Find idempotent elements in the ring $\mathbb{Q}[x]/(x^6-2x^3-3)$ 
As the title says I'm trying to determine the idempotent elements in the ring $R=\mathbb{Q}[x]/(x^6-2x^3-3)$. 

This is my attempt:
$p(x)=x^6-2x^3-3=(x^3-3)(x^2-x+1)(x+1)$ and $x^3-3$, $x^2-x+1$, $x+1$ are irreducible, so the only prime ideals containing $p(x)$ are $P_1=(x^3-3)$, $P_2=(x^2-x+1)$ and $P_3=(x+1)$. Let $I=(p(x))$ and suppose that $q(x)+I\in R$ is an idempotent element, so $q^2(x)-q(x)=q(x)(q(x)-1)\in I$. Since $q(x)(q(x)-1)\in I\subseteq P_i$ for $i=1,2,3$ and the $P_i$'s are prime we have that $q(x)\in P_i$ or $q(x)-1\in P_i$ for each $i$. If we work on each of the eight possibilities we would get all the possible classes of polynomials. 
Is this argument right? Is there any other (simpler) approach? Thanks in advance.
 A: Your argument is good. Perhaps I can point out a reformulation of it that you might find useful. 
Note that if $R$ is an integral domain, then the only idempotents of $R$ are $0$ and $1$. Indeed, if $x \in R$ such that $x^{2} = x$, then $x(x-1) = 0$, so we must have $x = 0$ or $x = 1$. Further, if $R_{1}, \ldots, R_{n}$ are commutative rings, the idempotents of $R_{1} \times R_{2} \times \cdots \times R_{n}$ are precisely tuples of the form $(e_{1}, \ldots, e_{n})$ for idempotents $e_{i} \in R_{i}$. This is straightforward to see, since $(e_{1}, \ldots, e_{n})^{2} = (e_{1}^{2}, \ldots, e_{n}^{2}) = (e_{1}, \ldots, e_{n})$ if and only if $e_{i}^{2} = e_{i}$ for each $i = 1, \ldots, n$.  
With this in mind, consider your example. As you note, $p(X) = X^{6} - 2X^{3}-3$ factorizes into irreducibles over $\mathbb{Q}[X]$ as $(X^{3}-3)(X^{2}-X+1)(X+1)$. With your notation, the ideals $P_{1}, P_{2}, P_{3}$ are therefore prime, and since $\mathbb{Q}[X]$ is a principal ideal domain, $P_{1}, P_{2}, P_{3}$ are pairwise comaximal. By the Chinese Remainder Theorem, we thus have an isomorphism
$$\mathbb{Q}[X]/p(X)\mathbb{Q}[X] \xrightarrow{~\sim~} \mathbb{Q}[X]/P_{1} \times \mathbb{Q}[X]/P_{2} \times \mathbb{Q}[X]/P_{3}$$
Since each $P_{i}$ is prime, $\mathbb{Q}[X]/P_{i}$ is a domain for each $i$, and hence has only trivial idempotents. Therefore, there are exactly $8$ idempotents of $\mathbb{Q}[X]/P_{1} \times \mathbb{Q}[X]/P_{2} \times \mathbb{Q}[X]/P_{3}$, namely tuples of the form $(\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3})$, where $\varepsilon_{i} \in \{\overline{0}, \overline{1}\} \subset \mathbb{Q}[X]/P_{i}$. You can determine the idempotents of $\mathbb{Q}[X]/p(X)\mathbb{Q}[X]$ by pulling back these idempotents under the isomorphism above.  
