3
$\begingroup$

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?

$\endgroup$
1
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thanks. What are the hypotesis on the field $K$ that let this work? Does it work on all fields? $\endgroup$
    – karmalu
    Sep 6 '15 at 15:39
  • 1
    $\begingroup$ Yes, this works for all fields. $\endgroup$
    – user26857
    Sep 6 '15 at 15:45
  • $\begingroup$ 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 $\endgroup$
    – karmalu
    Sep 6 '15 at 15:51
  • $\begingroup$ @user26857 don't we need $A$ to be atomic for Nagata? Why is this true for all fields? $\endgroup$
    – Arrow
    Jul 16 '16 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.