Determining some properties of $\mathbb{C}^3$ as a $\mathbb{C}[x]$ module

I'm wondering how I could answer the following questions about $N = \mathbb{C}^3$ with a $\mathbb{C}[x]$-module structure given by the linear map $A$ on $\mathbb{C}^3$:

$$A = \left(\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\right)$$

(i) What is the rank of $N$?

(ii) Is $N$ cyclic?

(iii) Write $N$ as the direct sum of 3 non-zero submodules

(iv) Is $N$ isomorphic to $\mathbb{C}[A]$ as a $\mathbb{C}[x]$-module

(v) Do your answers change if we replace $\mathbb{C}$ with $\mathbb{R}$ or $\mathbb{Z}_7$?

For (i), I think the rank must be zero since (as a $\mathbb{C}$ vector space) $\mathbb{C}[x]$ is infinite dimensional while $N$ is not. So the direct sum decomposition of $N$ according to the structure theorem can't contain $\mathbb{C}[x]$ as a summand. Is this correct? Intuitively it seems so, but going from vector spaces to modules a lot of intuition becomes wrong so I don't know.

For (ii), I think it is cyclic since I think $e_1$ is a cyclic vector, we can multiply by $A$ to get $e_2$ and $A^2$ to get $e_3$ so we can write any element as $p(A) e_1$ for some polynomial $p$.

I am confused about (iii) and (iv). I know that $\mathbb{C}[A] \cong \frac{\mathbb{C}[x]}{(m_A(x))}$ which I could then decompose further using the Chinese Remainder Theorem but how can I find what $N$ is isomorphic to? How do I find the invariant factors using $A$?

For (v) I believe that the main difference is that $m_A(x)$ will factorise differently in $\mathbb{R}$ and $\mathbb{Z}_7$.

You are correct about $(1)$ and $(2)$. To tackle $(3)$, note that a submodule of $N$ is just an $A$-invariant subspace of $N = \mathbb{C}^3$. The matrix $A$ is the companion matrix of the polynomial $x^3 - 1$ and so the characteristic and the minimal polynomials of $A$ is $x^3 - 1 = (x - 1)(x^2 + x + 1)$. Over $\mathbb{C}$, this polynomial has three distinct roots and so you have three distinct eigenvalues $\lambda_1,\lambda_2,\lambda_3$ (the cubic roots of $1$) with eigenvectors $v_1,v_2,v_3$ which you can find explicitly. Each $N_i = \operatorname{span} \{ v_i \}$ gives you a submodule and $N = \bigoplus_{i=1}^3 N_i$.
Regarding $(4)$, the module $N$ is isomorphic to $\mathbb{C}[x]/(x^3 - 1)$ via the isomorphism $$[1] \mapsto e_1, [x] \mapsto e_2, [x^2] \mapsto e_3.$$
Finally, the answer to $(3)$ will change based on whether $A$ is diagonalizable over $\mathbb{F}$ while the answer to $(1),(2),(4)$ won't change.