number of ideals in a set and determine the maximal ideals 
Let $f(X)=(X^2-2)(X^4-X)$ and $g(X)=(X^2-1)X\in \mathbb{Q}[X]$.
Let $I=(f,g)$ the ideal generated by $f$ and $g$.
1) How many ideals does $\mathbb{Q}[X]/I$ has?
2) What are the maximal ideals?

I have already computed that $I=X(X-1)$ and proved that $\mathbb{Q}[X]/I\cong Q\times Q$ where $Q$ is the field of fractions. But my knowledge of rings theory is a bit poor and I don't see how to answer the questions.
I have also thougth that $Q\times Q$ is not a field, then (f,g) is not maximal. But I don't know if it is useful.
Any hints will be appreciated
Note: In my notes I am asked to determine the number of ideals of $ \mathbb{Q}[X]/(fg) $, but I think that it is a misprint since in the same exercise I was asked to compute $ I $ and prove the isomorphism.
 A: Assuming you already have $\mathbb{Q}[x]/\langle x \rangle \cong \mathbb{Q}$ and $\mathbb{Q}[x]/\langle x-1 \rangle \cong \mathbb{Q}$. Your observation that $I$ is not a maximal ideal of $\mathbb{Q}[x]$ because $\mathbb{Q} \times \mathbb{Q}$ is not a field is correct. But I don't think the question is asking for the maximal ideals of $\Bbb{Q}[x]$. 
Observe that $\Bbb{Q}[x]$ is a PID (because $\mathbb{Q}$ is a field), so every ideal of $\Bbb{Q}[x]$ is of the form $\langle p(x) \rangle$ for some polynomial $p(x)$. Thus for ideals of $\Bbb{Q}[x]/I$ you should be looking at ideals of the form $\langle p(x) \rangle/ I$. This would mean that 
$$I \subseteq  \langle p(x) \rangle \quad \iff \quad  x(x-1) \in \langle p(x) \rangle \quad \iff \quad p(x) | x(x-1).$$
Thus all ideals of $\Bbb{Q}[x]/I$ are as follows:
$$\Bbb{Q}[x]/I, \quad \langle x-1 \rangle/I, \quad \langle x \rangle/I, \quad I/I$$
In all there are $4$ ideals of this ring: two trivial ones and two non-trivial ones.
The Third Isomorphism Theorem gives that for the ideals $I,J$ such that $I \subseteq J$, we have
$$(R/I)\big/(J/I) \cong R/J.$$
Thus we have,
$$\left(\Bbb{Q}[x]/I\right) \big/ \left(\langle x-1 \rangle/I\right) \cong \Bbb{Q}[x]/\langle x-1 \rangle \cong \Bbb{Q}.$$
Since $\Bbb{Q}$ is a field, this ideal is maximal. Likewise you can get the other non-trivial ideal to be maximal as well.
