# Hartshorne Proposition 4.3

I am trying to understand the following proof on page 25 of AG.

Let $Y$ to be a quasi-affine variety in $\mathbb{A}^n$, with $p\in Y$. Define $Z = \overline{Y} - Y$, this is a closed set in $\mathbb{A}^n$. Form the ideal $\mathfrak{a}\subseteq k[x_1,...,x_n]$ corresponding to $Z$. Choose a polynomial $f$such that $f \in \mathfrak{a}$ but $f(p) \not = 0$. Define $H$ to be the hypersurface of $f$. Hartshorne says (i) $Y - (Y\cap H)$ is a closed subset of $(\mathbb{A}^n - H)$ and (ii) therefore $Y-(Y\cap H)$ is affine since $(\mathbb{A}^n - H)$ is affine.

Where is (i) coming from? And how does (ii) follow, a closed subset of an affine set is not necessarily affine, is it?

(i) Check that $Y - H = \overline{Y} \cap (\mathbb{A}^n - H)$. Note that $H$ contains $Z$.
(ii) Closed subsets of closed subsets are closed! Maybe I'm misunderstanding you. Of course, when we say that $\mathbb{A}^n - H$ is affine we mean that it's isomorphic to a closed subset of $\mathbb{A}^{n+1}$.
• For number (i), I agree that $Y-H = \overline{Y}\cap (\mathbb{A}^n - H)$ but why does it follow that $Y-(Y-H)$ is closed in the complement of $\mathbb{A}^n$? – Nicolas Bourbaki Dec 7 '14 at 21:15
• I think I'm just using the definition of the subspace topology: I wrote the set as the intersection of a closed subset in the ambient space with the subspace. Where does $Y - (Y - H) = Y \cap H$ enter? – Hoot Dec 7 '14 at 21:42
• I understand everything you are writing I just do not see the conclusion. I know that the $Y-H$ is closed in $\mathbb{A}^n-H$ because it is written as an intersection of a closed set in the ambient space. I see that $Y - (Y-H) = Y\cap H$, and so $Y-(Y-H)$ is closed in $Y$. But I do not see how this implies that $Y-(Y-H)$ is closed in $\mathbb{A}^n - H$. – Nicolas Bourbaki Dec 7 '14 at 22:12
• I thought all we wanted was for $Y - (Y \cap H)$ [which is the same as $Y - H$; I just didn't want to type more characters] to be closed in $\mathbb{A}^n - H$. – Hoot Dec 7 '14 at 22:28
• An isomorphism is in particular a homeomorphism, so I hope closedness is clear. I'm not sure that irreducibility is important here, but $Y - H$ is a non-empty open subset of the irreducible space $Y$. – Hoot Dec 8 '14 at 0:06