I've just started reading about the relation between cyclic modules and Jordan Normal Form and, being honest, I've quite a doubt. The text I am using says that "clearly", the following assumption is true:
Let $A=K[T]$, where $K$ is a field and $M=K^n$ is a cyclic $A$-module and $\phi: M \rightarrow M$ such that $\phi(\sum a_iv_i) = U.(a_i)$, where $(a_i)$ is a row vector and $U$ is a $r\times r$ matrix with coefficients in K, then $U$ has a jordan normal form with only one Jordan block.
Could, anyone, help me understanding why this is to "trivial"?
We know that, by the proof of Theorem (Jordan Normal Form) that, for an endomorphism $\varphi: M \rightarrow M$, where $\varphi$ acts as $T$, $\varphi$ leaves the submodules $M_i = K[T]/\langle (T-\lambda_i)^{e_i})\rangle$ of $M$ invariant. Thus, to find the Jordan Normal Form we can concentrate on finding appropriate matrix representations $U_i$ for each restriction $\varphi_i: M_i \rightarrow M_i$ of $\varphi$ up to the submodules $M_i$ of $M$. Then, putting the bases $\mathcal{B}_i = \{ [1],[T -\lambda_i], ... , [(T-\lambda_i)^{e_i-1}]\}$ of $M_i$ together to a basis $\mathcal{B} = \bigcup \mathcal{B}_i$ of $M$, yields to a matrix representation for $\varphi$ as desired.
Since $U$ has only one Jordan block, there is only one $\varphi := \varphi_1: M \rightarrow M$ and only one eigenvalue $\lambda$, such that that $\mathcal{B} = \{ [1],[T-\lambda], ... , [(T-\lambda)^{e})]\}$ is a basis of $M$. By the other hand, note that $\mathcal{B} = \langle T-\lambda \rangle$ and, thus, $M$ is cyclic, as desired.