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OExercise I am trying to solve:

Show that the ring $R = \mathbb{C}[x^{3}, x^{2}y, xy^{2}, y^{3}]$ is normal, but $R$ is not a unique factorisation domain.

Normal means that $R$ is integrally closed in its field of fractions, i.e., $$R= \{ s \in \text{Frac}(R):s \text{ is integral over }R\},$$ where the field of fractions $\mathrm{Frac}(R)=\{\frac{f(x,y)}{g(x,y)}:f,g \in R \text{ s.t. } g\neq0\}$ and $s$ is integral over $R$ if there is a monic $h(t)\in (\mathbb{C}[x^{3}, x^{2}y, xy^{2}, y^{3}])[t]$ s.t. $h(t)=0$.

I know how to show it is not a UFD: $x^{3}y^{3}=(x^{3})(y^{3})=(x^{2}y)(xy^{2})$.

I have tried showing that any $h(x,y) \in \mathrm{Frac}(R)$ which is integral is actually in $R$, but I am not getting anywhere with it because all the ways I know to show a ring is normal use the fact that $R$ is a UFD.

Any help would be appreciated, thank you!

user26857
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user1318
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  • That equality does not prove that $R$ is not a UFD. In $\mathbb Z$ we also have $2\cdot 6=3\cdot 4$. You need to show that those factors are irreducible. – user26857 Mar 27 '22 at 16:29
  • In order to show that $R$ is integrally closed I suggest you to take a look at this answer. – user26857 Mar 27 '22 at 16:48
  • @user26857 Thanks! That question cleared things up, I couldn't find it when searching as I was looking for the key word normal not just integrally closed. Your point about showing those factors are irreducible: does it not follow directly from the fact that they are adjoined to the ring '$\mathbb{C}$'? I guess my question is: How would I show they're irreducible other than saying it is obvious? – user1318 Mar 28 '22 at 15:26
  • Look at the possible decompositions of $x^3$ in the polynomial ring $k[x,y]$ and notice that lower powers of $x$ don't belong to $R$. – user26857 Mar 28 '22 at 16:27

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