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.
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.