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Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. Let $Y$ be a locally closed subset of $X$. We denote by $X_0$ the set of closed points of $X$. Suppose $Y \cap X_0$ is closed in $X_0$. Is $Y$ closed? If yes, how do you prove it?

Motivation Let $k$ be an algebraically closed field. Let $X$ be an integral scheme of finite typle ove $k$. Let $X_0$ be the set of closed points of $X$. We can regard $X_0$ as a prevariety over $k$(e.g. Mumford's red book). Let $\Delta_0$ be the diagonal subset of $X_0 \times X_0$. Suppose $X_0$ is a variety, i.e. $\Delta_0$ is closed in $X_0 \times X_0$. Let $\Delta$ be the diagonal subscheme of $X \times_k X$. Then $\Delta$ is a locally closed subset of $X \times_k X$. I would like to prove $\Delta$ is closed in $X \times_k X$, i.e. $X$ is separated over $k$.

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    $\begingroup$ It suffices (math.stackexchange.com/questions/244719) to prove that $Y$ is stable under specialization. $\endgroup$ Dec 22, 2013 at 1:36
  • $\begingroup$ @MartinBrandenburg I know that. So? $\endgroup$ Dec 22, 2013 at 1:44
  • $\begingroup$ @MakotoKato I deleted my answer cause I haven't yet thought through whether or not it works (as to the comment you posted, when I had originally posted my answer, I had forgotten that I needed to prove that). Also, I think there's an easier method. $\endgroup$ Dec 22, 2013 at 3:07
  • $\begingroup$ Notice that this reduces to the case of $Y$ being a dense open subset of $X$. $\endgroup$ Dec 22, 2013 at 3:08
  • $\begingroup$ @MakotoKato I undeleted my answer, after having significantly altered it to (hopefully) make it correct. $\endgroup$ Dec 22, 2013 at 3:53

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I'm pretty sure that the condition on $Y$ needs to be that $Y\cap X_0$ is closed and non-empty, so I'm going to assume that.

For any closed subset $Z$ of $X_0$, let $t(Z)$ be the set of non-empty closed irreducible subsets of $Z$. One can show that these sets form the closed sets of a topology on $X_0$. In fact, $t$ gives a 1-1 correspondence b/w closed sets.

Snce $X$ is of finite type over a field, it's Noetherian and hence a Zariski space. Therefore, the map $f\colon X\to t(X_0)$ (of topological spaces) taking a point of $x\in X$ to its closure (intersected with $X_0$) is in fact a homeomorphism.

It's enough to prove that $Y$ is equal to its closure. So, we may assume without loss of generality that $\overline{Y}=X$, so that we've reduced to the case of $Y$ a (dense) open subset of $X$.

First, suppose $X$ is irreducible. Then $t(X_0)$ is irreducible. Now, if $X_0$ is the union of two closed subsets $Z_1$ and $Z_2$, then $t(X_0)=t(Z_1)\cup t(Z_2)$, which means that $t(Z_1)$ or $t(Z_2)$ equals $t(X_0)$. Hence, $Z_1$ or $Z_2$ equals $X_0$. Thus, $X_0$ is irreducible. Hence, since $$ [(X-Y)\cap X_0]\cup (Y\cap X_0)=X_0,$$ either $X-Y\supseteq X_0$ or $Y\supseteq X_0$. Hence, since $Y\cap X_0$ is nonempty, $Y\supseteq X_0$. Thus, because $X$ is a Zariski space and $Y$ is open, $Y=X$. In particular, $Y$ is closed.

Next, suppose $X_1,\ldots,X_r$ are the irreducible components of $X$ which intersect $Y$. Then by the above, $Y\cap X_i=X_i$ for all $i$, i.e. $Y=X_1\cup\cdots\cup X_r$. So, $Y$ is closed.

