# An exercise in model theory about positive homomorphisms

A formula is called positive if is is built from atomic formulas using only $$\land, \lor, \exists$$ and $$\forall$$. A homomorphism $$f : M \to N$$ is positive if $$M \models \varphi(m_1, \cdots, m_n) \Rightarrow N \models \varphi(f(m_1), \cdots, f(m_n))$$ for all positive $$\varphi(x_1, \cdots, x_n)$$ and all $$m_1, \cdots, m_n \in M$$. Let $$T$$ be a satisfiable $$L$$-theory and $$T_0 = \{\varphi : \varphi \text{ is positive and } T \models \varphi \} \, .$$

Prove that for any model $$A \models T_0$$ there is a model $$C$$ such that there exists 1) a model $$B \models T$$ and a positive homomorphism $$g : B \to C$$; and 2) an elementary embedding $$h : A \to C$$ with $$\text{Im}(h) \subset \text{Im}(g)$$. [Where $$\text{Im}(\cdot)$$ stands for "image"]

I really don't know how to approach such a problem. The best I can think of is to prove that there is a model for $$\text{ElDiag}(A) \cup \{\varphi : B \models \varphi \text{ and } \varphi \text{ is positive}\}$$ for some model $$B \models T$$ to which we have added all the constants of $$L_A$$ ($$L_A$$ is $$L$$ with all the elements in $$A$$ added as constants; $$\text{ElDiag}$$ = elementary diagram). Maybe we can use compactness to prove the existence, utilising that we know $$A \models T_0$$ and we have freedom to choose $$B \models T$$?

Could you provide a hint? I have been at this exercise for several days, but am completely stuck; I have no strategy to approach it.

• What are the $>\to$ and $>\Rightarrow$ supposed to mean? And the $>$ in the definition of $T_0$? – tomasz Mar 11 '19 at 19:51
• @tomasz Nothing. I fixed it now; apparently the blockquote tool does not handle LaTeX well. – Joachim Mar 12 '19 at 10:12

Your idea to solve the problem by the method of diagrams and compactness is the right one. But the first thing you should ask yourself is: What is $$B$$, and what is its relationship to $$A$$? Can we take any model $$B\models T$$?

Well, no, we can't in general... Since we must have $$\text{Im}(h)\subseteq \text{Im}(g)$$, for all $$a\in A$$ there must be some $$a'\in B$$ such that it is consistent that $$g(a') = h(a)$$, where $$h$$ is some elementary embedding $$A\to C$$ and $$g$$ is some positive homomorphism $$B\to C$$. What does this consistency amount to? Suppose $$\varphi(x)$$ is a positive formula such that $$B\models \varphi(a')$$, where $$h(a) = g(a')$$. Then $$C\models \varphi(g(a'))$$, so $$C\models \varphi(h(a))$$, and $$A\models \varphi(a)$$. This is a constraint on $$B$$: If $$A\models \lnot \varphi(a)$$, then we must have $$B\models \lnot \varphi(a')$$.

So let's define a negative formula to be the negation of a positive formula, and consider the $$L_A$$-theory: $$T\cup \text{Diag}^-(A)\text{, where }\text{Diag}^-(A) = \{\psi(a)\mid \psi(x)\text{ is negative, and }A\models \psi(a)\}.$$

You can show by compactness that this theory is consistent, using our assumption that $$A\models T_0$$.

So $$B$$ be a model. $$B$$ is an $$L_A$$-structure, so when we form the language $$L_B$$, we reuse the constants naming the elements of $$A$$, i.e. $$L_A\subseteq L_B$$. Now consider the $$L_B$$-theory: $$\text{ElDiag}(A)\cup \text{Diag}^+(B)\text{, where }\text{Diag}^+(B) = \{\varphi(b)\mid \varphi(x)\text{ is positive, and }B\models \psi(b)\}.$$

It remains to show that this theory is consistent, using compactness and the fact that $$B\models \text{Diag}^-(A)$$.

• Thank-you, Alex. I knew I was not that far off. This was a very good tip. I have found the solution now, but it was still not quite trivial to me (especially how to deal with the constants). You left me enough work to learn something :) . Can I interpret your second paragraph as to how you (or someone in general) might have come to this solution? – Joachim Mar 11 '19 at 13:54
• @Joachim Yes, I had to think through the reasoning in the second paragraph to see which theory to write down when finding $B$. – Alex Kruckman Mar 11 '19 at 14:09