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The Question

I believe I have proven the following question, and would like feedback on my proof.

Specifically, I am uncertain on these points:

  • How to approach the fact that $A\cong k[x_1,\dots,x_n]/I$ for an ideal $I\subset k[x_1,\dots,x_n]$, instead of the familiar case when $A=k[x_1,\dots,x_n]$.
  • The difference between the Zariski topology on $\mathbb{A}^n$ versus on $\operatorname{m-Spec}(A)$.

Furthermore, I will bold sentences that I am unsure I am able to make.

Fix an algebraically closed field $k$. Consider the Zariski topology on $\operatorname{m-Spec}(A)$, where $A$ is a finitely generated, reduced $k$-algebra. Given $f\in A$, consider its closed zero set $Z(f)\subseteq \operatorname{m-Spec}(A)$ and its open complement $U_f\subseteq\operatorname{m-Spec}(A)$. These are called principal open sets.

Show that the principal open sets form a base for the Zariski topology.

The Proof

To show that these open sets form a base for the Zariski topology $\mathcal{T}$, we check the following condition:

  • Given an open set $U\in\mathcal{T}$, if $p\in U$ then there is some $f\in A$ such that $x\in U_f\subseteq U$.

Let $U\in\mathcal{T}$ be an open set in the Zariski topology, and consider an element $p\in U$. This implies that for some ideal $I\in\operatorname{m-Spec}(A)$, there exists some $f\in I$ such that $f(p)\ne 0$.

Consider the principal open set $U_f$. To show that $U_f\subseteq U$, we note that if $p'\in U_f$, then $f(p')\ne0$, and since $f\in I$, it follows that $p'\notin Z(I)$ and so $p'\in U$. Hence, $U_f\subseteq U$, and so the principal open sets form a base of the Zariski topology.

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1 Answer 1

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Actually, the conditions you've mentioned only show that the $U_f$ form a base for some topology on $X$. You can't be sure a priori that this topology will be the Zariski topology (notice that you never even used any properties of the Zariski topology in your proof!).

If you already have a topology on a space $X$, and you want to show a collection of open sets $\mathcal B$ form a base for your specific topology, then you just need to show that given an open set $U\subseteq X$ and a point $x\in U$, there exists some $B\in\cal B$ such that $x\in B\subseteq U$.

Edit: your new proof is correct, though maybe you'd wanna write more explicitly in the first paragraph that the complement of $U$ is closed, so of the form $Z(I)$, and this is the $I$ you're using through the rest of the proof.

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