Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 1 down vote accepted

The computation seems correct to me.

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.