Let $M$ be the model $\langle \mathbb{Z}, s^M\rangle$ where $s$ is a unary function symbol interpreted as the standard successor function ($s^M(x)=x+1$). Let $F$ be a nonprincipal ultrafilter on $\omega$ and let $N$ be the ultrapower $N:=M^{\omega}/F$. I'm struggling to answer the following question:
Prove that for every $i <\omega$ there is an injective homomorphism $f_i: M \to N$ such that $f_i(M) \cap f_j(M) = \emptyset$ for every $i \neq j$.
I really don't how to define the image of $n$ under $f_i$. The sequence should somehow codes $i$ and $n$. All the 'obvious' guesses ($ n \mapsto [\langle 0, \ldots, 0, n, n, \ldots\rangle]_F$ or $n \mapsto [\langle 0, \ldots 0, n, n+1, n+2, \ldots \rangle]_F$) are clearly wrong.
Let $d_i : n\mapsto i\cdot n$. Then $[d_i]_F$ is an element of $N$.
For $m\in\mathbb N$, let $\hat m : n \mapsto m$ be the constant map.
Note that $f_i : m\mapsto [d_i+\hat m]_F$ is an embedding of $M$ in $N$.
Now, let $i<j$. I claim that $f_i[M]$ is disjoint of $f_j[M]$.
Pick two arbitrary elements $[d_i+\hat m_1]_F\in f_i(M)$ and $[d_j+\hat m_2]_F\in f_i(M)$.
Note that $(d_i+\hat m_1)(n)= i\cdot n+m_1$ and $(d_j+\hat m_j)(n)= j\cdot n+m_2$.
Hence $(d_i+\hat m_1)(n)<(d_j+\hat m_j)(n)$ for almost all $n$. Therefore $[d_i+\hat m_1]_F<[d_j+\hat m_2]$.
As $m_1$ and $m_2$ are arbitrary, the claim follows.