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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What is the definitions of $\mathrm{Hom}(\mathbb{R})$ and $\mathrm{Mon}(\mathbb{R})$? – Krish Jan 21 '15 at 19:06
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Hint: takes the rationals to themselves, preserves order. – André Nicolas Jan 21 '15 at 19:07
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$\mathrm{Hom}(\mathbb{K})$ is a set of all homomorphisms $\mathbb{K}\rightarrow\mathbb{K}$. $\mathrm{Mon}(\mathbb{K})$ is set of all monomorphisms $\mathbb{K}\rightarrow\mathbb{K}$. – mikis Jan 21 '15 at 19:10
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@mikis Updated my answer to correct a small flaw in it – Olórin Jan 22 '15 at 14:35
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@GeorgesElencwajg I did as you do : erased my comment. ;-) Btw, did you erase your false answer from http://math.stackexchange.com/questions/1112421/elementary-proof-of-the-irreducibility-of-t4-a-t-1-in-mathbfqt-whe or do you leave it ? ;-) – 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 $r$ ]
Georges Elencwajg
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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.
Olórin
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Thanks for pointing out the mistake. I've deleted my answer. Nice answer. – Krish Jan 22 '15 at 04:51
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@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. – Olórin Jan 22 '15 at 12:59
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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. – Krish Jan 22 '15 at 16:59
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@mikis You're welcome, don't hesitate to upvote if you're ok with the answer ! – Olórin Jan 22 '15 at 17:14
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The statement Krish is asking for a proof of is false in general. Given a map $f: K\to F$ and a field $L$ algebraic over $K$, you'll only be able to extend $f$ to a map $L\to F$ if $F$ has all the right roots to polynomials over $f(K)$ (of course, if $L$ is algebraically closed, then it has all the algebraic elements you need, so the fact is true). – Alex Kruckman Jan 22 '15 at 17:15
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@RobertGreen In fact, $\text{Gal}(\mathbb{R}/\mathbb{Q})$ is trivial. You've already shown in your answer that any map $\mathbb{R}\to\mathbb{R}$ is order-preserving and maps $\mathbb{Q}\to\mathbb{Q}$. No need to add the assumption of continuity - this already implies that it's the identity on all of $\mathbb{R}$. – Alex Kruckman Jan 22 '15 at 17:16
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Since @mikis already knows that the only embedding $\mathbb{R}\to\mathbb{R}$ is the trivial one (writing $\text{Mon}(\mathbb{R}) = {\text{id}}$) and just wants to know why $\text{Hom}(\mathbb{R},\mathbb{R}) = {\text{id}}$, I think the first paragraph of your answer completely answers the question. – Alex Kruckman Jan 22 '15 at 17:20
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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). $|x-y|<\frac{1}{m} \Rightarrow |\phi(x) - \phi(y)|<\frac{1}{m}.$
Krish
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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 $\mathrm{Aut}(F)=\mathrm{Mon}(F)$ ? – mikis Jan 21 '15 at 19:20
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@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) http://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map (2)http://math.stackexchange.com/questions/276189/a-field-that-is-an-ordered-field-in-two-distinct-ways – Krish Jan 21 '15 at 19:52
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@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. – Olórin Jan 22 '15 at 01:00