Let $X = \mathbb A^2$. Say we blow up the origin $(0,0)$, and then blow up the intersection of $(x=0)$ with the exceptional divisor. The resulting space is the blow-up of $\mathbb A^2$ along what subscheme?

I guess that the first blow-up is at $(x,y)$, while the second extracts the same divisor as the blow-up of $(x,y^2)$. So the two-point blow-up should be the blow-up at the product ideal $(x,y)(x,y^2) = (x^2,xy,y^3)$. Is this right?

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    $\begingroup$ Yes, indeed you are right. The blowup of a product of ideals is the same as blowing up the first ideal, and then blowing up the inverse image of the second ideal. This is an easy exercise to write down by hand. $\endgroup$ Apr 5, 2015 at 5:59

1 Answer 1


The question isn't quite well-defined: if we blow up $\mathbb A^2$ along $\mathfrak m = \langle x,y\rangle$ or along $\mathfrak m^{k}$ for $k>1$, we get the same space.

It's handy that the spaces you're asking about are all toric varieties. So one can compute polyhedrally. Instead of $X = \mathbb A^2$, consider the monoid $\{(p,q)\ :\ p,q\in\mathbb N\}$ (whose monoid algebra has spectrum $X$). To blow it up the first time, chop off the corner, leaving only the elements of $\mathfrak m$, or maybe of $\mathfrak m^k$ (i.e. removing an isosceles right triangle of size $k$).

But now you want to blow up again, cutting off one of the two new corners. I'll be a bit sketchy here and just say that to have a complete description, we want the resulting shape to have $2+1+1$ edges (not fewer) and only integer vertices. That suggests that the first cut should give $\mathfrak m^2$, leaving vertices $(2,0)$ and $(0,2)$, and the second should leave vertices $(2,0),(1,1),(0,3)$. Which exactly correspond to your suggested generators.


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