Let K be a field and F denote the prime field in K. Let \tilde{K} denote the set of all r \in K for which there exists a finite set A® with {r} \subseteq A® \subseteq K such that each mapping f:A® \to K that satisfies: 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), satisfies also f®=r. Obviously, each field endomorphism of K is the identity on \tilde{K}. We prove: \tilde{K} is a countable subfield of K, if char(K) \neq 0 then \tilde{K}=F, \tilde{C}=Q, if each element of K is algebraic over F=Q then \tilde{K}={x \in K: x is fixed for all automorphisms of K}, \tilde{R} is equal to the field of real algebraic numbers, \tilde{Q_p}={x \in Q_p: x is algebraic over Q}.

生成讲解博客的中文版本

生成AI智能校验的中文版本