Jacobson radical of polynomial quotient ring.

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Let $F$ be a field, and $A=F[x]/(x(x-1)^2)$.
1. Find the ideals of $A$. Which of them are simple or maximal?
2. Find the Jacobson radical, $J(A)$, of $A$.
3. Find two composition series for $A$, as an $A$-module.
4. Show that, between four $2$-dimensional $A/J(A)$-modules, there exist two that are isomorphic.

1. Since $F[x]$ is a PID, we get that the ideals of $A$ are $J=(q(x))/(x(x-1)^2)$ where $q(x)=x, (x-1), x(x-1), (x-1)^2, x(x-1)^2.$ Using 3rd isomorphism theorem we can see that $(x)/(x(x-1)^2), (x-1)/(x(x-1)^2)$ are maximal ideals. For the simple modules, we can see that $(x(x-1))/(x(x-1)^2) \subsetneq (x)/(x(x-1)^2), (x-1)/(x(x-1)^2)$, so these cannot be simple.
If there existed an ideal such that $(p(x))/(x(x-1)^2) \subsetneq (x(x-1))/(x(x-1)^2)$ then we would have that $x(x-1)| p(x)$, giving a contradiction. Same for $((x-1)^2)/(x(x-1)^2)$.
2. can now be answered easily, by showing that $(x)/(x(x-1)^2) \cap (x-1)/(x(x-1)^2)=(x(x-1))/(x(x-1)^2)$
3. Using the data from 1, we can now construct two composition series.
$(1)$: $0 \subseteq (x(x-1))/(x(x-1)^2) \subseteq (x)/(x(x-1)^2) \subseteq A$
$(2)$: $0 \subseteq ((x-1)^2)/(x(x-1)^2) \subseteq (x-1)/(x(x-1)^2) \subseteq A$.
Supposing all that I wrote on the previous questions are correct, we come to the fourth question, which is the reason for me to make this post. I know that simple $A/J(A)$-modules can be identified as simple $A$-modules, so if I could write down a $2$-dimensional $A$-module as a sum of three simple modules, wouldn't it be enough? My problem is that I'm kind of stuck on how a $2$-dimensional $A$(or $A/J(A)$)-module would look like. Any words of wisdom really appreciated. Thanks.

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I think the simplest way to do that is to notice that $A/J(A)\simeq F^2$ given your expression of $J(A)$. Then you have to show that there are only $3$ $F^2$-modules that are $2$-dimensional over $F$, but that's pretty easy, because a $F^2$-module is of the form $X\times Y$ where $X$ and $Y$ are $F$-vector spaces.