Given $L$ and $M=\langle\Bbb Q,\cdot\rangle$ where $M$ is a model of $L$ .
I need to show that there is no model $N$ where $N$ is an elementary sub-model of $M$ other than $M$ itself.
Now, I don't see why this is true nor how can I prove it, more than that, if I pick a model $A$ such as defined below, isn't it a valid elementary sub-model of $M$?
For every $x$ in $\Bbb Q$, $x$ can be writen as $\frac {\prod_{i=1}^n p_i^{a_i}} {\prod_{j=1}^m q_j^{b_j}}$ where for all $1\le i\le n, 1\le j\le m$: $p_i\neq q_j$ and $p_i, q_j$ are primes.
Let's define $\Bbb Q^*$ as all such $x$s where:
- $1\le i\le n$ $p_i\neq2$
- $1\le j\le m$ $q_j\neq 2$
Let $A=\langle\Bbb Q^*, \cdot\rangle$
As far as I can see, your example is correct. Notice that, given any finitely many elements $a_1,\dots,a_k$ of $\mathbb Q^*$, there is an isomorphism between $\mathbb Q^*$ and $\mathbb Q$ that fixes those finitely many elements, namely the isomorphism induced by a bijection between the odd primes (used in $\mathbb Q^*$) and all the primes (used in $\mathbb Q$) that fixes all the primes involved in $a_1,\dots,a_k$. The existence of such an isomorphism implies that $a_1,\dots,a_k$ satisfy the same formulas in $\mathbb Q^*$ as in $\mathbb Q$.