# Elementary equivalence of models

I'm quite new to model theory, so please correct me if I'm using wrong terminology. I need help with an exercise from Smirnov's book "Varieties of algebras" (In Russian).

Problem: Assume that a model $\mathfrak{B}$ satisfies all formulas of the elementary theory $\mathcal{T}(\mathfrak{A})$ of a model $\mathfrak{A}$. Prove that these models are elementary equivalent.

I have no ideas concerning the proof of this statement. We know that $\mathcal{T}(\mathfrak{A}) \subseteq \mathcal{T}(\mathfrak{B})$ and we need to show the reverse inclusion. But I don't know how to proceed since this statement seems false to me (I suppose this is due to I'm thinking more in terms of universal algebra and identities but not model theory and first order language formulas). Any help will be appreciated. Thanks!

## 1 Answer

The theory $\mathcal{T}(\mathfrak{A})$ is complete, so the reverse inclusion is automatic.

• There was no such term as "complete theory" in the book I've mentioned, so it was not obvious to me. Thank you! – Random Jack Jul 25 '14 at 15:44
• You are welcome. It is a standard term in model theory, not to be confused with the Completeness Theorem. A theory $T$ is complete if for any sentence $\phi$ one of $\phi$ or $\lnot\phi$ is a theorem of $T$. The theory $\mathcal{T}(\mathfrak{A})$ is complete because a sentence is either true or false in $\mathfrak{A}$. A more interesting example of a complete theory is the theory of algebraically closed fields of characteristic $0$. Also the theory of real-closed fields. – André Nicolas Jul 25 '14 at 15:55