finding the closure when blowing a variety at a singularity

Question

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)?.

Answer 1

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.