3
$\begingroup$

$\textit{The proposition}$: On any variety $Y$, there is a base for the topology consisting of open affine sets.

$\textit{The proof}$: Assume $Y$ is quasi-affine in $\mathbb{A}^n$ and let $Z=\overline{Y}-Y$ with $\mathfrak{a}$ the ideal of $Z$ in $k[x_1,\dots, x_n]$. Let $P\in Y$. If $P\notin Z$ there is a polynomial $f\in \mathfrak{a}$ for which $f(P)\neq 0$. Let $H = Z(f)\subseteq \mathbb{A}^n$. Then $P\in Y-Y\cap H$ is open in $Y$ and closed in $\mathbb{A}^n - H$, the latter an affine variety. Thus $Y-Y\cap H$ is the desired open neighborhood.

$\textit{The problem}$: $Y-Y\cap H = (\mathbb{A}^n - H)\cap Y$. We know $Y$ is open in $\overline{Y}$, and if $Y-Y\cap H$ were closed in that subspace then we should be able to express it as the intersection with a closed set. I'm just not sure why it is closed. If $\overline{Y}=\mathbb{A}^n$, then $Y=\mathbb{A}^n - Z(T)$ is a standard open set, hence quasi-affine variety, and $Y-Y\cap H$ would therefore be open in $\mathbb{A}^n$ by defintion of the subspace topology.

$\endgroup$
0
$\begingroup$

Since $f\in\mathfrak{a}$, $Z\subseteq H$. So $(\mathbb{A}^n-H)\cap Y=(\mathbb{A}^n-H)\cap \overline{Y}$, which is closed in $\mathbb{A}^n-H$.

$\endgroup$

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.