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I got this question in a homework:

Determine all ring homomorphisms from $\mathbb{R} \to \mathbb{R}$. Also prove that the only ring automorphism of $\mathbb{R}$ is the identity.

I know that $\mathbb{R}$ is a field, so the only ideals are $\mathbb{R}$ and $\{0\}$. Therefore the homomorphisms must be the identity and the function $f(x)=0$ where $x \in \mathbb{R}$.

But how do I prove these are the only two homomorphisms?

Also, I was told to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, how can I use this hint?

user26857
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pjox
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    It doesn't seem necessary to use the hint. One way to show that is to reason that every ring homomorphism also has to be a linear map $\mathbb{R} \rightarrow \mathbb{R}$; then determine what "slopes" are allowed for such a map to also be multiplicative. – Christopher A. Wong Jun 04 '13 at 22:06
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    Beware! There could be non-identity automorphisms with trivial kernel. For exmaple with $\mathbb C$ instead of $\mathbb R$, the complex conjucation has trivial kernel but is not the identity. – Hagen von Eitzen Jun 04 '13 at 22:12
  • By the way, with $Q[\sqrt 2]$ instead of $\mathbb R$, there would also be more automorphisms - even though $\mathbb Q$ is dense! – Hagen von Eitzen Jun 04 '13 at 22:24
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    @ChristopherA.Wong How would one argue that a ring homomorphism ${\bf R}\to{\bf R}$ is in fact $\bf R$-linear without arguing continuity? – anon Jul 31 '13 at 11:15
  • See also http://math.stackexchange.com/questions/449404/is-an-algebraic-automorphism-of-the-field-of-real-numbers-the-identity-map. – lhf Jul 31 '13 at 11:17
  • @anon, you're right, perhaps I did not think it through. – Christopher A. Wong Jul 31 '13 at 20:32

1 Answers1

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$f\colon \mathbb R\to\mathbb R$ is uniquely determined by $f(1)$. Why? By induction, $f(n)=n\cdot f(1)$ for $n\in\mathbb N$. Then by additivity, $f(x)=x\cdot f(1)$ for $x\in \mathbb Z$ and finally also for $x\in\mathbb Q$. We can make use of the densitiy of $\mathbb Q$ if we show that $f$ is continuous. Indeed, if $x\ge0$ then $x=y\cdot y$ for some $y\in\mathbb R$, hence $f(x)=f(y)f(y)\ge 0$, therefore $f$ preserves $\ge$ and hence $|y-x|\le \frac 1n$ implies $|f(y)-f(x)|\le \frac1n|f(1)|$, that is $f$ is continuous. We conclude that $f(x)=x\cdot f(1)$ for all $x\in\mathbb R$.

What values of $f(1)$ are allowed? We must have $f(1)=f(1\cdot 1)=f(1)\cdot f(1)$, hence $f(1)=0$ or $f(1)=1$.

Hagen von Eitzen
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    One should be careful not to write "...$f(1)=1$ or $f(1)=0!$" – Pedro Jun 04 '13 at 22:25
  • @Hagen :I have cleared $f(x) = xf(1)$ for all $x \in \mathbb Q$ and $f$ is continous but I have some doubt in one point; how conclude that $f(x) = xf(1)$ for all $x \in \mathbb R$ – Struggler Dec 08 '15 at 15:37
  • @PedroTamaroff why? – Not Euler Apr 17 '18 at 20:33
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    @HritRoy See the original post. That was a very weak attempt at humour: $0!=1$. :) – Pedro Apr 17 '18 at 20:51
  • @PedroTamaroff Hah! That did cross my mind once. But I asked it anyway just to be sure :) – Not Euler Apr 18 '18 at 13:55
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    Note that in some textbooks, the ring homomorphism is defined to be mapping the identity to the identity, in which case, we only have $ f(1)=1 $ . – Bach Jan 10 '19 at 08:16
  • @HagenvonEitzen, isn't there a way to prove that $f$ is continuous using the density of the real numbers (i.e. between any two distinct real numbers is a rational number)? Is it similar to the proof you gave? –  Oct 16 '19 at 21:02
  • @HagenvonEitzen Sir, how can we show $f(x)=x f(1)$ for $x\in\mathbb{Q}$? I just tried like this: $f(q•{1/q})=f(1)$ that implies $f(q)f(1/q)=f(1)$ that implise $qf(1)f(1/q)=f(1)$ But from this we can't conclude that, $f(1/q)={1/q}•f(1)$ unless we must have $f(1)=1$..... please reply. – Akash Patalwanshi Aug 21 '21 at 04:45