# Finding number of elements in the preimage under a blow up

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 $((0,0),[0:1])$?

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Regarding the linear change of coordinates, you are trying to change coordinates so that the point $(1,0)$ (in the old coordinates) becomes the origin, i.e. $(0,0)$ (in the new coordinates). Rather than writing this in the form $x \mapsto x-1, y\mapsto y$, as you have, which has led you to an incorrect equation for $X$ in the new coordinates (your new equation for $X$ doesn't pass through the origin!), you may find it better to write the old coordinates as linear functions of the new ones (e.g. $x = x' + 1,$ $y = y'$, where $x',y'$ are the new coordinates) and now substitute these into ... – Matt E May 16 '12 at 11:19
... the equation for $X$. Once you've successfully done this, you can get rid of the primes. Regards, – Matt E May 16 '12 at 11:20
Also, why do switch from $u$ and $v$ to $s$ and $t$? – Matt E May 16 '12 at 11:21
@Matt E: thank you, just added something, can you please have a look? – user31509 May 17 '12 at 13:21

The point $((0,0),[0:1])$ lies in the affine chart of the blow up with coordinates $y,s$, so let's work in those coordinates. The equation of the blow up is then $$y-s^5y+s^4 = 0$$, which is smooth at $(0,0)$, since it contains a linear term. So to answer your third question, you have to answer the question as to what is the dimension of the tangent space to a curve at a smooth point?
Alternatively, you could just compute $\mathfrak m/\mathfrak m^2$ directly, with $\mathfrak m$ being the maximal ideal $(y,s)$ in the affine ring $\mathbb C[y,s]/(y-s^5y+s^4),$ which is a straightforward computation.