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A scheme $X$ is quasi separated if intersection of two quasi compact subsets is a quasi compact subset.

Question is to prove that :

A scheme $X$ is quasi separated if intersection of two affine open subsets is a finite union of affine open subsets.

Let $U,V$ be quasi compact subsets of $X$ and let $\{\text{Spec}(A_i)\}$ be an open cover for $U\cap V$.

We have $\text{Spec}(A_i)\subseteq U$ for every $i$. So, $\bigcup \text{Spec}(A_i)\subseteq U$. We complete this cover by adding some more open subsets namely $\{\text{Spec}(B_j)\}$. We thus have $$U=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(B_j)\right).$$ Similarly for $V$ we will add some more open subsets, namely $\{\text{Spec}(C_k)\}$ and get an open cover for $V$. We thus have $$V=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(C_k)\right).$$ As $U$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_1)\bigcup \text{Spec}(B_1).$$ As $V$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_2)\bigcup \text{Spec}(C_2).$$ We thus have, $$U\cap V=(\text{Spec}(A_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(A_1)\cap \text{Spec}(C_2))\bigcup(\text{Spec}(B_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(B_2)\cap \text{Spec}(C_2)).$$ clearly, first three terms in the union is contained in $$\text{Spec}(A_1)\bigcup \text{Spec}(A_2).$$ It is the last term which we have to thik about. As it is given that intersection of two affine open subsets is finite union of affine open subsets, we have $$\text{Spec}(B_2)\cap \text{Spec}(C_2)=\text{Spec}(D_1)\bigcup\text{Spec}(D_2).$$ Clearly, $\text{Spec}(B_2)\cap \text{Spec}(C_2)\subseteq U\cap V$ i.e., $\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq U\cap V=\bigcup \text{Spec}(A_i)$.

As $\text{Spec}(D_1)$ is quasi compact and $\text{Spec}(D_1)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that $$\text{Spec}(D_1)=\text{Spec}(A_3)\bigcup \text{Spec}(A_4).$$ As $\text{Spec}(D_2)$ is quasi compact and $\text{Spec}(D_2)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that $$\text{Spec}(D_2)=\text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ So, $$\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq \text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ Thus, $$U\cap V\subseteq \text{Spec}(A_1)\bigcup \text{Spec}(A_2)\text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ So, given an open cover $\text{Spec}(A_i)$ for $U\cap V$ we have a finite sub cover $\{\text{Spec}(A_i)\}_{i=1}^6$ for $U\cap V$.

Thus, $U\cap V$ is quasi compact.

I would like to know if this justification is correct.

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  • $\begingroup$ Oh, you changed your question just as I posted my answer. $\endgroup$ – Hamed Jul 21 '17 at 19:51
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A topological space is called quasi-separated if the intersection of any two quasi-compact open (not just any) subsets is quasi-compact. Check your definition.

Let $X$ be a scheme. Suppose the intersection of any two affine open subsets is a finite union of affine open subsets. Take two quasi-compact open subsets $U, V\subset X$. We want to show that $U\cap V$ is quasi-compact.

First prove the following easy lemmas:

Lemma 1 Every affine scheme is quasi-compact.

Lemma 2: If $Y$ is a topological space which is a finite union of quasi-compact spaces, then $Y$ is quasi-compact.

Lemma 3: Let $X$ be a scheme, then the affine open subsets form a base for Zariski topology.

Using the thrid lemma both $U,V$ are unions of affine opens. Using the fact that $U,V$ are quasi-compact, we find that $U,V$ are finite union of affine opens, say $U=\bigcup_{i=1}^n \mathrm{Spec}A_i$ and $V=\bigcup_{j=1}^m \mathrm{Spec}B_j$. Then $$ U\cap V= \bigcup_{i=1}^n \bigcup_{j=1}^m \mathrm{Spec}A_i\cap \mathrm{Spec}B_j $$ since the intersection of any two affine opens is a finite union of affine opens, then $U\cap V$ is a finite union of affine opens. Since every affine schme is quasi-compact (lemma 1), and since $U\cap V$ is a finite union of affine opens/quasi-compacts, then $U\cap V$ is quasi-compact (lemma 2).

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  • $\begingroup$ I think the user was asking if his justification is correct or not. That being said, your answer deserves a +1.. $\endgroup$ – user87543 Jul 21 '17 at 19:51
  • $\begingroup$ Yeah, OP changed the question. At first, the question was "how should I prove this" more or less. I'm now in this awkward position that I don't know whether to keep or remove this answer. $\endgroup$ – Hamed Jul 21 '17 at 20:03
  • $\begingroup$ you should not remove this answer if you ask me. This is a good answer. $\endgroup$ – user87543 Jul 21 '17 at 20:07
  • $\begingroup$ Thanks for the answer. Apologies for changing the question. I was just giving some try. Your answer seems to be more straight forward and easy to understand. Can you tell me if my justification is correct? yes, I mean to say intersection of two quasi compact open subsets is quasi compact, $\endgroup$ – user312648 Jul 21 '17 at 20:36
  • $\begingroup$ Forgive me but no matter how hard I try I can't understand what you are doing in your suggestion (I get lost). Hopefully, someone else will help you with the verification. $\endgroup$ – Hamed Jul 21 '17 at 20:44

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