Blow up $\{x_1^4+x_2^2+x_3^2+x_4^2=0\}\subset\mathbb{C}^4$ twice at the origin I know the definition of the blow up at a point, but have only computed extremely easy cases. I'm reading a paper where the variety
$$x_1^4+x_2^2+x_3^2+x_4^2=0$$
is blown up twice at the origin. Now I tried to compute what this would look like, but to be honest I have no real idea what to do here. So according to my definition the blow up we would get the variety
$$
\begin{equation*}
V(x_1y_2=x_2y_1,\\
x_1y_3=x_3y_1,\\
x_1y_4=x_4y_1,\\
x_2y_3=x_3y_2,\\
x_2y_4=x_4y_2,\\
x_3y_4=x_4y_3,\\
x_1^4+x_2^2+x_3^2+x_4^2=0
)
\end{equation*}\subset \mathbb{C}^4\times \mathbb{P}^3$$
But now what? This seem quite ugly and unworkable to me. And Now To be honest I don't even know what it would mean to blow up the 'origin' here again, since I don't know what the origin in $\mathbb{C}^4\times \mathbb{P}^3$ is?
It would be much appreciated if someone could work out this computation in some detail.
 A: The blow-up as a subscheme of $\mathbb{C}^4\times \mathbb{P}^3$ comes with standard affine charts, which are where one of the co-ordinates $y_j$ is non-zero. You can work this out for each $j$, but here the interesting case is $j=1$. Then the blow-up formulae that you wrote down say $$x_i = x_1 (y_i y_1^{-1}),$$ where $x_1$ and the $y_i y_1^{-1}$ are affine co-ordinates on this chart. Putting this into the equation of the hypersurface singularity gives $$0 = x_1^4 + x_2^2 + x_3^2 + x_4^2 = x_1^4 + x_1^2 (y_2 y_1^{-1})^2 + x_1^2 (y_3 y_1^{-1})^2 + x_1^2 (y_4 y_1^{-1})^2  = x_1^2 ( x_1^2 + (y_2 y_1^{-1})^2 + (y_3 y_1^{-1})^2 + (y_4 y_1^{-1})^2).$$ If we write say $z_i = y_i y_1^{-1},$ this equation is $$x_1^2(x_1^2 + z_2^2 + z_3^2 + z_4^2) = 0.$$ In this chart, the exceptional divisor is the locus of $x_1 = 0$ (because if $x_1 = 0$ then all $x_i = x_1 (y_i y_1^{-1})$ are zero too, so the whole hyperplane $x_1 = 0$ maps to the origin in $\mathbb{C}^3$), and the proper transform of our hypersurface singularity is $$x_1^2 + z_2^2 + z_3^2 + z_4^2 = 0.$$ That's the result of one blow-up, in this chart. You see that the thing is still singular at the origin, meaning the point $x_1 = z_2 = z_3 = z_4 = 0$. If you do another blow-up, the result looks the same in each affine chart; the proper transform is $$1 + w_2^2 + w_3^2 + w_4^2 = 0,$$ say. You might want to check that if you look at the other standard charts of the blow-up of the singularity we started with, $x_1^4 + x_2^2 + x_3^2 + x_4^2 = 0$, the result is non-singular at the origin. (You should get some equation like $w^2 ( x^4 w^2 + y^2 + z^2 + 1) = 0$ for the total transform in the other charts.)
