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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?

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    $\begingroup$ 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. $\endgroup$ Mar 5, 2020 at 21:25
  • $\begingroup$ A formatting tip: \setminus is best for writing something like $Y\setminus \{0,0\}$. I've updated your post with this. $\endgroup$
    – KReiser
    Mar 5, 2020 at 21:25
  • $\begingroup$ @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? $\endgroup$
    – mss
    Mar 5, 2020 at 21:36
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    $\begingroup$ If you're trying to find the strict transform, there are several other posts on this site with explanations. See here for example. $\endgroup$
    – KReiser
    Mar 5, 2020 at 22:07

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