Let $X=V(y^{5}-x(x-1)^{4}) \subset \mathbb{A}^{2}$ and let $B_{q}(X)$ be the blow-up of $X$ at the point $q=(1,0)$. Let $p: B_{q}(X) \rightarrow X$ be the natural map. I want to find the cardinality of $p^{-1}(q)$ and for each $t \in p^{-1}(q)$ the dimension of the tangent space $T_{t}(B_{q}(X))$.
Well I guess the first step is to compute explicitly the blow-up of $X$. OK to do this we look at the set $W=\{((x,y),[u:v]): xv - yu = 0\} \subset \mathbb{A}^{2} \times \mathbb{P}^{1}$ and then we need to find the preimage of $V$ under $p$ and intersect this with $W$ no?
The first question: we are given a point distinct from the origin, so we need to do a linear change of coordinates no? we map $x \mapsto x-1$ and $y \mapsto y$ so the equation of $X$ is $y^{5}-(x-1)(x-2)^{4}$, however this gets messy so I'm guess I'm doing something wrong.
Second question: to compute the blow-up we would need to work in two-affine pieces no? $s=1$ and then $t=1$ and take the union?
Could you please explain? I would really like to understand this. Thanks in advance
Using Matt's E hint:
We can write $X$ (in the new coordinates) as $V(y^5-x^5-x^4)$. So we have to look at the set $\{((x,y),[s:t]) \in \mathbb{A}^{2} \times \mathbb{P}^{1}: y^{5}=x^{5}+x^{4},xt-sy=0\}$. In the affine piece $s=1$ we obtain that $y^{5}=x^{5}+x^{4}$ and $xt=y$ so that $x(t^{5}-1)-1=0$.
In the affine piece $t=1$ we obtain that $y^{5}=x^{5}+x^{4}$ and $x=sy$ and we get that $y-s^{5}y+s^{4}=0$.
We conclude that the blow-up of $X$ at the origin is given by:
$\{((x,y),[1 : t]) \in \mathbb{A}^{2} \times \mathbb{P}^{1} : x(t^{5}-1) - 1=0,y=xt\} \cup \{((x,y),[s : 1]) | y-s^{5}y+s^{4}=0,x=sy\}$
First question: is this correct?
Now the original question asked to compute $p^{-1}(q)$ for $q$ equal $(1,0)$ but we made a change of coordinates so we need to find $p^{-1}((0,0))$ no? so I get that the only point is $((0,0),[0 : 1])$.
Second question: is this correct?
Third question: how do we compute the dimension of the tangent space of $B_{q}(X)$ at the point $((0,0),[0:1])$?