TITLE: An ideal of $\mathbb Q[X]/(X^2)$ that isn't free.
QUESTION [0 upvotes]: I'm currently studying for an exam and redoing my old Homework. I came across this task and don't really understand the solution to it anymore. Task: 

Let $R:=\mathbb{Q}[X]/(X^2)$. Find a strict $R$-submodule (not equal to the $0$-module) $I$ so that $I$ is not a free module. 

I already know a solution to that: $(X+(X^2))$. However I don't really understand the approach to this task and have a few general questions: 
I) What exactly are the ideals of $R$? How do they look like? 
II) What exactly is $(X+(X^2))$? Is that equal to $R(X+(X^2))$? 
III) How do I come up with this solution? I already have a proof that has been discussed in my course weeks ago, but I just don't understand it anymore. Perhaps someone has an idea on how to solve this task. Thank you in advance for answers!

REPLY [0 votes]: I) In general, let $R$ be a commutative ring with $1$ and $I$ and ideal in $R$. Then we have a bijective correspondence 
$$ \{J \subseteq R ~ \text{an ideal} ~: I \subseteq J\} \rightarrow \{ A  \subseteq R/I~ \text{an ideal}\} \\
J \mapsto \pi(J) $$
In particular, let $a \in J$, then $a + I \in J/I$ is the corresponding element in $R$.
II) $(X + (X^2))_R$ is the ideal generated in $R$ by the element $X + (X^2)=:b$ i.e. the smallest ideal containing $(X + (X^2))_R$. Since $b$ is a single element, $(b)_R$ is principal and thus $(b)_R = Rb = \{rb \vert r \in R\}$.
III) Finding an proper submodule of $R$ which is not free now amounts to finding an ideal $J \subseteq R$ containing $(X^2)$ s.t. $\pi(J)$ is not free. The only proper ideal in $R$ which contains $(X^2)$ is $(X)$. Hence if there exists an answer to the question it has to be $\pi((X))$. Since $(X)$ is a principal ideal the only possible generator of $(X + (X^2))$ is $b$. However, while $b$ does span the entire module, it is not linearly independent, since $b \cdot b = 0$. Hence there does not exist a basis for $(X + (X^2))$ and thus it is not free.
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