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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?

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

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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 $$\forall n\in\Bbb N\quad(n<a\iff n<b)$$ 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 $n\in\Bbb N$
  • $s_{<0}$: $r<0$
  • $s_n$: $n<r<n+1$ 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 $(n,b),b,(b,n+1)$.

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