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Is my proof to the question correct?

In $\mathbb{C}$ we have that $i$ is the solution to $x^2 + 1 = 0 $. Thus if a homomorphism exists from $\mathbb{C} \to \mathbb{R}$ there is a solution in $\mathbb{R}$ to

$ ϕ(x^2 + 1) = ϕ(0)$

$ ϕ(x^2)+ ϕ(1) = ϕ(0)$ by closure under addition property of homomorphisms

$ ϕ(x^2) + 1 = 0 $ since $ϕ(1) = 1$ and $ϕ(0) = 0 $ in all ring homomorphisms

$ (ϕ(x))^2 = -1$

which there isn't. Thus no ring homomorphism of this type exists.

user12938120
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1 Answers1

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In your proof, you use that $\phi(1)=1$ for any ring homomorphism, whether this is true depends on your definition for those (a matter of convention). As you stated this fact, I assume that this is the setting you are working with, in which case your proof is correct.

In the more general case, $\phi(1)=\phi(1\cdot 1) = \phi(1)\cdot \phi(1) = (\phi(1))^2$, which must always be non-negative as $\phi(1)$ is a real number. But then analogously to your proof, $(\phi(x))^2 + \phi(1)=0$ and both are non-negative, so $\phi(x)=0$ for all $x$. Therefore the ring homomorphism $0$ would then be the only one fulfilling your requirements (and it obviously does not send 1 to 1)

Markus Zetto
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  • A small observation that might be usefull: As $\phi(1)^2=\phi(1)$, we note that $\phi(1)^2-\phi(1)=\phi(1)\cdot (\phi(1)-1)=0$, so either $\phi(1)=0$, which means that $\phi(x)=\phi(x\cdot 1)=\phi(x)\cdot 0 = 0$ for all $x$, or $\phi(1)=1$ such that we have your definition of ring homomorphisms. That means that the more general definition I added only adds one new homomorphism, namely $0$. – Markus Zetto May 03 '20 at 13:38