0

Given $\pi:X\to \mathbb{A}^2$, the blowup of $\mathbb{A}^2$ at the origin, I am trying to calculate the strict transform of $Y=\mathbb{V}(y^2-x^2(x+1))$, which has been defined as the closure of $\pi^{-1}(Y\setminus\{0,0\})$ inside $X$.

What is meant by taking the closure here? If I am not mistaken, the preimage $\pi^{-1}(Y\setminus\{0,0\})$ is contained in a patch of $X$ that is isomorphic to $\mathbb{A}^2$ - can I work in this patch and take an affine closure? Do I have to take some sort of projective closure?

KReiser
  • 65,137
mss
  • 733
  • 3
    I'm not sure what your issue is. $X$ is a topological space, so we can take the closure of any subset. After that I guess the question could be what scheme structure you want to put on it, but the word "closure" should not be ambiguous. – Captain Lama Mar 05 '20 at 21:25
  • A formatting tip: \setminus is best for writing something like $Y\setminus {0,0}$. I've updated your post with this. – KReiser Mar 05 '20 at 21:25
  • @CaptainLama I see. Any hints on how to actually find the closure? Viewing points in $X$ as $((x,y),(z_0,z_1))$ the vanishing of $y^2-x^2(x+1)$ is closed and contains $Y$, but is it the closure? – mss Mar 05 '20 at 21:36
  • 1
    If you're trying to find the strict transform, there are several other posts on this site with explanations. See here for example. – KReiser Mar 05 '20 at 22:07

0 Answers0