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

Any hints are appreciated. Thanks.

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    $\begingroup$ endomorphism of vector spaces or endomorphism of rings? $\endgroup$ – Gennaro Pasquale Oct 6 '19 at 14:55
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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 $A:=\{q \in \mathbb{Q}: r<q \}$ and let $B:=\{q \in \mathbb{Q}:q <r\}$. 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)<f(q)=q$. Hence $f(r)\leq \inf A=r$. $(1)$

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

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    $\begingroup$ Many thanks!!!! $\endgroup$ – Fyhswdsxjj Oct 7 '19 at 2:11
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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.

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    $\begingroup$ 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 $\endgroup$ – badjohn Oct 6 '19 at 15:22
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    $\begingroup$ 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. $\endgroup$ – reuns Oct 6 '19 at 16:18
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    $\begingroup$ Many thanks!!!! $\endgroup$ – Fyhswdsxjj Oct 7 '19 at 2:20

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