# What elements does a map of rings fix if both rings contain the same algebraically closed field as a subring?

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.

• Choose $R=S=k=\Bbb C$ and consider $\varphi(z)=\bar z$. Commented Jan 29, 2019 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? Commented Jan 30, 2019 at 3:23

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.