I'm doing Exercise I.11.8 from textbook Analysis I by Amann/Escher.
Show that the identity function and $z \mapsto \overline{z}$ are the only field automorphisms of $\mathbb{C}$ which leave the elements of $\mathbb{R}$ fixed.
Could you please verify if my attempt contains logical gaps/errors?
My attempt:
Let $\phi:\mathbb{C} \to \mathbb{C}$ be an automorphism of $\mathbb{C}$ which leaves the elements of $\mathbb{R}$ fixed.
Since group homomorphism maps identity elements to identity elements, $\phi(1) = 1$ and $\phi(0) = 0$. Then $(\phi(i))^2 = \phi(i^2) = \phi (-1) = – \phi(1) = -1$, so $\phi(i) = \pm i$.
For $z = a + ib \in \mathbb{C}$, we have $$\begin{aligned}\phi(z) &= \phi (a + ib) \\ &= \phi(a) + \phi(i) \phi(b) \\ &= a + \phi(i)b \\ &= a \pm ib \\ & = z \text{ or } \overline{z}\end{aligned}$$
This completes the proof.
Best Answer
It's fine, though maybe the write-up is a little terse. Your working could potentially be interpreted that, for all $z \in \Bbb{C}$, $\phi(z) = z$ or $\overline{z}$, depending on the value of $z$. That is, it might be true that, for the same $\phi$, there could exist $z$ and $w$ such that $\phi(z) = z$ and $\phi(w) = \overline{w}$.
Of course, your logic doesn't actually allow for this possibility. You're actually showing that $\forall z \in \Bbb{C}, \phi(z) = z$ OR $\forall z \in \Bbb{C}, \phi(z) = \overline{z}$, as the exercise requires. But, I think your write-up should reflect this better.
To explain this better, split into two cases: $\phi(i) = i$ and $\phi(i) = -i$. In the first case, show $\phi(z) = z$ for all $z$. In the second case, show $\phi(z) = \overline{z}$ for all $z$.