A discrete form of the Beckman-Quarles theorem for mappings from R^2 (C^2) to F^2, where F is a subfield of a commutative field extending R (C)

Abstract

Let F be a subfield of a commutative field extending R. Let phi_n:F^n \times F^n ->F, phi_n((x_1,...,x_n),(y_1,...,y_n))=(x_1-y_1)^2+...+(x_n-y_n)^2. We say that f:R^n->F^n preserves distance d>=0 if for each x,y \in R^n |x-y|=d implies phi_n(f(x),f(y))=d^2. Let A_n(F) denote the set of all positive numbers d such that any map f:R^n->F^n that preserves unit distance preserves also distance d. Let D_n(F) denote the set of all positive numbers d with the property: if x,y \in R^n and |x-y|=d then there exists a finite set S(x,y) with {x,y} \subseteq S(x,y) \subseteq R^n such that any map f:S(x,y)->F^n that preserves unit distance preserves also the distance between x and y. Obviously, {1} \subseteq D_n(F) \subseteq A_n(F). We prove: A_n(C) \subseteq {d>0: d^2 \in Q} \subseteq D_2(F). Let K be a subfield of a commutative field Gamma extending C. Let psi_2: Gamma^2 \times Gamma^2->Gamma, psi_2((x_1,x_2),(y_1,y_2))=(x_1-y_1)^2+(x_2-y_2)^2. We say that f:C^2->K^2 preserves unit distance if for each X,Y \in C^2 psi_2(X,Y)=1 implies psi_2(f(X),f(Y))=1. We prove: if X,Y \in C^2, psi_2(X,Y) \in Q and X \neq Y, then there exists a finite set S(X,Y) with {X,Y} \subseteq S(X,Y) \subseteq C^2 such that any map f:S(X,Y)->K^2 that preserves unit distance satisfies psi_2(X,Y)=psi_2(f(X),f(Y)) and f(X) \neq f(Y).