For an exercise in model theory I have to state if $\sqrt{2}$ is a definable element of the structure $\mathcal{R}=(\mathbb{R},+,\cdot,0,1)$. I expect it is not, but I haven't been able to prove this.
I know I should be able to find an automorphism $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ that maps $\sqrt{2}$ to something not equal to $\sqrt{2}$. First, is it true that this image of $\sqrt{2}$ must be $-\sqrt{2}$, since the set $\left\{\sqrt{2},-\sqrt{2}\right\}$ is a set definable by the formula $\psi=\psi(x):=[x\cdot_\mathcal{R}x=1_\mathcal{R}+_\mathcal{R}1_\mathcal{R}]$?
But, when I assume that $\varphi(\sqrt{2})=-\sqrt{2}$, I can't find a suiting function $\varphi$. Is this because I haven't looked good enough or because I'm doing something wrong and $\sqrt{2}$ is actually definable (by what formula??)
There is no homomorphism $\phi : \mathbb{R} \to \mathbb{R}$ of rings such that $\phi(\sqrt{2})=-\sqrt{2}$. In fact, the only ring homomorphism $\mathbb{R} \to \mathbb{R}$ is the identity. The crucial step in the proof is that we can define $\mathbb{R}_{\geq 0}$ as the set of squares $\{x^2 : x \in \mathbb{R}\}$.
Now it is clear how to define $\sqrt{2}$: It is the unique real number $x$ such that $x^2=2$ (and yes, this is an abbreviation for $x*x=1+1$) and such that there is another real number $y$ with $x=y^2$.