Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.