# Amann/Escher, Analysis I, Exercise I.11.8: field automorphisms of $\mathbb{C}$ which leave the elements of $\mathbb{R}$ fixed

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.

• If you prove $\phi(1)=1$, then shouldn't you also prove $\phi(-z) = -\phi(z)$ before you use it? Or maybe just prove $\phi(-1)=-1$ without using $\phi(-z) = -\phi(z)$. Or even just prove $\phi(i) = \pm i$ directly without using $\phi(-1)=-1$. – GEdgar Jun 29 at 15:33

## 2 Answers

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 $$z$$.

It's not quite right. There is a typo at the end: when you wrote $$=\pm z$$, what was $$=z\vee=\overline z$$. But the main problem is this: it seems that what you proved is that, for each $$z\in\mathbb C$$, $$\phi(z)=z\vee\phi(z)=\overline z$$. But you are supposed to prove a much stronger statement:$$(\forall z\in\mathbb C):\phi(z)=z\vee(\forall z\in\mathbb C):\phi(z)=\overline z.$$That's not a serious problem, though. In fact:

• if $$\phi(i)=i$$, then $$(\forall z\in\mathbb C):\phi(z)=z$$;
• if $$\phi(i)=-i$$, then $$(\forall z\in\mathbb C):\phi(z)=\overline z$$.

And what you did actually proves this.