# Is the strict transform a finite morphism?

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?

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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. – James D. Taylor Aug 28 '10 at 23:48
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! – damiano Aug 28 '10 at 23:56
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 $C[1/(x-1)]$. – BCnrd Aug 29 '10 at 2:27