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Let $k$ be an algebraically closed field. Let $R,S$ be rings such that $R,S$ both contain $k$ as a subring. Let $\varphi:R\to S$ be a ring homomorphism. Then does $\varphi(a)=a$ for all $a\in k$?

So far I think: $\varphi:R\to S$ is a ring homomorphism $\implies \varphi|_{k}:k\to k$ is a field homomorphism $\implies \varphi(1_{k})=1_{k}$.

But I think this only shows that $\varphi$ fixes elements of $k$ generated by $1_{k}$. So let's say $k$ is $\mathbb{C}$ (I don't know any other algebraically closed field). Then I reckon that this would imply $\varphi$ fixes $\mathbb{Z}$ (and by the answer I got from this, $\mathbb{Q}$ as well), but I don't know what I can say about irrational or imaginary numbers.

anonanon444
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    Choose $R=S=k=\Bbb C$ and consider $\varphi(z)=\bar z$. – Berci Jan 29 '19 at 22:21
  • Thanks for your comment, but a question, can we have a homomorphism (besides the one that maps $z$ to $z$) not defined this way? So can something that sends $z$ to something besides $\overline{z}$ still be a homomorphism? – anonanon444 Jan 30 '19 at 03:23

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The case where all the rings are $\mathbb C$ is already interesting enough! As Berci mentioned in the comments, conjugation gives an example of an automorphism that only fixes the reals. One can show that this is the only non-trivial automorphism fixing the reals.

To answer your second question in the comments: Assuming choice, there are even more exotic counterexamples, so called wild automorphisms. You might find the discussion on this MSE post interesting.

ffffforall
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  • I see, thanks for your answer. I didn't know that it would be so "wild" to have an automorphism besides the identity/conjugation, lol! – anonanon444 Jan 30 '19 at 18:00