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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 $\mathbb{R} \rightarrow \mathbb{C}$?

haleh
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    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). – Eric Wofsey May 31 '16 at 07:00
  • how could you disprove it? Once $f(1)=1$ then $f(r) = rf(1)=r$ how might there be more morphism? – YannickSSE May 31 '16 at 07:41
  • @YannickSSE: No, $f(r) = f(r)f(1)$. – Asaf Karagila May 31 '16 at 07:43
  • 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. – YannickSSE May 31 '16 at 07:47
  • @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 – reuns May 31 '16 at 07:50
  • @YannickSSE: Yes, for continuous homomorphism this is easy. But assuming the axiom of choice, there are discontinuous ones too. But they cannot be defined explicitly. – Asaf Karagila May 31 '16 at 07:54
  • @user1952009: See my previous comment. Essentially, yes. – Asaf Karagila May 31 '16 at 07:54
  • @Asaf : the problem is that it seems hard defining $\mathbb{R}$ (or $\mathbb{C}$) differently as the completion of $\mathbb{Q}$ in some way, so I wonder if it is not self-contradictory to define an homomorphism that doesn't preserve at all how this completion was build – reuns May 31 '16 at 07:58
  • @user1952009: You cannot define a discontinuous homomorphism. Period. You can prove the existence of such homomorphism using the axiom of choice. That is not the same as defining it. – Asaf Karagila May 31 '16 at 08:00
  • @AsafKaragila : yes of course, I meant defining the existence of the homomorphism, not defining it explicitely. but it seems natural to wonder if it is self-contradictory. or not – reuns May 31 '16 at 08:02
  • @YannickSSE : Yes. Specifying the images of ${0,1}$ specifies the image of $\Bbb{Z}$ specifies the image of $\Bbb{Q}$ specifies the image of $\Bbb{A}$ (since every polynomial you care to write down survives the homomorphism). Now find a $\Bbb{Q}$-algebraic generating set for the transcendentals, apply any permutation you like to that set and sprinkle the (permuted) rest of $\Bbb{R}$ into the holes left by the construction so far. Alternatively, are you sure that the set of transcendentals generated by one generator doesn't spiral around the origin (or do something else crazy)? – Eric Towers May 31 '16 at 20:10

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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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Maik Pickl
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  • From $Gal(\mathbb{C}/\mathbb{R})$ we do not know what you said. We know there is a non-identity homomorphism that does fix $\mathbb R$. – GEdgar May 31 '16 at 16:15
  • $Gal(\mathbb{C}/\mathbb{R})$ 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. – Maik Pickl May 31 '16 at 16:27
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    $Gal(\mathbb{C}/\mathbb{Q})$ should be $Aut(\mathbb{C}/\mathbb{Q})$, since $\mathbb{C}$ is not a Galois extension of $\mathbb{Q}$. I changed it in the answer. – Maik Pickl May 31 '16 at 16:38