Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm having trouble finding the closure in the definition of a blow-up. For example, take the following example, with a node at $(0,0)$ (and at some other points) (it's not a homework question, just a concept that I'm stuck on!). $xy=x^6+y^6$. Then the blow-up should be the closure of this set, taken over all $(x,y) \neq (0,0)$: $ \{ ((u,v), (x,y)) \in \mathbb{A}^{2} \times \mathbb{P}^{1} | uy=vx, xy = x^6 + y^6 \}$.

How do I explicitly find the closure of this set? I understand the fibre of the projection map at the singular point should consist of two points (both of which are non-singular) - why is this so, and what are those two non-singular points (in the smooth variety that is the resolution)?.

share|improve this question
2  
This singularity is a node, not a cusp! –  quim Dec 14 '09 at 13:17

1 Answer 1

up vote 7 down vote accepted

Look at the affine pieces: over the open subset $u \neq 0$, you have a local coordinate $z = v/u$ and your equations can be written as $y = zx$ and $xy = x^6 + y^6$. Substituting $y$ in the second equation gives you $x^2z = x^6 + x^6 z^6$. Now this equation factors as $x^2 = 0$ and $z = x^4(1 + z^6)$; the locus where the first one vanishes is the exceptional divisor, while the second one gives you the closure you are looking for. Non-singularity of the point over $(x, y) = (0,0)$ (which is $(x,z) = (0,0)$) follows from $$ \frac{d}{dz} \big[z - x^4(1 + z^6)\big] \big|_{(x,z) = (0,0)} = 1$$ The other point shows up when considering the other affine piece, $v \neq 0$.

The reason why there are two points over $(x,y) = (0,0)$ is because at that point your curve has two branches. To see that, look at the lowest degree piece of the polynomial defining it (essentially, you are looking here at a neighborhood of the origin in the classical/analytic topology): this is $xy$, which is the union of the two axes. Blowing up pulls these two branches apart.

share|improve this answer
    
In the first line it should be z = v/u, since you are over the open subset u \neq 0. –  Grétar Amazeen Dec 13 '09 at 20:17
    
@Grétar: thanks for catching that typo. It's fixed now. –  Alberto García-Raboso Dec 13 '09 at 20:20
    
Thank you very much. One quick question: does the blow-up of a (irreducible) variety have to be irreducible? –  Vinoth Dec 20 '09 at 10:34
1  
If you are blowing up a reduced point like in this case, yes: the blowup is birational to the original variety. In other cases you might have to be careful. Check out the examples in Eisenbud and Harris' "The Geometry of Schemes". –  Alberto García-Raboso Dec 21 '09 at 3:36
    
I think the blowup along a positive codimension closed subscheme (whenever it is reduced or not) is always birational. Do you think about a specific example in Eisenbud-Harris ? –  Qing Liu Feb 23 '10 at 10:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.