I want to prove that the ideals $\frac{(x^2)}{(x^2(x-1))}, \frac{(x(x-1))}{(x^2(x-1))},\frac{(x^2(x-1))}{(x^2(x-1))}=(0)$ are not prime in the ring $R=\mathbb{Q}[x]/(x^2(x-1))$. So it is enough to show that for each ideal $I$ we have $R/I \simeq L$ where $L$ is not a domain. My tries so far:
$\frac{\mathbb{Q}[x]/(x^2(x-1))}{(0)}\simeq \mathbb{Q}[x]/(x^2(x-1))\simeq \mathbb{Q}[x]/(x^2) \times \mathbb{Q}$ not a domain
$\frac{\mathbb{Q}[x]/(x^2(x-1))}{(x^2)/(x^2(x-1))}\simeq \mathbb{Q}[x]/(x^2)$ which is a ring of dual numbers so it is not a domain (?)
$\frac{\mathbb{Q}[x]/(x^2(x-1))}{(x(x-1))/(x^2(x-1))}\simeq \mathbb{Q}[x]/(x(x-1))\simeq \mathbb{Q} \times \mathbb{Q} $ which is not a domain.
Can someone help me and check if my tries are correct? Thanks in advance!