# Let $T$ be a model complete theory and let $\mathfrak{M}$ be a model of $T$ which embeds into every model of $T$. Show that $T$ is complete.

I need to prove the statement but am not sure about my proof. Here it is:

First we recall some definitions and basic results that we will use to prove the statement. At theory is model complete if for all models $$\mathfrak{M}^{1}$$ and $$\mathfrak{M}^{2}$$ of $$T$$, if $$\mathfrak{M}^{1} \subset \mathfrak{M}^{2}$$ then $$\mathfrak{M}^{1} \prec \mathfrak{M}^{2}$$. A theory $$T$$ is complete if and only if all models of $$T$$ are elementarily equivalent. Whenever $$\mathfrak{M}^{1}$$, $$\mathfrak{M}^{2}$$ are models of $$T$$ with a common $$\mathcal{L}$$-substructure $$\mathfrak{A}$$, then Th($$\mathfrak{M}_{A}^{1}$$)=Th($$\mathfrak{M}_{A}^{2}$$). We can now say that $$\mathfrak{M}$$ is a common elementary substructure since it embeds into every other model of $$T$$. Now since all those models have a common $$\mathcal{L}$$-substructure, the equivalence of each theory of each model follows. Hence $$T$$ is complete.

Could anyone validate it or give me a hint on how to tackle it?

Whenever $$\mathfrak{M}^{1}$$, $$\mathfrak{M}^{2}$$ are models of $$T$$ with a common $$\mathcal{L}$$-substructure $$\mathfrak{A}$$, then $$\text{Th}(\mathfrak{M}_{A}^{1})=\text{Th}(\mathfrak{M}_{A}^{2})$$.

This is false in general! It's equivalent to saying that $$T$$ has quantifier elimination, and there are model complete theories which do not have quantifier elimination.

Now since all those models have a common $$\mathcal{L}$$-substructure, the equivalence of each theory of each model follows.

Instead of using the fact that $$\mathfrak{M}^1$$ and $$\mathfrak{M}^2$$ have a common $$\mathcal{L}$$-substructure, use the (much stronger) fact that they have a common elementary substructure.

Suppose a sentence $$\varphi$$ is true in $$\mathfrak{M}^1$$. Then it's true in $$\mathfrak{M}$$. Then it's true in $$\mathfrak{M}^2$$. Do you see why? And do you see why this implies $$T$$ is complete?

• I think because $\mathfrak{M}$ embeds into every model of T? So the sentence would be satisfied for every model of T actually. SInce now $\mathfrak{M}$ is an elementary substructure oft those, that implies that T is complete, because in this case the theories of all models are equivalent? – craft Nov 20 '20 at 17:18
• @craft Yes, that's right. – Alex Kruckman Nov 20 '20 at 18:11
• Great, thanks so much – craft Nov 20 '20 at 18:23