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Probably the following question is trivial, but somehow I do not see a proof or a counterexample.

Let $k$ be a field of characteristic zero, and let $x,y$ be two commuting variables over $k$.

Is $k[y,xy]$ a UFD?

Clearly it is an integral domain, since it is contained in $k[x,y]$.

Edit: Let $k \subsetneq R \subseteq k[x,y]$ be a $k$-subalgebra of $k[x,y]$, with $\dim(R) \in \{1,2\}$.

Is the following claim true: If $R$ is a UFD, then $R[y,xy]$ is a UFD? Notice that $y,xy$ are not algebraically independent over $R$ so we cannot apply this known result.

Examples: (1) $R=k[y^2,y^3,x]$ is not a UFD, but $R[y,xy]=k[y^2,y^3,x,y,xy]$ is a UFD.

(2) $R=k[y^2,y^3]$ is not a UFD, but $R[y,xy]=k[y^2,y^3,y,xy]$ is a UFD (isomorphic to $k[y,xy]$).

(3) $R=[x^2,y^2]$ is a UFD and so is $R[y,xy]$, I guess..

(4) $R=k[x^2,x^3]$ is not a UFD, $R[y,xy]$ is also not a UFD?

If $\dim(R)=2$ and $Q(R)=k(x,y)$ (the field of fractions of $R$ equals $k(x,y)$), is it true that necessarily $R$ is a UFD? No, as example (1) shows. Could we add a 'mild' condition that will guarantee that $R$ is a UFD?

Any hints are welcome; thank you!

user237522
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    Isn't $k[y,xy]\cong k[x,y]$? – Angina Seng Sep 01 '20 at 19:25
  • Yes, you are right..this is because $y,xy$ are algebraically independent over $k$. Actually, I had a different question in mind..I will soon explain. – user237522 Sep 01 '20 at 19:31
  • $\dim R$ is necessarily $\le 2$. – user26857 Sep 01 '20 at 22:11
  • @user26857, thank you; I thought of $k \subseteq R \subseteq k[x,y]$ with $\dim(R) \neq 0$ (hence $\dim(R) \in {1,2}$). I have written $k \neq R$ so I agree with you that condition $\dim(R) \in {1,2}$ could be deleted. Please, any ideas about my question? – user237522 Sep 02 '20 at 01:51

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