3
$\begingroup$

Woe is me! I'm again resorting to this forum to ask a silly question.

Here is the example I had in mind: observe the (complex) curve $y^3=x^2(x-1)$. In attempt to normalize this curve, I've begun by blowing it up once at the origin. Of the two affines resulting from the blow up, the $x=ys$ affine is the one that will meet the strict transform. Indeed the strict transform will be $\mathbb{C}[x,y,s]/x=ys,y=s^2(x-1)$. I've always assumed the natural morphism from the strict transform to the original curve is a finite one (otherwise using blow-ups to desingularize would be an odd concept!). This would imply that the above ring is integral over $\mathbb{C}[x,y]/y^3-x^2(x-1)$. But I've been staring at this, and staring at this, and by the life of me I can't come up with a monic polynomial that $s=\frac{x}{y}$ would satisfy over this ring.

Is the strict transform a finite morphism?

$\endgroup$
3
  • $\begingroup$ Really? This perplexes me. This curve has only one tangent at the origin, and it is x=0. Indeed, the other affine would give (say v=1/s) y(v^2)=x-1 - which has no points above x=y=0. $\endgroup$ Aug 28, 2010 at 23:48
  • $\begingroup$ You are right: I read your equation as being $y^2=...$! Sorry for the confusion. It is a sign that I should not try to say anything else! $\endgroup$
    – damiano
    Aug 28, 2010 at 23:56
  • 3
    $\begingroup$ Sure, strict transforms are always proper by construction, hence finite when quasi-finite as for curves. But your chart is integral over its image in the base curve $C$, not the entirety of $C$ necessarily (unless that is its image). Your error is that you have got to remove the image of the points outside of your chart, which amounts to removing the point $(1,0)$. This is the unique point where $C$ meets $x=1$, so invert $x-1$. There's an integral relation $s^3 = x/(x-1)$ over $C[1/(x-1)]$. $\endgroup$
    – BCnrd
    Aug 29, 2010 at 2:27

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.