Dividing under ultraproduct

Question

Let $\{M_i\}_{i<\omega}$ be a family of structures and let $\phi(x)$ be a formula which $k$-divides in $M_i$ for each $i<\omega$. Let $M^*$ be an ultraproduct of $M_i$'s.

Question. Does $\phi(x)$ k-divide (or just divide) in $M^*$? In other words; is dividing preserved under ultraproduct?

Answer 1

We need to be a little bit careful about the parameters here - recall that a formula without parameters trivially does not divide (unless it is inconsistent)!

So I'll assume we're given a formula $\varphi(x,y)$ and a tuple $b_i\in M_i$ such that $\varphi(x,b_i)$ $k$-divides (over $\varnothing$) in $M_i$ for all $i<\omega$. Then, writing $b^* = [b_i]_{i<\omega}$ for the equivalence class of $(b_i)_{i<\omega}$ in the ultraproduct, we have that $\varphi(x,b^*)$ $k$-divides (over $\varnothing$) in $M^*$.

This follows pretty easily from Łoś's Theorem. The $k$-dividing is witnessed by, in each $M_i$, a sequence $(b_{i,j})_{j<\omega}$ such that:

• $\text{tp}(b_{i,j}) = \text{tp}(b_i)$ for all $j<\omega$. This is witnessed by $M_i\models \psi(b_{i,j})\leftrightarrow \psi(b_i)$ for all $j<\omega$ and all formulas $\psi(y)$.
• $\{\varphi(x,b_{i,j})\mid j<\omega\}$ is $k$-inconsistent. This is witnessed by $M_i\models \lnot \exists x\, (\varphi(x,b_{i,j_1})\land \dots \varphi(x,b_{i,j_k}))$ for all $j_1<\dots<j_k<\omega$.

Now define $b^*_j = [b_{i,j}]_{i<\omega}$. By Łoś's Theorem, we have:

• $M^*\models \psi(b^*_j)\leftrightarrow \psi(b^*)$ for all $j<\omega$ and all formulas $\psi(y)$. So $\text{tp}(b^*_j) = \text{tp}(b^*)$ fro all $j<\omega$.
• $M^*\models \lnot \exists x\, (\varphi(x,b^*_{j_1})\land \dots \varphi(x,b^*_{j_k}))$ for all $j_1<\dots<j_k<\omega$. So $\{\varphi(x,b^*_j)\mid j<\omega\}$ is $k$-inconsistent.

Thus $(b_j^*)_{j<\omega}$ witnesses $k$-dividing of $\varphi(x,b^*)$ in $M^*$.

Note that is was important that the formula $\varphi$ and the number $k$ was uniform across the $M_i$! It is not true in general that if $\text{tp}(a_i)$ divides (over $\varnothing$) in each $M_i$, then $\text{tp}(a^*)$ divides (over $\varnothing$) in $M^*$, where $a^* = [a_i]_{i\in \omega}$. To lift dividing to the ultraproduct, we need to know that the same formula witnesses $k$-dividing for the same $k$ in all $M_i$ (or at least in a large set of the $M_i$, with respect to the ultrafilter).