# The only endomorphism of $\mathbb{R}$ is identity [duplicate]

How to prove that the only endomorphism of $$\mathbb{R}$$ is identity?

Any hints are appreciated. Thanks.

• endomorphism of vector spaces or endomorphism of rings? – Gennaro Pasquale Oct 6 '19 at 14:55

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 $$r < s$$. 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 $$f(r)< f(s)$$. Hence $$f$$ is strictly increasing.

Now, let $$r\in \mathbb{R}$$ and let $$A:=\{q \in \mathbb{Q}: r and let $$B:=\{q \in \mathbb{Q}:q . Then $$r=\inf A$$ and $$r=\sup B$$.

As $$f$$ is strictly increasing, for every $$q \in A$$ it is the case that $$f(r). Hence $$f(r)\leq \inf A=r$$. $$(1)$$

Moreover, as $$f$$ is strictly increasing, for every $$q \in B$$ it is the case that $$q=f(q). Hence $$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.

• Many thanks!!!! – Fyhswdsxjj Oct 7 '19 at 2:11

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.

• That $\mathbb{Q}$ is fixed is simple. Assuming continuity allows the extension to the whole of $\mathbb{R}$ easily but it is not needed as Gennaro shows. The situation for $\mathbb{C}$ is different – badjohn Oct 6 '19 at 15:22
• Concretely it is because the continuity, ie. the order on $\Bbb{R}$, can be algebraized : $s \ge 0$ iff $x^2-s$ has a real root. – reuns Oct 6 '19 at 16:18
• Many thanks!!!! – Fyhswdsxjj Oct 7 '19 at 2:20