Let $A:=k[x]$ (a $k$-algebra). We know every finite representation $(\rho, V)$ of $A$ is determined by $\rho(x):V\to V$. By the Jordan Normal Form theorem, $\rho(x)$ can be written as a Jordan normal matrix under some basis. Let $J_1, \dots, J_r$ be Jordan blocks of this matrix.
Each Jordan block $J_i$ determines an invariant subspace $V_i$ and $(\rho_{V_i},V_i)$ is a subrepresentation. Now I have two questions:
(1) How to prove that each $(\rho_{V_i},V_i)$ is an indecomposable(not irreducible) representation?
(2) Are different $(\rho_{V_i},V_i)$ mutually nonisomorphic? If so, how to prove it?
One only has a decomposition into Jordan blocks if the characteristic or minimum polynomial factors into linear factors over $k$.
The characteristic polynomial of a Jordan block matrix equals its minimum polynomial and is $(x-\lambda)^n$ where $\lambda$ is the eigenvalue and $n$ is the size of the block. If this were decomposable as a representation, then each factor's minimum polynomial divides $(x-\lambda)^n$, so the factors have minimum polynomials $(x-\lambda)^a$ and $(x-\lambda)^b$ for some $a$ and $b$. Each of $a$ and $b$ is $<n$ as each is at most the dimension of the appropriate invariant subspace. But that means the original representation has minimum polynomial $(x-\lambda)^{\max(a,b)} \ne(x-\lambda)^n$, a contradiction.