# $\mathbb{Q}[X,Y]/(Y^2-X^3)$ 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.

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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?

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You are talking about cosets? ($Y^2+I =X^3+I$). Does your comment mean that $\bar{Y} \cdot \bar{Y} =\bar{X} \cdot \bar{X} \cdot \bar{X}$ are the two desired factorisation? Could you tell me if and why (not) $\bar{X}$ and $\bar{Y}$ are irreducible? –  Koenraad van Duin Aug 14 '13 at 20:41
Try and factor them in the quotient. Lift this to an expression in Q[x,y] By degree considerations show that the element of the ideal must be zero, and so you've actually factored them upstairs. –  Alex Youcis Aug 14 '13 at 22:29
@KoenvanDuin: Dear Koen, Yes, I am working in the quotient, so I am talking about cosets, although in my experience it doesn't normally clarify matters to make the cosets explicit. You can certainly put bars over the elements, as you did, if it helps, but I prefer to just remember that whenever I see $Y^2$ I can replace it by $X^3$. You might like to prove that every coset has a unique representative of the form $f(X) + g(X) Y$ (where $f(X)$ and $g(X)$ are two polynomials in $X$, uniquely determined by the coset). This makes computations in the quotient ring fairly straightforward, and ... –  Matt E Aug 14 '13 at 22:33
... once you've done this you should have no trouble proving that each of $\overline{X}$ and $\overline{Y}$ is irreducible. Speaking from the perspective of someone who's thought about and taught this material for many years, I hope you won't mind the following advice: I think it would be good practice for you to try and do prove this the way I'm suggesting, and to work out the details of the computations I'm suggesting, and not just rely on the indirect approach you outline in your OP (although it would be good to pursue the details of that approach as well!) --- it will help to develop ... –  Matt E Aug 14 '13 at 22:36
... your comfort level for working in quotient rings of this general form, i.e. polynomial rings modulo an ideal (which are some of the most common commutative rings that one encounters in applications of algebra, since they arise all the time in algebraic geometry). Regards, –  Matt E Aug 14 '13 at 22:37

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.

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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.

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Your approach is fine. Here is a hint to help you find your polynomials $f,g,$ and $p$.

$$\bar{Y}^2\in(\bar{X})$$

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