# Is every projection birational morphism between integral schemes a blowing up morphism?

I apologize if this question is too basic, but I figure this should be an easy question to answer for experts.

Theorem 8.1.24 in Qing Liu's "Algebraic Geometry and Arithmetic Curves" says:

"Theorem 1.24: Let $f : Z\rightarrow X$ be a projective birational morphism of integral schemes. Suppose that $X$ is quasi-projective over an affine Noetherian scheme. Then $f$ is the blowing-up morphism of $X$ along a closed subscheme."

I'm having some difficulty following the proof, though even without looking at the proof, I'm wondering if there's a problem with the statement:

Let $X$ be a regular integral Noetherian surface, and let $\tilde{X}\rightarrow X$ be the blowup along a closed point $x\in X$. Let $\tilde{\tilde{X}}\rightarrow\tilde{X}$ be the blowup of a closed point $\tilde{x}$ on the exceptional divisor of $\tilde{X}$, then my understanding is that the composition $$\tilde{\tilde{X}}\rightarrow\tilde{X}\rightarrow X$$ should be an isomorphism above $X - \{x\}$, and the fiber over $x$ should be two projective lines intersecting transversally. Letting $f$ be this composition, then $f$ certainly satisfies the hypotheses of the theorem, but can't be a blowing up morphism, since any blowup morphism has fibers either singleton points, or projective lines.

The only solution I can imagine is that perhaps by "blowing up morphism", he means a sequence of blowups, but it seems that at the end of the proof he literally concludes that $f$ is the blowup of $X$ along $V(\mathcal{I})$ for a certain ideal sheaf $\mathcal{I}$ on $X$.

Where does the problem lie?

EDIT: As pointed out by aginensky, the morphism $f$ may be the blow-up along a nonreduced closed subscheme. Since one can only conclude that the exceptional fibers of a blowup morphism are projective lines if the blowup locus is a regular closed subscheme, this resolves the problem.

New question - If $X = \text{Spec }k[x,y]$, is it clear what ideal of $k[x,y]$ one should blow up at to obtain the double-blowup $\tilde{\tilde{X}}$? (The obvious thing to test, namely $(x,y)^2 = (x^2,xy,y^2)$ doesn't work since that doesn't change the blowup algebra.

• i think the answer is 'it depends what you mean by blowing up a closed subscheme'. If you look in Hartshorne you will see proved that every birational morphism is the blowing up a coherent sheaf . However it may not be what you think. For example the sheaf of ideals could be non-reduced etc. – meh Apr 18 '17 at 20:07
• @aginensky AH! I see... Indeed an earlier theorem of his which shows that the blowup has fibers which are projective lines assumed that the closed subscheme was regular, and of course nonreduced schemes are not regular. Alright time to compute some examples... – stupid_question_bot Apr 18 '17 at 20:16
• my favorite example is the map from $Sym^d(C) \to Jac(C)$ where C is a smooth curve. Once d is big enough so that there are special divisors of degree d you get special fibers which are projective spaces. Simple case is d=3 and trigonal curve- it's explained in Mumford's "Curves and their Jacobians". An amusing case- C a generic curve of genus 9 and d=8. General fiber is a point, but over the 4 dimensional variety $\Theta_{sing}$ the fibers are $P^1$'s except over 42 points where the fiber is a $P^2$. – meh Apr 19 '17 at 13:36
• Oops - I wrote that EARLY this morning. My numbers are off by one. s/b 6 & 5, not 5 & 4 – meh Apr 19 '17 at 16:07

You should blow up the product of two ideals --- one is $(x,y)$, the ideal of the origin, and the other is $(x,y^2)$, the ideal corresponding to the center of your second blowup. So the answer is the ideal $(x^2,xy,y^3)$.
• Hmm, so what is the reason behind choosing $(x,y^2)$? I suppose intuitively the nonreduced point should correspond to an ideal, say $(x,y)$ together with a tangent direction of the affine plane at $x = 0,y = 0$. What does choosing the ideal $(x,y^2)$ mean in this context? – stupid_question_bot Apr 19 '17 at 1:03
• The relative Picard group of $\tilde{\tilde{X}}$ over $X$ is of rank 2, hence the relative Mori cone has two generators. One corresponds to the blowup of $(x,y)$, another one is the blowup of $(x,y^2)$ (equivalently, it can be obtained by the contraction of the $(-2)$-curve in $\tilde{\tilde{X}}$). – Sasha Apr 19 '17 at 5:16
• I'm not familiar with the "Mori cone". How is the relative Picard group relevant? Surely there must be a simpler way to explain why $(x,y)\cdot(x,y^2)$ is the right blowup? – stupid_question_bot Apr 19 '17 at 18:51