A valued field is said to be algebraic maximal if all its algebraic extension have either a bigger value group or a bigger residue field.
Do you know for which field this is a first order property ? Tanks.
2
-
1$\begingroup$ Could you clarify which formal language you intend? $\endgroup$– Joel David HamkinsCommented Sep 18, 2012 at 23:28
-
$\begingroup$ Thanks for the answer. I didn't think of any language in particular, I'm interested in any example of language in which being algebraic maximal is first order. $\endgroup$– nadeauCommented Sep 19, 2012 at 7:35
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The property of being algebraically maximal is a first-order property in the language of valued fields, $\{+,-,\times, 0, 1, \mid\}$, where `$x\mid y$' if and only if '$v(x)\leq v(x)$'. This is proved in Quelques propriétés des corps values en théorie des modèles (1982), by F.Delon.
More specifically, it is proved that
$(K,V)$ is algebraically maximal, if and only if, for each positive number $n$, every polynomial $P\in K[X]$ of degree not larger than $n$ has the property that, there exists an $x$ such that for all $y$, $v(P(y))\leq v(P(x))$.
