0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

(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 $\mathbb{A}^{n+1}$.

$\endgroup$
  • $\begingroup$ 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 $\mathbb{A}^n$? $\endgroup$ – Nicolas Bourbaki Dec 7 '14 at 21:15
  • $\begingroup$ 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? $\endgroup$ – Hoot Dec 7 '14 at 21:42
  • $\begingroup$ 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 $\mathbb{A}^n - H$. $\endgroup$ – Nicolas Bourbaki Dec 7 '14 at 22:12
  • 1
    $\begingroup$ 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 $\mathbb{A}^n - H$. $\endgroup$ – Hoot Dec 7 '14 at 22:28
  • 2
    $\begingroup$ 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 $Y$. $\endgroup$ – Hoot Dec 8 '14 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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