arXiv math/06023102006-02math.NT

数的算术邻域

Arithmetic neighbourhoods of numbers

Apoloniusz Tyszka

备注：: 15 pages, LaTeX2e
发表信息：: International Mathematical Forum 3 (2008), no. 8, pp. 349-362

摘要：

设K是环，A是K的子集.我们说一个映射f：A \to K是算术的，如果它满足以下条件：如果1 \in A那么f（1）=1，如果a，b \in A和a+b \in A那么f（a+b）=f（a）+f（b），如果a，b \in A和a \cdot b \in A那么f（a \cdot b）=f（a）\cdot f（b）。我们称元素r为\in K算术固定的，如果存在有限集A \subseteq K（r在K内的算术邻域），其中r为\in A，使得每个算术映射f：A \to K固定r，即f（r）=r。我们证明：对于无穷多个整数r，对于r在Z内的某个算术邻域，该邻域是r在R内的邻域，而不是r在Z内的邻域[ \sqrt{-1} ];对于无穷多个整数r，对于r在Z内的某个算术邻域，该邻域不是r在Q内的邻域;如果K=Q（\sqrt{5}）或K=Q（\sqrt{33}），则对于无穷多有理数r，对于r在Q内的某个算术邻域，这个邻域不是r在K内的邻域;对于每个n个\in（Z \cap [3，\infty））\setminus {2^2，2^3，2^4，.}存在有限集J（n）\subseteq Q使得J（n）是n在R内的算术邻域，而J（n）不是n在C内的算术邻域.

英文摘要：

Let K be a ring and let A be a subset of K. We say that a map f:A \to K is arithmetic if it satisfies the following conditions: if 1 \in A then f(1)=1, if a,b \in A and a+b \in A then f(a+b)=f(a)+f(b), if a,b \in A and a \cdot b \in A then f(a \cdot b)=f(a) \cdot f(b). We call an element r \in K arithmetically fixed if there is a finite set A \subseteq K (an arithmetic neighbourhood of r inside K) with r \in A such that each arithmetic map f:A \to K fixes r, i.e. f(r)=r. We prove: for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is a neighbourhood of r inside R and is not a neighbourhood of r inside Z[\sqrt{-1}]; for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is not a neighbourhood of r inside Q; if K=Q(\sqrt{5}) or K=Q(\sqrt{33}), then for infinitely many rational numbers r for some arithmetic neighbourhood of r inside Q this neighbourhood is not a neighbourhood of r inside K; for each n \in (Z \cap [3,\infty)) \setminus {2^2,2^3,2^4,...} there exists a finite set J(n) \subseteq Q such that J(n) is an arithmetic neighbourhood of n inside R and J(n) is not an arithmetic neighbourhood of n inside C.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

摘要：

英文摘要：

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