I'm studying Model theory using David Marker's "Model Theory: An Introduction".
In the first chapter he presents the following proposition:
Proposition 1.3.5:
Let $\cal M$ be an $\cal L$-structure. If $X\subseteq M^n$ is $A$-definable, then every $\cal L$-automorphism of $\cal M$ that fixes $A$ pointwise fixes $X$ setwise.
and after this, he gives an example that, in the languange of rings $\mathcal{ L }= \{+,-,\cdot, 0,1\}$, the only automorphism of $\mathcal{ M}=(\mathbb R,+,-,\cdot,0,1)$ is the indentity. The argument is as follows:
"Any automorphism of the real field must fix the rational numbers. Because the order is $\emptyset$-definalbe it must be preserved by the automorphism. Because the rationals are dense, the only automorphism is the identity".
I understand why the order must be preserved, and I understand why "order being preserved" + "rationals being fixed" $\implies$ "any automorphism is the identity", but I'm having some trouble understanding why the rational numbers must be fixed by the automorphism.
Why is this the case?
If it's a field automorphism it must fix $0$ and $1$ as those are the only two numbers equal to their own square. That in turn fixes any two integers $n$ and $m$ as $1+1$ $n$ or $m$ times, which in turn fixes any rational $n/m$.