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arXiv math/0502565math.LOmath.NT
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论域中的\emptyset可定义元素

On \emptyset-definable elements in a field

Apoloniusz Tyszka

摘要:

设K是域,\tilde{K}表示所有r个\in K的集合,对于该集合存在有限集A(r),其中{r} \subseteq A(r)\subseteq K使得每个映射f:A(r)\to K满足:如果1 \in A(r)则f(1)=1,如果a,b \in A(r)和a+b \in A(r)则f(a+b)=f(a)+f(b),如果a,b \in A(r)和a \cdot b \in A(r)则f(a \cdot b)=f(a)\cdot f(b),也满足f(r)=r。我们证明:\tilde{K}是K的子字段,\tilde{K} ={x \in K:{x}在无参环语言中是存在一阶可定义的},若K的某个子域是代数闭的,则\tilde{K}是K中的素域,\tilde{K}的某些元素在Q上是超越的(在R上,在Q_p上)对于在Q上生成的一大类域K(推广R,推广Q_p),若K是R的勾股子域,t是K上的超越子域,且r是\in K的递归逼近,则{r}是\emptyset -可定义于(K(t),+,\cdot,0,1),如果实数r是递归可逼近的,那么对于某个一元谓词U,{r}是在(R,+,\cdot,0,1,U)中存在\emptyset可定义的,而U是在(R,+,\cdot,0,1)中隐式\emptyset可定义的.

英文摘要:

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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