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I searched a lot but I couldn't solve my problem! I know that $$ f(1) = f(1.1)= f(1).f(1) \Longrightarrow f(1) = 0 \quad or \quad f(1) = 1 $$ I know that if we suppose that $f(1) = 0$ then $f$ is trivial ,I don't have any problem to prove this ,but if $f(1) = 1$ I don't know how to make contradiction!can any one help me please!I become confused because I know that there is trivial and identity homomorphism $f:\mathbb{R} \rightarrow \mathbb{R}$ so is there only one homomorphism from $\mathbb{R} \rightarrow \mathbb{C}$?

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    $\begingroup$ Does your definition of "ring homomorphism" require homomorphisms to send $1$ to $1$? If not, then you have already disproven the statement. If so, the statement is still false, but is much harder to disprove (you must use the axiom of choice). $\endgroup$ – Eric Wofsey May 31 '16 at 7:00
  • $\begingroup$ how could you disprove it? Once $f(1)=1$ then $f(r) = rf(1)=r$ how might there be more morphism? $\endgroup$ – YannickSSE May 31 '16 at 7:41
  • $\begingroup$ @YannickSSE: No, $f(r) = f(r)f(1)$. $\endgroup$ – Asaf Karagila May 31 '16 at 7:43
  • $\begingroup$ Yeah ok but $f(n)=nf(1)$ for all $n\in\mathbb{N}$ hence also $f(q)= q$ for all $q\in\mathbb{Q}$ and then I msut admit it's not that easy anymore. $\endgroup$ – YannickSSE May 31 '16 at 7:47
  • $\begingroup$ @AsafKaragila : do you mean $\mathbb{R} \to \mathbb{C}$ is not the same as $\mathbb{Q} \to \mathbb{Q}(i)$ ? so if we add the usual metric $|.|$ and ask for the homomorphism to preserve it, everything becomes simple and obviously $f(x) = x$ even when $x$ is irrational $\endgroup$ – reuns May 31 '16 at 7:50
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We know that $Aut(\mathbb{C}/\mathbb{Q})$ has infinitly many elements but the Galois group $Gal(\mathbb{C}/\mathbb{R})$ has 2 elements. So there has to be a ring homomorphism $\varphi:\mathbb{C} \to \mathbb{C}$ that doesn't fix $\mathbb{R}$. Now take $i:\mathbb{R}\to \mathbb{C}$ our usual inclusion. Now $\varphi\circ i$ is a ring homomorphism that is not $i$.

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  • $\begingroup$ From $Gal(\mathbb{C}/\mathbb{R})$ we do not know what you said. We know there is a non-identity homomorphism that does fix $\mathbb R$. $\endgroup$ – GEdgar May 31 '16 at 16:15
  • $\begingroup$ $Gal(\mathbb{C}/\mathbb{R})$ has precisely 2 elements. The identity and complex conjugation. These 2 elements are also in $Gal(\mathbb{C}/\mathbb{Q})$, but there are more. Take one of these as the $\varphi$ I proposed. This doesn't fix $\mathbb{R}$. Other than that I really don't now what you try to say with your comment. $\endgroup$ – Maik Pickl May 31 '16 at 16:27
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    $\begingroup$ $Gal(\mathbb{C}/\mathbb{Q})$ should be $Aut(\mathbb{C}/\mathbb{Q})$, since $\mathbb{C}$ is not a Galois extension of $\mathbb{Q}$. I changed it in the answer. $\endgroup$ – Maik Pickl May 31 '16 at 16:38

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