# A criterion for a finite union of locally closed subsets of an affine space to be closed in Zariski topology

Let $k$ be a field. Let $n > 0$ be an integer. Let $K$ be an algebraically closed extension of $k$. Suppose the trancendence dimension of $K/k \ge n$. Let $A = k[x_1,\dots,x_n]$ be a polynomial ring. Let $E$ be a subset of $A$. We denote by $V(E)$ the common zeros of $E$ in $K^n$. It is easy to see that by defining subsets of the form $V(E)$ as closed sets, we can define topology in $K^n$.

Let $C$ be a finite union of locally closed subsets of $K^n$. Suppose the closure of $\{x\}$ is contained in $C$ for every $x \in C$. Is $C$ closed?

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In $\mathbb{C}^n$, $\{p\}$ is closed for every point $p$, and I think the same is true for any alg. closed field $K$: consider the variety of the ideal generated by $x_1 - p_1, x_2 - p_2, ..., x_n - p_n$ where $p = (p_1, p_2, ..., p_n)$. This seems to say that the criterion that the closure of $\{x\}$ is in $C$ for all $x \in C$ is trivial, i.e. true for any subset $C$ of $K^n$. – smackcrane Nov 26 '12 at 3:56
@smackcrane How does $k$ play a part? – Makoto Kato Nov 26 '12 at 4:24
it's not clear to me that $k$ does play a part ... where does this question come from? – smackcrane Nov 26 '12 at 4:32
@smackcrane In our definition, a closed set $V(E)$ is the common zeros of $E$, where $E$ is a set of polynomials over $k$, not over $K$. For example, we may consider the case where $k = \mathbb{Q}$ and $K = \mathbb{C}$. – Makoto Kato Nov 26 '12 at 5:44
ah, you're quite right, I misread your question. I was assuming that our polynomials were over $K$ rather than $k$. My mistake! – smackcrane Nov 26 '12 at 6:33

The answer is yes, and follows from the fact that a constructible set in a Noetherian scheme is closed if and only if it is closed under specialization.

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The following is a detailed proof of the fact stated in the answer by Matt E.

Notation Let $X$ be a topological space. Let $E$ be a subset of $X$. We denote by $cl(E)$ the closure of $E$ in $X$.

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

Definition 2 Let $X$ be a topological space. Let $x \in X$. If $cl(\{x\}) = X$, $x$ is called a generic point of $X$.

Lemma 1(A slight generalization of Hartshorne, Ex. 3.18 (b), II) Let $X$ be an irreducible topological space. Suppose $X$ has a (not necessarily unique) generic point $\eta$. Let $C$ be a constructible subset of $X$. Suppose $C$ is dense in $X$. Then $C$ contains $\eta$. Moreover $C$ contains non-empty open subset of $X$.

Proof: Let $C = \bigcup_i (U_i \cap F_i)$ be a finite union of locally closed subsets of $X$, where $U_i$ is open and $F_i$ is closed. Since $cl (C) = \bigcup_i cl(U_i \cap F_i)$, $\eta \in cl(U_i \cap F_i)$ for some $i$. Since $cl(\{\eta\}) \subset cl (U_i \cap F_i)$, $X = cl(U_i \cap F_i)$. Since $\eta \in U_i$, $\eta$ is a generic point of $U_i$. Since $cl(U_i ∩ F_i) \cap U_i = U_i$, $U_i \cap F_i$ is dense in $U_i$. Since $U_i \cap F_i$ is a closed subset of $U_i$, $U_i \cap F_i = U_i$. Hence $\eta \in U_i \cap F_i \in C$. Since $U_i \cap F_i = U_i$, $U_i \subset C$. QED

Lemma 2(A slight generalization of Hartshorne, Ex. 3.18 (c), II) Let $X$ be a Noetherian topological space. Suppose every irreducible closed subset of $X$ has a generic point. Let $C$ be a constructible subsets of $X$. Suppose $cl(\{x\}) \subset C$ for every $x \in C$. Then $C$ is closed.

Proof: Since $X$ is Noetherian, the subspace $C$ is also Noetherian. Hence $C = C_1 \cup \cdots \cup C_n$, where each $C_i$ is closed and irreducible in the subspace $C$. Since $C_i = C \cap F_i$ for some closed subset $F_i$ of $X$, $C_i$ is constructible. Since $cl(C) = cl(C_1) \cup \cdots\cup cl(C_n)$ and each $cl(C_i)$ is irreducible, $C_i$ contains a generic point of $cl(C_i)$ by Lemma 1. Hence, by the assumption, $cl(C_i)$ is contained in $C$. Hence $cl(C) = C$. QED

Lemma 3 Let $K/k$ be as in the question. Then $X = K^n$ is Noetherian with the topology defined in the question and every irreducible closed subset of $X$ has a generic point.

Proof: Since $A = k[x_1,\dots, x_n]$ is a Noetherian ring, it is easy to see that $X$ is Noetherian. Let $Y$ be an irreducible closed subset of $X$. Then $Y = V(P)$, where $P$ is a prime ideal of $A$. Let $L$ be the field of fractions of $A/P$. Since the trancendence dimension of $L/k \le n$, there exists a $k$-homomorphism $\psi\colon L \rightarrow K$. Let $\bar x_i$ be the image of $x_i$ by the canonical homomorphism $A \rightarrow A/P$ for each $i$. Let $\psi(\bar x_i) = y_i$. Let $y = (y_1, \dots, y_n)$. Then $\psi$ induces a $k$-isomorphism $A/P \rightarrow k[y_1,\dots,y_n]$. Hence $f(y) = 0$ if and only if $f \in P$, where $f \in A$. Hence $y$ is a generic point of $Y$. QED

Proof of the assertion in the question This is immediate from Lemma 2 and Lemma 3.

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