I'm trying to show that $R = \mathbb{Q}[X,Y]/(Y^2-X^3)$ is not a UFD, but I got stuck.
To prove this, I could try to find two "different" factorisations for one element, but I am not familiar with this, so I tried to use a lemma and one of the previous exercises. If my syllabus is right, in every UFD counts $$ x \ \text{is irreducible} \quad \iff \quad x \ \text{is prime}$$ My syllabus also states that the elements $\bar{X}, \bar{Y} \in R$ are irreducible. So I tried to show that at least one of the elements $\bar{X}, \bar{Y} \in R$ does not generate a prime ideal.
This would mean that I should find two polynomials $f,g \in \mathbb{Q}[X,Y]$, such that $$\exists p \in \mathbb{Q}[X,Y], \quad fg - pX \in (Y^2-X^3)$$ and $$\forall q \in \mathbb{Q}[X,Y], \quad f-qX \notin (Y^2-X^3) \ \wedge \ g-qX \notin(Y^2-X^3)$$ or the same thing but then for $Y$.
I hope that you can tell me if this approach is correct, and provide me a hint. I'd appreciate it if you told me the solution, but please start your answer with a hint, clearly separated from the rest.