Counting the number of $F[x]$-submodules Let $F = \mathbb{Z} / p \mathbb{Z}$ be the finite field with $p$ elements and consider the $F[x]$-module $V = F[x]/(x^2) \oplus F[x]/(x^3)$.

Question 1.  How many  $F[x]$-submodules with $p$ elements does $V$ have?

I think the answer is 5 submodules which I will list explicitly:
$\{ v \in V : v = (a,0),  \,\, a \in \mathbb{Z}_p \}$
$\{ v \in V : v = (0,b),  \,\, b \in \mathbb{Z}_p \}$
$\{ v \in V : v = (ax,0),  \,\, a \in \mathbb{Z}_p \}$
$\{ v \in V : v = (0,bx),  \,\, b \in \mathbb{Z}_p \}$
$\{ v \in V : v = (0,bx^2),  \,\, b \in \mathbb{Z}_p \}$
Is this correct or did I miss something?

Question 2.  How many cyclic $F[x]$-submodules with $p^2$ elements does $V$ have and how many noncyclic $F[x]$-submodules with $p^2$ elements does $V$ have?

This is where I get confused in constructing modules with $p^2$ elements.
Should I consider sets of the form $\{ v \in V : v = (a+bx,0),  \,\, a,b \in \mathbb{Z}_p \}$ will this still be cyclic?
$\{ v \in V : v = (ax,bx^2),  \,\, a ,b\in \mathbb{Z}_p \}$
$\{ v \in V : v = (ax,bx),  \,\, a,b \in \mathbb{Z}_p \}$ will these have $p^2$ elements and not be cyclic?
Thanks for your help.
 A: I think it's appropriate in this case to work out most of the first problem. Hopefully this will get the wheels turning.
We have projections $p_1\colon V \to F[x]/(x^2)$ and $p_2\colon V \to F[x]/(x^3)$. If $W$ is a submodule of $V$, then $p_1(W)$ and $p_2(W)$ are submodules of $F[x]/(x^2)$ and $F[x]/(x^3)$, respectively. Some thought will reveal that an $F[x]$-submodule of $F[x]/(x^2)$ is the same thing as an ideal of the ring $F[x]/(x^2)$. The latter sort correspond to ideals of $F[x]$ containing $(x^2)$, which are easy to list because $F[x]$ is a principal ring: $(x^2), (x), (1)$. These correspond to submodules of $F[x]/(x^2)$ of sizes $1$, $p$, and $p^2$, respectively.
Similarly, $F[x]/(x^3)$ has submodules corresponding to the $F[x]$-ideals $(x^3), (x^2), (x), (1)$; these submodules have sizes $1$, $p$, $p^2$, and $p^3$.
Allow me to commit the mild sin of identifying $x$ with its image in these quotient rings/modules. Now, the size of $p_1(W)$ gives us a lower bound on the size of $W$. So $p_1(W) = 0 \text{ or } (x)$, and $p_2(W) = 0 \text{ or } (x^2)$. Clearly $(x) \oplus 0$ and $0 \oplus (x^2)$ are two possibilities.
We need to find out whether we can have at once  $p_1(W) = (x)$ and $p_2(W) = (x^2)$. I claim that this can happen in multiple ways. Each of the $p$ elements of $(x)$ (remember that I'm writing this for $(x)/(x^2) = {ax + (x^2) : a \in F}$). will correspond to a unique element of $(x^2)$, and it follows that we merely need to specify an element $(x, ax^2) \in W$, where $a \in F^*$. This gives us $p - 1$ more possibilities. There are certainly some assertions to check, here!
