Theory vs complete diagram of a model

Question

$\mathcal M$ is a structure in language $\mathcal L$. The expanded language $\mathcal L_\mathcal M$ is obtained from $\mathcal L$ by adding a new constant $c_a$ for each $a \in |\mathcal M|$.

The theory of $\mathcal M$ is the set of sentences $\phi$ of $\mathcal L$ such that $\mathcal M \models \phi$.

The complete diagram of $\mathcal M$ is the set of sentences in the expanded language $\mathcal L_\mathcal M$ which are true in $\mathcal M$, that is the set of sentences $\phi(c_{a_1}, \dots, c_{a_n})$ such that $\mathcal M \models \phi(a_1, \dots, a_n)$.

What is the difference between the two theories? To define $\mathcal M \models \phi$ don't we have to use the expanded language to support quantifier sentences? My understanding of satisfaction / modeling is that we have to work in the expanded language.

Answer 1

Let us say that $\mathcal{L}$ is the language of rings, and $\mathcal{M}$ is $\mathbb{R}$.

Then the theory of $\mathcal{M}$ contains (non-trivial) formulas such as $$\forall x,\forall y, \exists z, x^2+y^2=z^2$$ (a sum of squares is a square). This already contains meaningful information about $\mathbb{R}$, which is definitely not true in general structures.

But you can't refer to any specific real number in those formulas, except possibly those which are definable in the language of rings (like the rational numbers).

On the other hand, in the extended language, you can write formulas such as $$\exists x, \pi + e = x^2$$ with explicit mention of any real numbers. You can see that this formula would of course make no sense for other structures of $\mathcal{L}$, such as $\mathbb{Z}$.

Answer 2

There is already a very good answer, so this answer is just meant as a supplement.

The theory $\operatorname{Th}(\mathcal{M})$ of any structure $\mathcal{M}$ is a complete theory. So any structure $\mathcal{N}$ will be a model of that theory precisely when it is elementarily equivalent to $\mathcal{M}$. For example, if $\mathcal{M} = (\mathbb{R}, <)$, the reals with just the order symbol, then $\operatorname{Th}(\mathcal{M})$ will be the theory of dense linear orders without endpoints (DLO). Then $(\mathbb{Q}, <)$ is a model of this theory.

If we consider the complete diagram $\operatorname{Diag}(\mathcal{M})$ of $\mathcal{M}$, we get something much stronger. The point of this theory is that now any model $\mathcal{N} \models \operatorname{Diag}(\mathcal{M})$ will not only be elementarily equivalent to $\mathcal{M}$, but it will also be an elementary extension of $\mathcal{M}$. Since $\operatorname{Diag}(\mathcal{M})$ contains a constant $c_a$ for every $a \in |\mathcal{M}|$, we can define a function $f: \mathcal{M} \to \mathcal{N}$ by $f(a) = c_a^{\mathcal{N}}$. Here $c_a^{\mathcal{N}}$ is the interpretation of $c_a$ in $\mathcal{N}$. Then $$ \mathcal{M} \models \varphi(a_1, \ldots, a_n) \quad \Longleftrightarrow \quad \varphi(c_{a_1}, \ldots, c_{a_n}) \in \operatorname{Diag}(\mathcal{M}) \quad \Longleftrightarrow \quad \mathcal{N} \models \varphi(c_{a_1}^{\mathcal{N}}, \ldots, c_{a_n}^{\mathcal{N}}), $$ and this last expression is just $\mathcal{N} \models \varphi(f(a_1), \ldots, f(a_n))$. So $f$ is an elementary embedding.

Back to our example of $\mathcal{M} = (\mathbb{R}, <)$. Any model $\mathcal{N} \models \operatorname{Diag}(\mathcal{M})$ must contain a copy of $\mathbb{R}$. So $(\mathbb{Q}, <)$ can no longer be a model of the complete diagram of $\mathcal{M}$, even though it was a model of the theory of $\mathcal{M}$.