Let $\tau$ be a signature and $\mathcal{A}$ and $\mathcal{B}$ are two $\tau$-structures with domains $A$ and $B$. $\mathcal{A}$ is called an elementary substructure of $\mathcal{B}$ if $A\subseteq B$ and if for every $a_1,\ldots,a_n\in A$ and every $\tau$-formula (First-order logic) $\varphi(x_1,\ldots,x_n)$ it holds that
$$\mathcal{A}\models \varphi[a_1,\ldots,a_n]\Longleftrightarrow \mathcal{B}\models\varphi[a_1,\ldots,a_n].$$
Now let's look at the signature $\tau:=\{0,1,+,\cdot\}$ and the $\tau$-structure $\mathcal{Q}=(\mathbb{Q},0,1,+,\cdot)$. I want to show, that $\mathcal{Q}$ has no proper elementary substructures.
I do have some approaches, but it's a little bit weird... Here's my thought process about this:
So my idea is to assume that there's a proper elementary substructure $\mathcal{P}=(P,0,1,+,\cdot)$ of $\mathcal{Q}$ and build a $\tau$-formula $\varphi(x_1,\ldots,x_n)$ such that there are $q_1,\ldots,q_n\in\mathbb{Q}$ with $\mathcal{Q}\models\varphi[q_1,\ldots,q_n]$ but $\mathcal{P}\not\models\varphi[q_1,\ldots,q_n]$. It might be possible, that $\varphi(x_1,\ldots,x_n)$ has no free variable... Not sure yet...
To find a good formula for this, I was thinking about what I know about $\mathcal{Q}=(\mathbb{Q},0,1,+,\cdot)$. I know, that this is a model for the theory of fields. It is also a model for the theory of commutative unitary rings. In addition to that, we know that there are proper commutative unitary subrings of $\mathbb{Q}$ (e.g. $\mathbb{Z}$) but there are no proper subfields of $\mathbb{Q}$. And this seems kind of interesting for this task... The difference between a commutative unitary ring and a field is that a commutative unitary ring does not have to have multiplicative inverses. That's why I first tried
$$\varphi:=\forall x\exists y(x\cdot y\equiv 1).$$
Obviously, $\mathcal{Q}\models \varphi$ since $\mathcal{Q}$ is a field. But what happens if we remove one element from $\mathcal{Q}$, that is not $0$ or $1$ and call it $\mathcal{P}_1$. Then $\mathcal{P}_1\not\models \varphi$ since there's an element (the inverse of the element we just removed) in $\mathcal{P}_1$ that has no inverse. But then we could just remove that element and call it $\mathcal{P}_2$, and we have $\mathcal{P}_2\models \varphi$ again... So this formula does not work for this... This does not really trigger a chain reaction that throws out all the elements.
Maybe I should try completeness regarding addition or multiplication. But this alone does not work either, since we have substructures like $\mathbb{Z}$ that satisfy this condition.
So maybe I need completeness regarding addition AND existence of a multiplicative inverse. That would be
$$\varphi := (\forall x\exists y(x\cdot y\equiv 1)) \land (\forall x\forall y\exists z (x+y=z)).$$
If we now have a structure $\mathcal{P}=(P,+,\cdot,0,1)$ with $P\subsetneq \mathbb{Q}$ there is just no way we can have $\mathcal{P}\models \varphi$ because of the fact that there are no subfields of $\mathbb{Q}$. But maybe there are still proper subsets of $\mathcal{Q}$ with the condition $\varphi$. So maybe I have to conjugate every axiom of theory of fields...
I'm not sure if my reasoning works that way... Maybe one of you has an idea or a tip that could provide some clarity in the chaos.
Thank you very much. Kind regards, Max.