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I want to show that $\mathbb{A}^n$ is quasi-compact. I'm kind of stuck, I really don't know where to go with my proof, so I'll show what I have

Proof: So suppose that $\cup U_i$ was an open cover for $\mathbb{A}^n$, then we look at $\mathbb{A}^n - (U_1 \cup \dots \cup U_i)$ which closed.

I'm stuck here, I wanna use the fact that the Zariski topology has the Noetherian property but I can't really see how to do it in this case.

Alex
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1 Answers1

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A space is noetherian if and only if every ascending chain of open subspaces stabilize.¹

For any open cover $(U_i)_{i∈I}$ of $\mathbb A^n$, look at the collection of finite unions of its cover members $$\bigcup_{j ∈ J} U_j;~\text{$J ⊂ I$ is finite}.$$ Use the noetherian property to show that every chain in this collection has an upper bound in this collection. Apply Zorn’s Lemma and examine what you got. Then you’re done.

This works for all noetherian spaces. I’m not sure that you have to use Zorn’s Lemma, though.

¹This is because the closed subsets of a space are dual to the open subsets of the space by taking complements. So the descending chain condition on closed sets directly translates to the ascending chain condition on open sets.

k.stm
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  • You don't need Zorn's lemma. Every chain has an upper bound, and any such upper bound must be $\mathbb{A}^n$, or you could add one more open set to have a bigger finite union. – Kevin Carlson Jan 21 '15 at 21:58
  • @KevinCarlson But $U_1 ⊂ U_1 ⊂ U_1 ⊂ …$ might be a chain with an upper bound which is not all of $\mathbb A^n$? Don’t you need something guaranteeing the existence of a chain running through suffiently many open sets? Oh, one can try to construct strictly increasing chains and fail to do so, right? – k.stm Jan 21 '15 at 22:03
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    You often hear it said that you don't need the axiom of choice to get maximal ideals in Noetherian rings... – Hoot Jan 21 '15 at 22:27
  • Right, it's only strictly increasing chains that we're interested in bounding here. – Kevin Carlson Jan 21 '15 at 23:41
  • Ok, I’m confused now: Does or doesn’t one need the axiom of choice? @KevinCarlson – k.stm Jan 22 '15 at 07:51
  • No, one doesn't. That's a crucial aspect of the noetherian property. – Kevin Carlson Jan 22 '15 at 16:22
  • @k.stm : How do you show that $\Bbb A^n(k)$ is a noetherian topological space? Does it follow from Hilbert's basis theorem (in order to say that $k[x_1, \dots, x_n]$ is noetherian, and then $\Bbb A^n(k)$ is a subspace of the spectrum of $k[x_1,\dots,x_n]$, which is noetherian, and any subspace of a noetherian space is noetherian)? Ok, I see… This is proved here. – Watson Nov 05 '16 at 09:56