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Elementary equivalence versus isomorphism. (English) Zbl 1162.12302

From the introduction: The aim of this note is to give new evidence for a long standing open question concerning the relation between elementary equivalence and isomorphism in the class of finitely generated fields. More precisely, we give a positive answer to this question in the “general” case.
Recall that two fields \(K\) and \(L\) are called elementarily equivalent, if for every sentence \(\psi\) in the language of fields one has: \(\psi\) is true in \(K\) if and only if \(\psi\) is true in \(L\).
Theorem A (Arithmetic variant). Let \(K\) and \(L\) be finitely generated fields which are elementarily equivalent. Let \(\kappa\) and \(\lambda\) be their absolute subfields. Then one has:
(1)
\(\kappa\) and \(\lambda\) are isomorphic, and \(\text{td}(K|\kappa)\) equals \(\text{td}(L|\lambda)\) (td= transcendence degree).
(2)
Moreover, there exists an embedding \(\iota:K\to L\) such that \(L\) is finite separable over \(\iota(K)\). Furthermore, if \(K\) is of general type, then \(K\cong L\) as fields.
Theorem B (Geometric variant). Let \(K|\kappa\) and \(L|\lambda\) be function fields over algebraically closed fields \(\kappa\), respectively \(\lambda\). Suppose that \(K\) and \(L\) are elementarily equivalent as fields. Then one has:
(1)
\(\kappa\) and \(\lambda\) are elementarily equivalent, and \(\text{td}(K|\kappa)\) equals \(\text{td}(L|\lambda)\).
(2)
Suppose \(K|\kappa\) is of general type. Then there exist function subfields \(K_0|\kappa_0\hookrightarrow K|\kappa\) and \(L_0|\lambda_0\hookrightarrow L|\lambda\) such that \(K=K_0\kappa\) and \(L=L_0\lambda\), and \(K_0|\kappa_0\cong L_0|\lambda_0\) as function fields.
In particular, if \(\kappa\cong\lambda\) are isomorphic, then \(K|\kappa\cong L|\lambda\) are isomorphic as function fields.

MSC:

12L12 Model theory of fields
03C60 Model-theoretic algebra
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