Completeness of Algebraically Closed Valued Fields(ACVF) Theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:29:37Z http://mathoverflow.net/feeds/question/58979 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58979/completeness-of-algebraically-closed-valued-fieldsacvf-theory Completeness of Algebraically Closed Valued Fields(ACVF) Theory Santiago 2011-03-20T15:34:25Z 2011-03-20T20:21:20Z <p>One can prove Elimination of Quantifiers of ACVF finding an extension of any partial embedding of a model $K$ into a $|K|^+$ Saturated one using the language $\mathcal{L} = ( 0,1,+,*, U, \mid )$. In this Language $U$ is the unary predicate standing for the Valuation Ring of the model, and $\mid$ is a binary relation such that $x\mid y \leftrightarrow \exists z\in U \ x*z=y$. How do you prove the completeness of this theory in that language?</p> http://mathoverflow.net/questions/58979/completeness-of-algebraically-closed-valued-fieldsacvf-theory/58985#58985 Answer by Dave Marker for Completeness of Algebraically Closed Valued Fields(ACVF) Theory Dave Marker 2011-03-20T17:35:38Z 2011-03-20T17:35:38Z <p>Just like with algebraically closed fields, for completeness you need to specify the characteristic of the field and the characteristic of the residue field. </p> <p>There is a general trick if you have a theory $T$ with quantifier elimination and a structure $A$ that is embedded in every model of $T$, then $T$ is complete. Let $M$ and $N$ be models of $T$ and let $\phi$ be any sentence. There is a quantifier free sentence $\psi$ that is equivalent to $\phi$ in models of $T$. But, since $\psi$ is quantifier free, $$M\models \psi\Leftrightarrow A\models\psi\Leftrightarrow N \models\psi.$$ Thus $M\models\phi\Leftrightarrow N\models \phi$.</p> <p>Now, suppose you are looking at characteristic 0 fields with characteristic p residue field. The rationals with the p-adic valuation are a substructure of any model of ACVF of characteristic $0$ with characteristic $p$ residue field. </p>