f)$$
Additionaly $p_f=p_1^{n(1,s_1)}\cdot … \cdot p_r^{n(r,s_r)}$ is the minimal polynomial of $f$ and one has
$$\dim_K V_{ij}=n(i,j)\cdot \deg p_i$$
and
$$\dim_KV=\sum_{i=1}^r\sum_{j=1}^{s_i}n(i,j)\cdot \deg p_i$$
I have the feeling that the bold part is important for the proof but I cannot see why.
My second question is that the author claims that every $V_{ij}$ is $f$-indecomposable.
I really don't understand the proof and help for the two questions would be much appreciated.
Best Answer
For your first question, it is important to recall (from your title) that $p_f$ is a power of an irreducible polynomial, say $p_f=P^m$ where $P$ is irreducible. Then assume $V=A\oplus B$. You can apply the theorem on $A$ and on $B$ since they are $f$-stable, so you will have decompositions $A=\bigoplus A_{ij}$ and $B=\bigoplus B_{ij}$. This gives a decomposition $V=\bigoplus A_{ij} \oplus \bigoplus B_{ij}$, with each factor isomorphic to some $F[T]/p_i^{n_i}$.
But you already have a decomposition with one factor: $V\simeq F[T]/P^m$. So using the uniqueness in the theorem, those two decompositions have to be the same, so actually in $V=\bigoplus A_{ij} \oplus \bigoplus B_{ij}$ there can be only one factor, so either $A$ or $B$ is $0$.
Your second question is a special case of the first one: each $V_{ij}$ is a $f$-cyclic subspace, with the minimal polynomial of $f$ restricted to $V_{ij}$ being a power of an irreducible polynomial, so it is indecomposable. (Unless the fact that the $V_{ij}$ are indecomposable is used in the proof of the theorem ? That is not really clear in your question.)