# Exercise about ultraproducts

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 $$i. I claim that $$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.

• @Mockingbird Alex Kruckman has been quicker than me. Note that the answer is exactly the same. Jan 17 at 15:05
• Yes @PrimoPetri he was quicker you provided a little more detail. I take some time to choose the answer to accept Jan 17 at 15:46

How about $$f_i(n)=[\langle n, i+n, 2i+n, 3i+n,\dots\rangle]_F$$?

It's worth verifying for yourself that this works, and also spending some time thinking about how I could have come up with this solution.