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This question is related to another question ($\mathbb{Q}[X,Y]/(Y^2-X^3)$ is not a UFD), but mine is of another order: I don't see what I'm supposed to do here.

The question is as follows: Let $R=\mathbb{Q}[X,Y]/(Y^2-X^3)$. Show that the cosets of $X$ and $Y$ are irreducible elements in $R$. Furthermore: show that $R$ is a domain, and prove that it is not a UFD.

Well: everything I know of irreducible elements is related to UFDs, but I can't apply that, since $R$ isn't. Please help.

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Misunderstood Observer
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  • Your question is not an exact duplicate of that question, since it is about specific elements of that ring, and not about UFD at all. – Thomas Andrews Oct 16 '13 at 20:06
  • Start by examining the definitions: what is an irreducible element of a ring? (try to see why the question is applicable to non-UFD's) – Jonathan Y. Oct 16 '13 at 20:13

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It may be easier to picture if you see that there is an injective map $f : R \to \Bbb Q[T]$ induced by $f(X) = T^2$ and $f(Y) = T^3$. Since $\Bbb Q[T]$ is a domain, so is $R$.

The image of $f$ is $\Bbb Q[T^2,T^3] = \Bbb Q \oplus T^2 \Bbb Q[T]$ : a polynomial in $\Bbb Q[T]$ is in the image of $f$ if and only if the coefficient of $T$ is $0$.

The possible factorisations in $\Bbb Q[T]$ of $f(X) = T^2$ are $(xT)(x^{-1}T)$ for a nonzero rational $x$, but these factors are not in the image of $f$. Hence $X$ is irreducible in $R$. Similarly, the possible factorisations in $\Bbb Q[T]$ of $f(Y)$ are $(xT)(x^{-1}T^2)$, but the first factor is not in the image of $f$.

mercio
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