# What closure for strict transform of affine variety?

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?

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