On \emptyset-definable elements in a field

Details On \emptyset-definable elements in a field On \emptyset-definable elements in a field

2005-02-27

arxiv.org/abs/math/0502565v35

Collectanea Mathematica 58 (2007), no. 1, pp. 73-84

Mathematics

Logic

15 pages, LaTeX2e, the version which will appear in Collectanea Mathematica

Scientific paper

Let K be a field and \tilde{K} denote the set of all r \in K for which there exists a finite set A(r) with {r} \subseteq A(r) \subseteq K such that each mapping f:A(r) \to K that satisfies: if 1 \in A(r) then f(1)=1, if a,b \in A(r) and a+b \in A(r) then f(a+b)=f(a)+f(b), if a,b \in A(r) and a \cdot b \in A(r) then f(a \cdot b)=f(a) \cdot f(b), satisfies also f(r)=r. We prove: \tilde{K} is a subfield of K, \tilde{K}={x \in K: {x} is existentially first-order definable in the language of rings without parameters}, if some subfield of K is algebraically closed then \tilde{K} is the prime field in K, some elements of \tilde{K} are transcendental over Q (over R, over Q_p) for a large class of fields K that are finitely generated over Q (that extend R, that extend Q_p), if K is a Pythagorean subfield of R, t is transcendental over K, and r \in K is recursively approximable, then {r} is \emptyset-definable in (K(t),+,\cdot,0,1), if a real number r is recursively approximable then {r} is existentially \emptyset-definable in (R,+,\cdot,0,1,U) for some unary predicate U which is implicitly \emptyset-definable in (R,+,\cdot,0,1).

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