On \emptyset-definable elements in a field
| dc.creator | Tyszka, Apoloniusz | |
| dc.date | 2005-02-27 | |
| dc.date | 2006-04-18 | |
| dc.date.accessioned | 2026-07-07T06:39:29Z | |
| dc.date.available | 2026-07-07T06:39:29Z | |
| dc.description | 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). | |
| dc.description | 15 pages, LaTeX2e, the version which will appear in Collectanea Mathematica | |
| dc.identifier | https://arxiv.org/abs/math/0502565 | |
| dc.identifier | http://arxiv.org/abs/math/0502565 | |
| dc.identifier | Collectanea Mathematica 58 (2007), no. 1, pp. 73-84 | |
| dc.identifier.uri | http://salesiana.dossiersoluciones.com/handle/123456789/101097 | |
| dc.subject | Logic | |
| dc.subject | Number Theory | |
| dc.subject | 03C60, 12L12 | |
| dc.title | On \emptyset-definable elements in a field | |
| dc.type | text |