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  • $\begingroup$ Isn't $Y \cap X_0$ non-empty whenever $Y$ is non-empty? $\endgroup$ Dec 22, 2013 at 5:06
  • $\begingroup$ @Makoto Possibly, but I'm not entirely sure. $\endgroup$ Dec 22, 2013 at 6:28
  • $\begingroup$ I've just posted a question on this matter. math.stackexchange.com/questions/615390/… $\endgroup$ Dec 22, 2013 at 6:46
  • $\begingroup$ I believe the crux is that closed points are dense in every closed set. It follows that the closed points of a locally closed set Y are dense in the closure of Y. If there were a point of the closure of Y that is not in Y, there would be a closed such point, so Y would not be closed in X0. This is essentially your argument, and I deluded myself that I could say it more simply, but I seem to have only made it less clear. $\endgroup$
    – roy smith
    Nov 3, 2019 at 21:06
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We denote by $\mathbb{2}$ the two element field $\mathbb{Z}/2\mathbb{Z} =\{0, 1\}$. Let $X$ be a set. We denote by $\mathbb{2}^X$ the set of maps $X \rightarrow \mathbb{2}$. We make $\mathbb{2}^X$ into a ring with pointwise addition and multiplication. We denote by $\chi_A$ the characteristic function with respect to a subset $A$ of $X$. We regard $\chi_A$ as an element of $\mathbb{2}^X$ in the obvious way. Let $A, B$ be subsets of $X$. We denote by $A\Delta B$ the symmetric difference of $A$ and $B$, i.e. $(A - B) \cup (B - A)$. It is easy to see that $\chi_{A\Delta B} = \chi_A + \chi_B$. It is also easy to see that $\chi_{A\cap B} = \chi_A\chi_B$.

Lemma 1 Let $X$ be a set. Let $A, B, C$ be subsets of $X$. Then $(A\Delta B)\cap C = (A\cap C) \Delta (B\cap C)$.

Proof: $\chi_{(A\Delta B)\cap C} = \chi_{A\Delta B}\chi_C = (\chi_A + \chi_B)\chi_C = \chi_A\chi_C + \chi_B\chi_C = \chi_{A\cap C} + \chi_{B\cap C} = \chi_{(A\cap C) \Delta (B\cap C)}$. Hence $(A\Delta B)\cap C = (A\cap C) \Delta (B\cap C)$. QED

Lemma 2 Let $X$ be a set. Let $A, B$ be subsets of $X$. Then $A\Delta B = \emptyset$ if and only if $A = B$.

Proof: Suppose $A\Delta B = \emptyset$. Then $\chi_{A\Delta B} = 0$. Hence $\chi_A + \chi_B = 0$. Hence $\chi_A = -\chi_B = \chi_B$. Hence $A = B$. The converse is clear. QED

Definition 1 Let $X$ be a topological space. A subset of $X$ is called quasi-constructible if it is a finite union of locally closed subsets of $X$.

Lemma 3 Let $X$ be a topological space. A finite union of quasi-constructible subsets of $X$ is quasi-constructible. A finite intersection of quasi-constructible subsets of $X$ is quasi-constructible. The complement of a quasi-constructible subset of $X$ is quasi-constructible.

Proof: Clear.

Lemma 4 Let $k$ be a field. Let $X$ be a $k$-scheme locally of finite type. Let $X_0$ be the set of closed points of $X$. Let $Y$ be a non-empty locally closed subset of $X$. Then $Y \cap X_0$ is non-empty$.

Proof: This can be proved exactly in the same way as my answer to this question.

Lemma 5 Let $k$ be a field. Let $X$ be a $k$-scheme locally of finite type. Let $X_0$ be the set of closed points of $X$. Let $Z$ be a non-empty quasi-constructible subset of $X$. Then $Z \cap X_0$ is non-empty$.

Proof: This follows immediately from Lemma 4.

Lemma 6 Let $k$ be a field. Let $X$ be a $k$-scheme locally of finite type. Let $X_0$ be the set of closed points of $X$. Let $Y_1, Y_2$ be locally closed subsets of $X$. Suppose $Y_1 \cap X_0 = Y_2 \cap X_0$. Then $Y_1 = Y_2$.

Proof: By Lemma 1, $(Y_1\Delta Y_2)\cap X_0 = (Y_1\cap X_0) \Delta (Y_2\cap X_0) = \emptyset$. Since $Y_1\Delta Y_2$ is quasi-constructible, $Y_1\Delta Y_2 = \emptyset$ by Lemma 5. Hence $Y_1 = Y_2$ by Lemma 2. QED

Proposition Let $k$ be a field. Let $X$ be a $k$-scheme locally of finite type. Let $X_0$ be the set of closed points of $X$. Let $Y$ be a locally closed subsets of $X$. Suppose $Y \cap X_0$ is closed in $X_0$. Then $Y$ is closed in $X$.

Proof: There exists a closed subset $Z$ of $X$ such that $Y \cap X_0 = Z\cap X_0$. Then $Y = Z$ by Lemma 6. QED

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