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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 $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.

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  • $\begingroup$ What is the definitions of $\mathrm{Hom}(\mathbb{R})$ and $\mathrm{Mon}(\mathbb{R})$? $\endgroup$ – Krish Jan 21 '15 at 19:06
  • $\begingroup$ Hint: takes the rationals to themselves, preserves order. $\endgroup$ – André Nicolas Jan 21 '15 at 19:07
  • $\begingroup$ $\mathrm{Hom}(\mathbb{K})$ is a set of all homomorphisms $\mathbb{K}\rightarrow\mathbb{K}$. $\mathrm{Mon}(\mathbb{K})$ is set of all monomorphisms $\mathbb{K}\rightarrow\mathbb{K}$. $\endgroup$ – mikis Jan 21 '15 at 19:10
  • $\begingroup$ @mikis Updated my answer to correct a small flaw in it $\endgroup$ – Olórin Jan 22 '15 at 14:35
  • $\begingroup$ @GeorgesElencwajg I did as you do : erased my comment. ;-) Btw, did you erase your false answer from math.stackexchange.com/questions/1112421/… or do you leave it ? ;-) $\endgroup$ – Olórin Jan 22 '15 at 20:20
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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 $r$ ]

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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.

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  • $\begingroup$ Thanks for pointing out the mistake. I've deleted my answer. Nice answer. $\endgroup$ – Krish Jan 22 '15 at 4:51
  • $\begingroup$ @Krish Np ! And thx ! Don't hesitate to upvote if it's nice ;-) More seriously yes, a lot of things (even nastier that this one) can happen in $\textrm{Gal}(\mathbf{R}/\mathbf{Q})$, really. $\endgroup$ – Olórin Jan 22 '15 at 12:59
  • $\begingroup$ @Krish Updated my answer to correct it, as there was a flaw in it. $\endgroup$ – Olórin Jan 22 '15 at 14:26
  • $\begingroup$ @RobertGreen, Thank you very much! $\endgroup$ – mikis Jan 22 '15 at 16:07
  • $\begingroup$ In the first line of the construction, there is a typo: "transcendence basis of $\mathbb R$ over $\mathbb Q$" not "over $\mathbb R$". Also can you give a reference/proof for the claim: "Now, as $\mathbb R$ is algebraic over $K,$ one can (classic fact) extend $F:K \to \mathbb R$ to $\mathbb R.$" I know if $K/k$ is an algebraic extension and there is a map $k \to C,$ where $C$ is an algebraically closed field then it can be extended to a map $K \to C.$ But here $\mathbb R$ is not algebraically closed. May be I'm not thinking correctly or missing something trivial. $\endgroup$ – Krish Jan 22 '15 at 16:59
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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). $|x-y|<\frac{1}{m} \Rightarrow |\phi(x) - \phi(y)|<\frac{1}{m}.$

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  • $\begingroup$ Right, it can be proved as $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$. So, can it be generalize for dense-ordered field $(F,<)$ ? What condition we must have to prove that $\mathrm{Aut}(F)=\mathrm{Mon}(F)$ ? $\endgroup$ – mikis Jan 21 '15 at 19:20
  • $\begingroup$ @mikis: I'm sorry, but I don't know much about ordered field. You can see some information about it in the answers by Pete L. Clark in the following two links: (1) math.stackexchange.com/questions/449404/… (2)math.stackexchange.com/questions/276189/… $\endgroup$ – Krish Jan 21 '15 at 19:52
  • $\begingroup$ @Krish $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ is false, as you may see in my answer. OP never wrote that morphisms were continuous, and even restated - without mention to continuity - what Mon and Hom mean. $\endgroup$ – Olórin Jan 22 '15 at 1:00

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