Submodules of $F[x]/(g)$ Let $g$ be the polynomial $x^3(x-1)$ over a field $F$. What are the $F[x]$-submodules of $F[x]/(g)$? How are they contained in one another? 
 A: Denote $p:F[x]\twoheadrightarrow F[x]/(g)$. Let $M$ be an $F[x]$-submodule of $F[x]/(g)$. Then $p^{-1}(M)$ is an ideal of $F[x]$. Since $F[x]$ is principal, $p^{-1}(M)=\langle f_M\rangle$ for some $f_M\in F[x]$, thus $M=\langle f_M+(g)\rangle$.
Also, for two submodules $M,N$, $M\subseteq N$ if and only if $p^{-1}(M)\subseteq p^{-1}(N)$, if and only if $f_N\mid f_M$.
Moreover, we can show that $\langle f_M+(g)\rangle=\langle\gcd(f_M,g)+(g)\rangle$, so we are concerned only with the factors of $g$.
As to why the equality holds, first since $\gcd(f_M,g)\mid f_M$, it follows that $\langle f_M+(g)\rangle\subseteq\langle\gcd(f_M,g)+(g)\rangle$.
Then by Bézout identity, $\gcd(f_M,g)=af_M+bg$ for some $a,b\in F[x]$. Therefore $\gcd(f_M,g)+(g)=a\cdot(f_M+(g))$, and hence $\langle f_M+(g)\rangle\supseteq\langle\gcd(f_M,g)+(g)\rangle$.

Hope this helps.
A: In general, if $M$ is an $R$-module and $N$ is a submodule of $M$, there is a bijection between the set of submodules of $M$ containing $N$ and the submodules of $M/N$.
If $p\colon M\to M/N$ is the canonical projection, the bijection is $X\mapsto L/N$, with inverse $Y\mapsto p^{-1}(Y)$.
Therefore, the submodules of $M/N$ are uniquely expressed as $L/N$, where $L$ is a submodule of $M$ containing $N$. Note that this bijection preserves inclusion.
In your case, the submodules of $F[x]$ containing $(g)$ are ideals, so they are principal. Every nonzero ideal is generated by a unique monic polynomial and $(f)\supseteq(g)$ is the same as $f\mid g$. So you just have to find the monic divisors of $g$.

 They are $g(x)=x^3(x-1)$, $x^3$, $x^2(x-1)$, $x^2$, $x(x-1)$, $x$, $x-1$ and $1$.

You can find for yourself the inclusions between the ideals generated by these polynomials.
