The only endomorphism of $\mathbb{R}$ is identity How to prove that the only endomorphism of $\mathbb{R}$ is identity? 
Any hints are appreciated. Thanks.
 A: As you wrote "field theory", I assume you are talking about endomorphisms of rings.
Let $f \colon \mathbb{R}\to \mathbb{R}$ be an endomorphism of rings. Then, knowing that $f(1)=1$, you can prove that $f(q)=q$ for every $q \in \mathbb{Q}$.
Now, suppose that $r,s\in \mathbb{R}$ are such that $r \leq s$. Then there is $t \in \mathbb{R}$ such that $r+t^2=s$ (just pick $t:=\sqrt{s-r}$). Then $f(s)=f(r+t^2)=f(r)+f(t)^2$. In particular $f(r) \leq f(s)$. Hence $f$ is increasing.
Now, let $r\in \mathbb{R}$ and let $A:=\{q \in \mathbb{Q}: r \leq q \}$ and let $B:=\{q \in \mathbb{Q}:q \leq r\}$. Then $r=\inf A$ and $r=\sup B$.
As $f$ is increasing, for every $q \in A$ it is the case that $ f(r) \leq f(q)=q$. Hence $f(r)\leq \inf A=r$. $(1)$
Moreover, as $f$ is increasing, for every $q \in B$ it is the case that $q=f(q) \leq f(r)$. Hence $r=\sup B\leq f(r)$. $(2)$
By $(1)$ and $(2)$ it is the case that $f(r)=r$ and $r \in \mathbb{R}$ is arbitrary.
A: if $f$ a continuous endomorphism (of rings) of $\mathbb{R}$, we have $f(1)=1$ and then $f(x)=f(x\cdot 1)=xf(1)=x$ for every real $x$.
Remark: to prove $f(x\cdot 1)=xf(1)$ you can prove it for $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and pass to $\mathbb{R}$ by density and continuity.
