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Suppose X is dimension two locally Noetherian scheme. Y is a closed subscheme of X, with codimension 1. Denote X' to be the blow up of X along Y. Prove that the structure morphism f:X'-->X is a finite morphism.

It suffices to show it's quasi-finite according to Zariski's main theorem. But I can't exclude the possibility that an irreducible component of $f^{-1}(Y)$ maps to a closed point of Y.

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2 Answers 2

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I think it's not true :

Let $X=Spec(A)$ with $A=k[x,y,z]/(x^2-y^2-z^2)$ be a quadratic cone. Let $Y$ be a line through the origin of the cone : its ideal is $I=(z,x-y)$. We calculate :

$$X'=Proj_{A}A[t,u]/(zt-(x+y)u,(x-y)t-zu),$$ [EDIT : THE FORMULA HAS BEEN CORRECTED]

where, in the graded $A$-algebra $A+I+I^2+....$ we denoted $t$ and $u$ the degree one generators corresponding to $z$ and $x-y$. Now, quotienting by $x$, $y$, and $z$, we calculate the fiber over the origin of this blow-up It is Proj(k[t,u]), which is a positive-dimensional projective line !

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  • $\begingroup$ Here I take $Proj (\sum_{n>=0}I^n)$ as the definition of blowing up along V(I) on Spec A. So it is not always equal to $Proj_{A}A[t,u]/(zt-(x+y)u)$. More precisely, it is equal to $Proj_{A}A[t,u]/(zt-(x+y)u)$ when $Proj_{A}A[t,u]/(zt-(x+y)u)$ is integral. But here this expression is not an integral scheme. So this counterexample does not work under our definition of blowing up! $\endgroup$
    – TJCM
    Commented Jan 6, 2010 at 21:53
  • $\begingroup$ We do have the same definition ! And $Proj_A A[t,u](zt−(x+y)u)$ is just the concrete expression of $Proj(\sum_{n\geq 0} I^n)$ in my example ! $\endgroup$
    – Olivier Benoist
    Commented Jan 6, 2010 at 21:59
  • $\begingroup$ And after calculation, if you take $Proj (\sum_{n>=0}I^n)$ as the definition of blowing up, then you will see it's indeed quasi-finite in this example. $\endgroup$
    – TJCM
    Commented Jan 6, 2010 at 22:00
  • $\begingroup$ No, unfortunately. After localizing on the open set D(u), one would get (k[x,y,z,t]/(z^2-x^2-y^2, zt-(x-y)))_z = (k[y,z,t]/(z^2-y^2-(zt+y)^2))_z = k[z,z^{-1},t], and the fiber of the origin point is indeed finite. As you have seen, the point is that you need to take localization at D(z) if you take the definition as $Proj (\sum_{n>=0}I^n)$, but you don't localize if you take definition as $Proj_A A[t,u](zt−(x+y)u)$. $\endgroup$
    – TJCM
    Commented Jan 6, 2010 at 22:07
  • $\begingroup$ No, it's not. It is the union of two lines, the strict transform of $Y$, and another contracted to the origin by the blow-up as you can see from : $$Proj(\sum_{n\geq 0}I^n)=Proj (A/I)[t,u]/((x+y)u).$$ $\endgroup$
    – Olivier Benoist
    Commented Jan 6, 2010 at 22:17
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How do you define blow-up? It should be straightforward to show that an explicit construction has relative dimension 0 over $X$.

Update: In the comments I suggest to take the formal two-dimensional ring $k[[x_1, x_2]]$ and work with it. This assumes that $X$ is smooth. If it's not, then Olivier gave an example of quadratic cone where blowup has relative dimension 1.

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  • $\begingroup$ It can be resated algebraically as follows: A is Noetherian ring, I is a height one ideal, prove that $Proj (\sum_{n>=0} I^n/I^{n+1})$ --> $Spec A/I$ is a finite morphism. $\endgroup$
    – TJCM
    Commented Jan 6, 2010 at 20:46
  • $\begingroup$ I'm not good in commutative algebra, but you can reduce to $A$ being a formal ring $k[[x_1, x_2]]$ and $I = (f)$, where $f$ is homogenious and then everything should follow. $\endgroup$
    – Ilya Nikokoshev
    Commented Jan 6, 2010 at 21:08
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    $\begingroup$ No, you can't unless Y is a Cartier divisor. in this case X'=X... which is not very interesting ! $\endgroup$
    – Olivier Benoist
    Commented Jan 6, 2010 at 21:17
  • $\begingroup$ Yes, just sa Oliver has said, the point is that the Weil divisor is not necessarily a Cartier divisor; and because we are not only considering algebraically varieties, so we cannot even have a good expectation of the normal locus on X. $\endgroup$
    – TJCM
    Commented Jan 6, 2010 at 21:22

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