Infinite dimensional representation such that every subrepresentation is reducible 
Let $V$ be a nonzero finite dimensional representation of an algebra $A$. 
a) Show that
  it has an irreducible subrepresentation. 
b) Show by example that this does not always hold for infinite dimensional representations.

I did not have any problems with part a), but I'm struggling to find an example for part b). Any help?
 A: Let $A=\mathbb{C}[x]$ and consider the regular representation of $A$ acting on itself. Every nonzero element generates a subrepresentation isomorphic to $A$, so this definitely does not have any irreducible subrepresentations.
A: Try the following. $A$ needs to be infinite dimensional, so let's use $A=\Bbb{C}[C_\infty]$, the group algebra of the infinite cyclic group $C_\infty=\langle c\rangle$. Let $V=\Bbb{C}[T,T^{-1}]$ be the ring of Laurent polynomials, i.e.
$$
V=\{a(T)=\sum_{i=m}^n a_iT^i\mid m,n\in\Bbb{Z}, m\le n, a_i\in\Bbb{C}\}.
$$
We can turn $V$ into an $A$-module by letting the element $c^j, j\in\Bbb{Z}$, act by shifting $j$ positions. IOW,
$$c^j\cdot a(T)=T^ja(T).$$
I then claim that $V$ has no irreducible submodules. This is seen as follows. Assume contrariwise that $W\neq\{0\}$ is an irreducible submodule. If $a(T)=\sum_{i=m}^na_iT^i\neq0$ has $a_m\neq0\neq a_n$, then let's call $(n-m+1)$ the width of $a(T)$. So all the non-zero elements of $V$ have a width that is a positive integer. Therefore any submodule of $V$ has elements of a minimal width. But if $W$ is an irreducible submodule then the space $(T-1)W=\{(T-1)a(T)\mid a(T)\in W\}$ is a subspace of $W$. Because the minimum width of elements of $(T-1)W$ is one higher than the minimum width of elements of $W$, it follows that $(T-1)W$ is a proper subspace of $W$.
But $(T-1)W$ is clearly also an $A$-submodule of $W$ contradicting the assumption that $W$ is irreducible. Therefore $V$ has no irreducible $A$-submodules.
A: Here is my try at this problem: 
Let $k$ be a field. $A:=k[x_1, x_2]$, which is Noetherian by Hilbert Basis Theorem. Hence, if $I \triangleleft k[x_1, x_2]$, $I= \langle f_1, \ldots, f_j \rangle$. Then $I' := \langle f_1 g, \ldots, f_j g \rangle \subsetneq I$ for some $g \not| f_1$.  If we had an equality,$$f_1 = \sum \alpha_i f_ig \Rightarrow  g | f_1 $$
 Contradiction (where I assumed $f_i$s are nonzero).  
A: I'm not quite sure if this contraption is correct, but I'll give it a go, because I'm also interested in this question.

b) Consider an algebra $A = \mathbb{Z}$ and its infinite-dimensional representation $V\subset\mathbb{Z^\infty}$ defined by recurrence relation $c_{i+1} = c_{i} + 1$ together with a homomorphism of algebras $\rho: A \rightarrow EndV$ defined by an operator $T$ such that:

$$\rho(a) = T_a: (c_i,c_{i+1},c_{i+2},\dots) \rightarrow (c_{i+a},c_{i+a+1},c_{i+a+2},\dots)$$
Irreducibility of this representation follows immediately from the observation that every $\rho(a)$, $a \in A$ has an infinite image.
