I am working on a paper, which states that the dual of a finite dimensional cocyclic module is cyclic. I tried to write down a proof, but I failed and I do not know if this is true in full generality or of it is true just in my case.
Let $A$ be a complex Hopf algebra and $M$ a finite $\mathbb{C}$-dimensional cocyclic $A$-module with cocyclic vector $m \in M\setminus \lbrace 0 \rbrace$, i.e. every nontrivial submodule $U \subset M$ contains $m$. Let $M^* = \operatorname{Hom}(M,\mathbb{C})$ be the dual module. The statement is, that $M^*$ is cyclic with cyclic vector $m'$ corresponding to $m$.
What I tried so far, is the following. Let $e_1, \dots, e_n$ be a $\mathbb{C}$-basis of $M$ and let $\delta_1, \dots, \delta_n$ be the dual basis of $M^*$, i.e. $\delta_i(e_j) = \delta_{ij}$.
We can assume that $m = e_1$, in particular $m' = \delta_1$. Now since $m = e_1$ is cocyclic, there exists for every $e_i$ an element $a_i \in A$ such that $a_ie_i = e_1$. To show that $M^*$ is cyclic with cyclic vector $\delta_1$, it suffices to find elements $\tilde a_i$ such that $\tilde a_i \delta_1 = \delta_i$ for every $1 \le i \le n$.
The first obvious idea is to try to act with $a_i$ on $\delta_1$ and look what happens. It is clear that \begin{align*} (a_i \delta_1)(e_i) = \delta_1(a_i e_i) = \delta_1(e_1) = 1, \end{align*} but if we write $a_i e_j = \sum_k a_k^{ij} e_k$, we have \begin{align*} (a_i \delta_1)(e_j) = \delta_1(a_i e_j) = a_1^{ij} \end{align*} which a priori can be nonzero.
I tried to correct the elements $a_i$ to make $a_1^{ij}$ be $0$, but this only works if $n = 2$, and moreover I think, that there should be a more intrinsic reason why the dual of a cocyclic is cyclic.
In the paper I am working on, my algebra is the universal enveloping algebra $U(\mathfrak{g} \otimes \mathbb{C}[t])$ of the current algebra, where $\mathfrak{g}$ is a semisimple finite dimensional Lie algebra. My module $M$ is in fact graded, but I do not think, that this is important.
It's not true that cyclic or cocyclic modules have to be simple (e.g. Verma modules are cyclic but not always simple).
A module is cyclic if (not only if) it has unique maximal proper submodule, and a cocyclic module is one with a unique minimal nonzero submodule ($Am$, in your notation - let's call it $S$).
To prove $M^*$ is cyclic, we show it has a unique maximal submodule. Consider the proper submodule $S^\perp=\{f\in M^* : f(S)=0\} \leqslant M^*$ which I claim is unique maximal. To see this, let $T < M^*$. Then $0 \neq T^o = \{ x \in M : \forall t \in T, t(x)=0\} \leqslant M$, so $S \leqslant T^o$, so $T^{o \perp} \leqslant S^\perp$, so $T \leqslant S^\perp$ as $T \leqslant T^{o \perp}$. The only reason you need Hopf algebras is so that all of these things are submodules.