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It is well known that we can define $\mathbb{N}$ in $(\mathbb{Z},+,\cdot)$ via an existentially quantified equality, as follows. Letting $n$ be an integer parameter $$ n\in \mathbb{N} \Longleftrightarrow \exists w,x,y,z\in \mathbb{Z},\ n=w^2+x^2+y^2+z^2. $$ On the other hand, it is conjectured that $\mathbb{Z}$ has no diophantine definition in $(\mathbb{Q},+,\cdot,0,1)$. However, $\mathbb{Z}$ can be defined using a $\forall_{10}$ formula (see https://arxiv.org/abs/2301.02107 for the argument, as well as additional historical information).

Question: Does $\mathbb{Z}_{>0}$ have a diophantine definition in $(\mathbb{Q}_{>0},+,\cdot,1)$?

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    $\begingroup$ I'm probably missing something: $\mathbb{Q}_{>0}$ has a Diophatine definition in $\mathbb{Q}$, namely $q>0$ if and only if $q (a^2+b^2+c^2+d^2)=1$ has a solution in $\mathbb{Q}$. So, if $\mathbb{Z}_{>0}$ were diophantine in $\mathbb{Q}_{>0}$, wouldn't that mean $\mathbb{Z}_{>0}$ was diophantine in $\mathbb{Q}$? And then you can define $x \in \mathbb{Z}$ as $x = a-b$ for $a$, $b \in \mathbb{Z}>0$? $\endgroup$ Commented Oct 22 at 20:29
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    $\begingroup$ @DavidESpeyer I don't think you are missing anything. I should have given it more thought! That's a nice solution. (The conjunction of two equalities $a=b$ and $c=d$ can be expressed as a single equality $(a-b)^2+(c-d)^2=0$. So everything you did can be expressed by a single equality.) $\endgroup$ Commented Oct 22 at 20:48
  • $\begingroup$ [And of course the squares can be expanded, and the negatives moved to the other side, if we want to avoid negations.] $\endgroup$ Commented Oct 22 at 20:51
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    $\begingroup$ I see that at the moment you have reputation of exactly 19,000. Congrats! $\endgroup$ Commented Oct 23 at 1:38

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Recording the comments as an answer:

$\mathbb{Z}$ has a Diophantine definition in $(\mathbb{Q}, +, \cdot, 0, 1)$ iff it has a Diophantine definition in $(\mathbb{Q}^+, +, \cdot, 1)$.

Proof: If $z\in \mathbb{Z}$ can be defined by $$(\exists x_1, \ldots x_n \in \mathbb{Q})\bigwedge_{i=1}^k P_i(z,x_1,\ldots,x_n)=0$$ then it can also be defined replacing $x\in\mathbb{Q}$ by $v,w\in\mathbb{Q}^+$ with
$$(\exists v_1, \ldots v_n, w_1, \ldots, w_n \in \mathbb{Q}^+)\bigwedge_{i=1}^k P_i(z,v_1-w_1,\ldots,v_n-w_n)=0$$

Likewise, if $z\in \mathbb{Z}$ can be defined by $$(\exists x_1, \ldots x_n \in \mathbb{Q^+})\bigwedge_{i=1}^k P_i(z,x_1,\ldots,x_n)=0$$ then it can also be defined replacing $x\in\mathbb{Q}^+$ by $s,t,u,v,w\in \mathbb{Q}$ with $$(\exists s_1, \ldots, w_n \in \mathbb{Q})\bigwedge_{i=1}^k P_i(z,\frac{1+s_1^2+t_1^2+u_1^2+v_1^2}{1+w_1^2},\ldots,\frac{1+s_n^2+t_n^2+u_n^2+v_n^2}{1+w_n^2})=0$$

In both cases, we assume that the $P$’s are polynomials with integer coefficients, and can add terms to both sides to cancel out any negative terms, or multiply both sides by denominators to leave positive polynomials.

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  • $\begingroup$ Not that David needs the reputation, but I think it might be appropriate, when recording someone else's comments as an answer, to give them the opportunity to post it first (I believe David originally posted it as a comment to make sure he wasn't missing something), or at least to make the answer CW. $\endgroup$ Commented Oct 23 at 14:29

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