$tp^\mathbb{Q}(a/\mathbb N) = tp^\mathbb{Q}(b/\mathbb N)$ iff there is an automorphism $\sigma$ of $\mathbb Q$ fixing $\mathbb N$ and $\sigma(a)=b$

If $$a, b \in \mathbb Q$$, then $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)$$ if and only if there is an automorphism $$\sigma$$ of $$\mathbb Q$$ fixing $$\mathbb N$$ pointwise with $$\sigma(a) = b$$.

I have proven that backward implication, and I am stuck on proving the forward implication. I have shown that for any natural number $$n \in \mathbb N$$ we have that $$n < a \leq n+1$$ iff $$n < b \leq n+1$$. However, I wasn't able to go further from here.

EDIT: thanks for your helpful suggestions and comments. So the underlying language is $$\{ < \}$$ and the automorphism is a model isomorphism from $$\mathbb Q$$ to itself. Upon thinking a bit I figured that I could define an automorphism from $$\mathbb Q$$ to itself by defining it to be the identity on anywhere outside $$(n,n+1]$$, where $$n$$ is the (unique) integer such that $$n < a \leq n+1$$ and do a back-and-forth argument between $$(n,a)$$ and $$(n,b)$$, and similarly do a back-and-forth argument between $$(a,n+1]$$ and $$(b,n+1]$$.

The reason that I argued $$n < a ≤ n+1$$ iff $$n < b ≤ n+1$$ is as follows: consider the formula $$c_i < x$$ in the language $$\mathcal L_{\mathbb N}$$, where $$i \in \mathbb N$$ and where the interpretation of $$c_i$$ is $$c_i^{\mathbb Q} = i \in \mathbb N$$. If $$\mathbb Q \models c_i < x [a]$$, then this means that $$i < a$$ (we are just subsituting $$a$$ for $$x$$ and interpreting $$c_i$$ as $$i$$. But if $$\mathbb Q \models c_i < x [a]$$ means that the sentence $$c_i < x$$ is in $$\text{tp}^{\mathbb Q}(a/\mathbb N)$$, and since we are assuming that $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)$$, we also have (after interpretation) that $$i < b$$ also.

We can do a similar thing for $$x < c_i \lor (x = c_i)$$ to get that $$a \leq i$$ iff $$b \leq i$$. Combining this with the previous argument, we have that, for any natural number $$i$$, $$i < a \leq i+1$$ iff $$i < b \leq i+1$$.

Now let $$j$$ be a negative integer such that $$j < a$$. Then this means that $$0 < a + (-j)$$ where $$-j$$ is a natural number. Then $$c_0 < x + c_{-j}$$ is a formula in the language $$\mathcal L_{\mathbb N}$$, and if this sentence is in $$\text{tp}^{\mathbb Q}(a/\mathbb N)$$, i.e. (after interpretation of this sentence) $$0 < a + (-j)$$, then we have that $$j < a$$. We also have that this sentence is in $$\text{tp}^{\mathbb Q}(b/\mathbb N)$$, which after interpretation means that $$0 < b + (-j)$$, i.e. $$j < b$$. By similar argument for the case $$a < j \lor (b = j)$$ we get that $$a \leq j$$ iff $$a \leq j$$ for a negative integer $$j$$.

That is how I got that $$n < a \leq n+1$$ iff $$n < b \leq n+1$$ for any integer $$n$$.

Again, as I have mentioned, I thought of now defining an automorphism from $$\mathbb Q$$ to itself by defining it to be the identity on anywhere outside $$(n,n+1]$$, where $$n$$ is the (unique) integer such that $$n < a \leq n+1$$ and do a back-and-forth argument between $$(n,a)$$ and $$(n,b)$$, and similarly do a back-and-forth argument between $$(a,n+1]$$ and $$(b,n+1]$$.

So I wanted to ask: is my argument valid?

• Please define $\text{tp}^{\mathbb Q}(x/\mathbb N)$ (in which language?) and "automorphism" (of ring or what?), and show how you proved $n < a \leq n+1\iff n <b \leq n+1.$ Nov 27, 2023 at 14:55
• You should also clarify what the structure is on $\mathbb{Q}$. Nov 27, 2023 at 14:57
• Are these linear orderings? It may help to look at the proof that every countable dense linear ordering without endpoints is isomorphic to $\mathbb{Q}$. Nov 27, 2023 at 15:00
• What does $\text{tp}^\mathbb{Q}(x/\mathbb{N})$ mean? Nov 28, 2023 at 4:27
• "the underlying language is $\{ < \}$" (which excludes $+$) seems to contradict "$c_0 < x + c_{-j}$ is a formula in the language $\mathcal L_{\mathbb N}$". Nov 28, 2023 at 14:19

Let $$a,b\in\Bbb Q$$ such that $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)$$.

Your proof that $$\forall n\in\Bbb N\quad(n is correct, and similarly $$\forall n\in\Bbb N\quad(n>a\iff n>b)$$ and $$\forall n\in\Bbb N\quad(a=n\iff b=n)$$ but this does not extend to $$n\in\Bbb Z,$$ because $$+$$ does not belong to the language, so your claim "$$c_0 < x + c_{-j}$$ is a formula in the language $$\mathcal L_{\mathbb N}$$" is false.

Assuming that your conventions consider $$0$$ as an element of $$\Bbb N,$$ the types $$\text{tp}^{\mathbb Q}(r/\mathbb N)$$ are:

• $$t_n$$: $$r=n$$ for some $$n\in\Bbb N$$
• $$s_{<0}$$: $$r<0$$
• $$s_n$$: $$n for some $$n\in\Bbb N.$$

If $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)=t_n,$$ we can choose $$\sigma={\rm id}_{\Bbb Q}$$.

If $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)=s_{<0},$$ we can choose an automorphism $$\sigma$$ of $$(\Bbb Q,<)$$ fixing every point of $$\Bbb Q_{\ge0}$$ and sending respectively $$(-\infty,a),a,(a,0)$$ to $$(-\infty,b),b,(b,0)$$.

If $$\text{tp}^{\mathbb Q}(a/\mathbb N) = \text{tp}^{\mathbb Q}(b/\mathbb N)=s_n,$$ we can choose an automorphism $$\sigma$$ of $$(\Bbb Q,<)$$ fixing every point of $$\Bbb Q\setminus(n,n+1)$$ and sending respectively $$(n,a),a,(a,n+1)$$ to $$(n,b),b,(b,n+1)$$.