首页> 外文期刊>Journal of Geometry >A discrete form of the Beckman-Quarles theorem for mappings from ${mathbb{R}}^{{text{2}}} {left( {{mathbb{C}}^{{text{2}}} } right)}$ to ${mathbb{F}}^{{text{2}}},$ where ${mathbb{F}}$ is a subfield of a commutative field extending ${mathbb{R}}({mathbb{C}})$
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A discrete form of the Beckman-Quarles theorem for mappings from ${mathbb{R}}^{{text{2}}} {left( {{mathbb{C}}^{{text{2}}} } right)}$ to ${mathbb{F}}^{{text{2}}},$ where ${mathbb{F}}$ is a subfield of a commutative field extending ${mathbb{R}}({mathbb{C}})$

机译:用于$ {mathbb {R}} ^ {{text {2}}} {{left({{mathbb {C}} ^ {{text {2}}}}右)的贝克曼夸勒定理的离散形式} $到$ {mathbb {F}} ^ {{text {2}}},$,其中$ {mathbb {F}} $是交换字段的子字段,扩展了$ {mathbb {R}}({mathbb {C }})$

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Let ${mathbb{F}}$ be a subfield of a commutative field extending ${mathbb{R}}.$ Let $varphi _{n} :{mathbb{F}}^{n} times {mathbb{F}}^{n} to {mathbb{F}},varphi _{n} {left( {{left( {x_{1} , ldots x_{n} } right)},{left( {y_{1} , ldots y_{n} } right)}} right)} = {left( {x_{1} - y_{1} } right)}^{2} + ldots + {left( {x_{n} - y_{n} } right)}^{2}$ . We say that $f:{mathbb{R}}^{n} to {mathbb{F}}^{n}$ preserves distance d ≥ 0 if |x − y| = d implies $varphi _{n} {left( {f(x),f(y)} right)} = d^{2}$ for each $x,y in {mathbb{R}}^{n} $ . Let $A_{n} ({mathbb{F}})$ denote the set of all positive numbers d such that any map $f:{mathbb{R}}^{n} to {mathbb{F}}^{n}$ that preserves unit distance preserves also distance d. Let $D_{n} ({mathbb{F}})$ denote the set of all positive numbers d with the property: if $x,y in {mathbb{R}}^{n}$ and |x − y| = d then there exists a finite set S xy with ${ x,y} subseteq S_{{xy}} subseteq {mathbb{R}}^{n}$ such that any map $f:S_{{xy}} to {mathbb{F}}^{n}$ that preserves unit distance satisfies |x − y|2 = φ n (f (y), f (y)). Obviously, ${ 1} subseteq D_{n} ({mathbb{F}}) subseteq to A_{n} ({mathbb{F}})$ . We prove: $A_{n} ({mathbb{C}}) subseteq { d > 0:d^{2} in {mathbb{Q}}} subseteq D_{2} ({mathbb{F}})$ . Let ${mathbb{K}}$ be a subfield of a commutative field Γ extending ${mathbb{C}}$ . Let $psi _{2} :Gamma ^{2} times Gamma ^{2} to Gamma ,,,,psi _{2} {left( {(x_{1} ,x_{2} ),(y_{1} ,y_{2} )} right)} = (x_{1} - y_{1} )^{2} + (x_{2} - y_{2} )^{2}$ . We say that $f:{mathbb{C}}^{2} to {mathbb{K}}^{2}$ preserves unit distance if ψ2(X,Y) = 1 implies ψ2(f (X), f (Y)) = 1 for each $X,Y in {mathbb{C}}^{2}$ . We prove: if $X,Y in {mathbb{C}}^{2} ,psi _{2} (X,Y) in {mathbb{Q}}$ and X ≠ Y, then there exists a finite set S XY with ${ X,Y} subseteq S_{{XY}} subseteq {mathbb{C}}^{2}$ such that any map $f:S_{{XY}} to {mathbb{K}}^{2}$ that preserves unit distance satisfies ψ2(X, Y) = ψ2(f (X), f (Y)) and f (X) ≠ f (Y).
机译:令$ {mathbb {F}} $是扩展$ {mathbb {R}}的交换字段的子字段。$令$ varphi _ {n}:{mathbb {F}} ^ {n}乘以{mathbb {F} } ^ {n}至{mathbb {F}},varphi _ {n} {left({{left({x_ {1},ldots x_ {n}} right)},{left({y_ {1}, ldots y_ {n}}}}} right)} = {left({x_ {1}-y_ {1}} right)} ^ {2} + ldots + {left({x_ {n}-y_ {n }} right)} ^ {2} $。我们说$ f:{mathbb {R}} ^ {n}到{mathbb {F}} ^ {n} $如果| x − y |保留距离d≥0。 = d表示$ varphi _ {n} {left({f(x),f(y)} right)} = d ^ {2} $,{mathbb {R}} ^ {n}中的每个$ x,y $。令$ A_ {n}({mathbb {F}})$表示所有正数d的集合,这样任何映射$ f:{mathbb {R}} ^ {n}到{mathbb {F}} ^ {n保留单位距离的} $也保留距离d。让$ D_ {n}({mathbb {F}})$表示具有该属性的所有正数d的集合:if $ x,y在{mathbb {R}} ^ {n} $中| x − y | = d,则存在一个带有$ {x,y}子集S _ {{xy}}子集{mathbb {R}} ^ {n} $的有限集S xy ,这样任何映射$ f:S _ {{ xy}}到{mathbb {F}} ^ {n} $保持单位距离满足| x − y | 2 (f(y),f(y))。显然,$ {1}子集D_ {n}({mathbb {F}})到A_ {n}({mathbb {F}})$$的子集。我们证明:$ A_ {n}({mathbb {C}})个子集{d> 0:d ^ {2}在{mathbb {Q}}}子集D_ {2}({mathbb {F}})$中。令$ {mathbb {K}} $是扩展$ {mathbb {C}} $的交换字段Γ的子字段。让$ psi _ {2}:Gamma ^ {2}乘以Gamma ^ {2}到Gamma ,,,, psi _ {2} {left({{(x_ {1},x_ {2}),(y_ {1 },y_ {2})} right)} =(x_ {1}-y_ {1})^ {2} +(x_ {2}-y_ {2})^ {2} $。我们说,如果ψ2(X,Y)= 1表示ψ2,则$ f:{mathbb {C}} ^ {2}至{mathbb {K}} ^ {2} $保留单位距离。 {mathbb {C}} ^ {2} $中的每个$ X,Y(f(X),f(Y))= 1。我们证明:如果{mathbb {C}} ^ {2}中的$ X,Y,{mathbb {Q}} $中的psi _ {2}(X,Y)且X≠Y,则存在一个有限集S具有$ {X,Y}子集S _ {{XY}}子集{mathbb {C}} ^ {2} $的XY ,这样任何映射$ f:S _ {{XY}}到{mathbb {K}保留单位距离的} ^ {2} $满足ψ2(X,Y)=ψ2(f(X),f(Y))和f(X)≠f(Y)。

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