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.
Best Answer
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 $(x^2,xy,y^3)$.