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 $\mathbb{R} \rightarrow \mathbb{C}$?
2025-01-13 08:02:53.1736755373
Prove or Disprove: there is only one ring homomorphism $f:\mathbb{R} \rightarrow \mathbb{C}$
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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 $i$.