Elementary extension

If M ⊆ N is an inclusion of structures, we say that N is an elementary extension of M, or M is an elementary substructure of N, denoted $$M \preceq N$$, if for every formula $$\phi(x)$$ and every tuple $$b$$ in M, we have
 * $$ M \models \phi(b) \iff N \models \phi(b) $$.

In other words, if b satisfies $$\phi(x)$$ in one of M and N, then it must satisfy it in the other.

Specializing to the case where x is a tuple of length 0, we recover elementary equivalence: if N is an elementary extension of M then M ≡ N. However, $$M \preceq N$$ is a strictly stronger condition than $$M \subseteq N$$ plus $$M \equiv N$$.

The elementary inclusion relation is transitive: if $$M \preceq M' \preceq M$$, then $$M \preceq M$$.

Non-examples
The rationals are not an elementary substructure of the reals (as rings): $$\mathbb{Q} \not\preceq \mathbb{R}$$. For example, if $$\phi(x)$$ is the formula asserting that x is a square:
 * $$\phi(x) \iff \exists y : x = y^2 $$,

then
 * $$ \mathbb{Q} \not \models \phi(2)$$ but $$\mathbb{R} \models \phi(2)$$

because 2 is a square in the reals, but not the rationals.

It is possible for M ⊆ N and M ≡ N to hold without N being an elementary extension of M. For example, let N be the natural numbers {0,1,2,...} with the structure of an ordered set. Let M be the subset {1,2,3,...}. Then M is a substructure of N and M is elementarily equivalent to N (because M and N are isomorphic). However, $$M \not \preceq N$$. Indeed, if $$\phi(x)$$ is the formula
 * $$ \exists y : y < x$$,

then $$\phi(1)$$ holds in N, but not M, because 1 is not the least element in N, but it is the least element in M.

Examples
If M ⊆ N, and b is a tuple from M, and $$\phi(x)$$ is a quantifier-free formula, then
 * $$ M \models \phi(b) \iff N \models \phi(b)$$

is automatic. Consequently, if T is a theory with quantifier elimination, then any inclusion of models is an elementary inclusion. Indeed, if M ⊆ N, $$\psi(x)$$ is an arbitrary formula, then we can find an equivalent quantifier-free formula $$\phi(x)$$. Then for a tuple b from M,
 * $$ M \models \psi(b) \iff M \models \phi(b) \iff N \models \phi(b) \iff N \models \psi(b)$$.

More generally, if T is model complete, then any inclusion of models of T is an elementary inclusion. In fact, this is one way of characterizing model completeness.

From this, one gets many examples of elementary extensions:
 * The algebraic numbers are an elementary substructure of the field of complex numbers (because ACF is model complete).
 * The real algebraic numbers are an elementary substructure of the field of real numbers (because RCF is model complete).
 * Any extension of non-trivially valued algebraically closed valued fields is an elementary extension, because ACVF is model complete.

The Tarski-Vaught Criterion
Theorem: Let M be a structure, and S be a subset of M. Then S is an elementary substructure of M if and only if the following property holds: for every formula $$\phi(x;y)$$ and every tuple b ∈ S, if $$\phi(M;b)$$ is non-empty, then it intersects S. In other words, every non-empty subset of M definable over S intersects S. (This is called the Tarski-Vaught Criterion.)

Proof: First suppose that S is an elementary substructure of M. Let $$\phi(x;y)$$ be a formula. Suppose that $$\phi(M;b)$$ is non-empty. Then
 * $$ M \models \exists x : \phi(x;b) $$.

In particular, b satisfies the formula $$\exists x : \phi(x;y) $$. Because $$S \preceq M$$, we conclude that
 * $$ S \models \exists x : \phi(x;b) $$

Therefore there is some a in S such that
 * $$ S \models \phi(a,b) $$

Again, because $$S \preceq M$$, we conclude
 * $$ M \models \phi(a,b) $$.

Therefore $$a \in \phi(M;b)$$, so $$\phi(M;b)$$ intersects S.

Conversely, suppose the Tarski-Vaught criterion holds. We prove by induction on n that if $$\psi(x)$$ is a formula in prenex form with n quantifiers, then
 * $$ \psi(S) = \psi(M) \cap S $$,

which is exactly the condition necessary for $$S \preceq M$$ to hold. Since every formula can be written in prenex form, this will suffice.

The base case where n = 0 is the case where $$\psi(x)$$ is quantifier-free. In this case, it is automatically true that $$\psi(S) = \psi(M) \cap S$$, without any assumptions on S or M.

For the inductive step, write $$\psi(x)$$ as $$\exists y : \chi(x;y)$$ or $$ \forall y : \chi(x;y)$$ for some formula $$\chi(x;y)$$ with one fewer quantifier. The existential and universal case are related by negation, so it suffices to consider the existential case. The inductive hypothesis says that if a and b are tuples from S, then
 * $$S \models \chi(a;b) \iff M \models \chi(a;b)$$

Let a be a tuple from S, we want to show that
 * $$ S \models \exists y : \chi(a;y) \iff M \models \exists y : \chi(a;y)$$ (*)

Suppose the right hand side holds. Then there is b in S such that
 * $$ S \models \chi(a;b)$$

By the inductive hypothesis
 * $$ M \models \chi(a;b)$$ and therefore $$ M \models \exists y : \chi(a;y)$$.

So the right hand side of (*) implies the left hand side.

Conversely, suppose that $$M \models \exists y : \chi(a;y)$$. Then $$\chi(a;M)$$ is non-empty, so it intersects S; let b be some point of intersection. Thus
 * $$ M \models \chi(a;b) $$

and b is in S. By the inductive hypothesis,
 * $$ S \models \chi(a;b)$$ and therefore $$ S \models \exists y : \chi(a;y) $$.

So the two sides of (*) are equivalent. QED.

Elementary Chains
An elementary chain is a chain of models
 * $$ M_1 \subset M_2 \subset \cdots $$

such that $$M_i \preceq M_j $$ for i ≤ j. (The chain could have transfinite length). The Tarski-Vaught Theorem on unions of elementary chains says that the union structure
 * $$ \bigcup_i M_i $$

is an elementary extension of Mj for each j.

Elementary Amalgamation Theorem
An elementary embedding is an injective map f: M -> N which induces an isomorphism between M and an elementary substructure of N. For example, any inclusion of an elementary substructure into an elementary extension is an elementary embedding.

The Elementary Amalgamation Theorem says that if f1: M -> N1 and f2: M -> N2 are two elementary embeddings, then there is a structure N3 and elementary embeddings g1 : N1 -> N3 and g2 : N2 -> N3 such that everything commutes:
 * $$g_1 \circ f_1 = g_2 \circ f_2$$

In other words, if M can be elementarily embedded into two structures N1 and N2, then there is an elementary extension of M into which both N1 and N2 can be embedded over M.

This theorem can be proven by extending tp(N1/M) to N2 and realizing the resulting type in an elementary extension of N2.