$\mathbb{Q}[X,Y]/(Y^2-X^3)$ is not a UFD 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.
 A: Just for funsies, here is another approach.
Claim: $A:=\mathbb{Q}[X,Y]/(Y^2-X^3)$ isn't integrally closed (it has a singularity at the origin). 
To see this, note that $\displaystyle\frac{Y}{X}\in\text{Frac}(A)$, but that $\displaystyle \frac{Y}{X}\notin A$. Indeed, if $\displaystyle \frac{Y}{X}\in A$ then there exists  polynomials $f(X,Y),g(X,Y)\in\mathbb{Q}[X,Y]$ such that $Y=Xf(X,Y)+g(X,Y)(Y^2-X^3)$. This is clearly impossible though. Note though that $\displaystyle \frac{Y}{X}$ satisfies $T^2-X\in A[X]$. Thus, $A$ isn't integrally closed, and so can't be a UFD.
A: As Jared notes, your approach is fine.  You also consider directly the relation $Y^2 - X^3 = 0$, which implies that $Y^2 = X^3$.  Does this give you any hints for an element which has two distinct factorizations?
A: If $\overline{X}$ was prime, then it would generate a prime ideal, but the quotient of $\dfrac{\mathbb{Q}[X,Y] }{(Y^2-X^3)}$ by $(\overline{X})$ is $\mathbb{Q}[Y]/(Y^2)$ which is not an integral domain since $Y \cdot Y =0$ in that ring.
A: Your approach is fine.  Here is a hint to help you find your polynomials $f,g,$ and $p$.
$$\bar{Y}^2\in(\bar{X})$$
A: Counterexample
$(y-x)(y+x) = y^2-x^2 = x^3-x^2 = x^2(x-1)$.
