An ideal of $\mathbb Q[X]/(X^2)$ that isn't free.

Question

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!

Answer 1

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.

Answer 2

Every element of $R$ can be written in the form $a + b X$, with $a, b \in \mathbb Q$. (Strictly speaking, the class of $a+bX$ in the quotient $\mathbb Q[X] / (X^2)$.)

So, suppose now that $I$ is a proper nonzero ideal that contains $a+bX$.

Because $I$ is proper, $a+bX$ must not be invertible in $R$. Note that $X$ is nilpotent, and recall that a nilpotent element plus a unit is again a unit. What does this tell about $a, b$? How many nonzero proper ideals does $R$ have?

You should find that $(X)$ is the only nonzero proper ideal of $R$. Is it free? It must not be, because otherwise the question has no answer. Can you show that a single element of $(X)$ is not linearly independent?