# $(K[x,y,z]/(x^2+y^3+z^7))_{(x,y,z)}$ is a UFD.

How can I show given a field $K$ that $(K[x,y,z]/(x^2+y^3+z^7))_{(x,y,z)}$ is a UFD?

I found that statement in a Wikipedia page so i'm not 100% sure it's true, maybe it's true for some field only?

• Sep 6 '15 at 14:33

## 1 Answer

Set $A=K[X,Y,Z]/(X^2+Y^3+Z^7)$ and denote by $x,y,z$ the residue classes of $X,Y,Z$ modulo $(X^2+Y^3+Z^7)$. We have that $A/zA\simeq K[X,Y]/(X^2+Y^3)$ is an integral domain, so $z\in A$ is a prime element.

Now let's look at $A[z^{-1}]$. We have $A[z^{-1}]=K[x,y,z][z^{-1}]$. Furthermore, if $x'=x/z^3$ and $y'=y/z^2$ we get $z=-(x'^2+y'^3)$. This shows us that $A[z^{-1}]=K[x',y'][z^{-1}]$ and this is a UFD since $x',y'$ are algebraically independent over $K$. Now use Nagata's criterion for factoriality.

• Thanks. What are the hypotesis on the field $K$ that let this work? Does it work on all fields? Sep 6 '15 at 15:39
• Yes, this works for all fields. Sep 6 '15 at 15:45
• I ask this because Samuel to prove a similar statement (a particular case in effect) in this paper takes a different and very long road and i wonder if there is a reason Sep 6 '15 at 15:51
• @user26857 don't we need $A$ to be atomic for Nagata? Why is this true for all fields? Jul 16 '16 at 18:54