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

Question

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?

Related: http://math.stackexchange.com/questions/1418478/a-ufd-ring-for-which-the-related-formal-power-series-ring-is-not-a-ufd — user26857, Sep 06 '15 at 14:33

user26857 · Accepted Answer · 2017-11-15T08:31:59.403

2

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.

edited Nov 15 '17 at 08:31

answered Sep 06 '15 at 15:13

user26857

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Thanks. What are the hypotesis on the field $K$ that let this work? Does it work on all fields? – karmalu Sep 06 '15 at 15:39
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Yes, this works for all fields. – user26857 Sep 06 '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 – karmalu Sep 06 '15 at 15:51
@user26857 don't we need $A$ to be atomic for Nagata? Why is this true for all fields? – Arrow Jul 16 '16 at 18:54
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@Arrow The ring $A$ is a Noetherian domain and hence atomic. – Dhruv Goel Aug 07 '24 at 15:59

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

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