Field homomorphisms of $\mathbb{R}$ How to prove that $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ ?
(We treat it as field homomorphisms. )
I know that $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$ and $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.
 A: Let us show that every morphism of fields $f:\mathbb R\to \mathbb R$ is the identity, namely $f(r)=r$ for all real $r$.    
0) Trivially $f(q)=q$ for all $q\in \mathbb Q$    
1) Notice that $f$ preserves the order relation in $\mathbb R$ : $x\leq  y \implies f(x)\leq f(y)$.
Indeed if $r\geq 0$ we can write $r=\rho^2$ for some $\rho\in \mathbb R$ and then $f(r)=f(\rho^2)=(f(\rho))^2$ .
Since the  square of a real number is nonnegative ,  we see that $f(r)\geq 0$.
But then $x\leq  y \implies y-x\geq 0 \implies f(y-x)\geq 0\implies f(y)-f(x)\geq 0\implies f(x)\leq f(y)$
and the order relation in $\mathbb R$ is preserved, as claimed.     
2) Given $r\in \mathbb R$  we have for any rational $q,q' \in \mathbb Q$  :
$$q\leq r\implies f(q)=q\leq f(r) \quad\operatorname {and} \quad r\leq q'\implies f(r)\leq f(q')$$ 
Since $r$ and $f(r)$ have the same position relative to rational numbers they are equal: $f(r)=r$ and we have proved that, as announced,  $f$ is the identity.     
[By the way   the definition of real numbers by Dedekind cuts, a definition not very popular nowadays, is based on the very idea that a real number $r$ is characterized by the rational numbers smaller (resp. larger) than $r$ ] 
A: Let $\varphi : \mathbf{R}\to \mathbf{R}$ be a morphism of fields. The kernel of $\varphi$ is an ideal of the field $\mathbf{R}$, and a (commutative) field $k$ has only two ideal, $(0)$ and $k$, and the kernel of $\varphi$ can't obviously be equal to whole $\mathbf{R}$, as then we would have $\varphi(1)=0$ and at the same time (as $\varphi$ is a morphism of fields) $\varphi (1) = 0$. This shows that the kernel of $\varphi$ is $(0)$, that is, that $\varphi$ is injective.
A: Hint: Let $\phi: \mathbb R \to \mathbb R$ is a ring homomorphism. Prove the followings: (1). $\phi(r) = r, \forall r \in \mathbb Q.$ (2). $\phi(x) > 0, \forall x >0.$ (3). $|x-y|<\frac{1}{m} \Rightarrow |\phi(x) - \phi(y)|<\frac{1}{m}.$
