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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)?.

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This singularity is a node, not a cusp! – quim Dec 14 '09 at 13:17
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.

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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
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

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