TITLE: When does a permutation of the roots define an automorphism?
QUESTION [5 upvotes]: I often see it stated that the map $i \mapsto -i$ defines an automorphism of $\mathbb Q(i)$ as though this were obvious (and it seemed obvious to me). However it's not true in general that every permutation of an irreducible polynomial's roots leads to an automorphism of the corresponding splitting field (that fixes the first field).
Eg. Let $K$ be the splitting field of the irreducible polynomial $x^3-10$ over $\mathbb Q(\sqrt{-3})$ (edit: this originally mistakenly said over $\mathbb Q$). Then we know the Galois group is $\mathbb Z / 3 \mathbb Z$, so clearly every possible permutation of the roots is not forming an automorphism.
What exactly is going wrong? As in the first example ($i$ to $-i$) I would have thought defining $\sigma$ to fix $\mathbb Q$ while permuting the roots any way we like would work.

REPLY [3 votes]: You don't need to know anything about permuting roots to show that $i \mapsto -i$ defines an automorphism of $\mathbb{Q}(i)$. The point is that $\mathbb{Q}(i) \cong \mathbb{Q}[x]/(x^2 + 1)$, so a homomorphism from $\mathbb{Q}[x]/(x^2 + 1)$ to another field of characteristic $0$ (or any other $\mathbb{Q}$-algebra) is uniquely determined by where it sends $x$, which is subject only to the constraint that $x^2 = -1$. 
So there are exactly two homomorphisms $\mathbb{Q}(i) \to \mathbb{Q}(i)$ because $\mathbb{Q}(i)$ has exactly two elements, namely $i$ and $-i$, which square to $-1$. I'm not picking a permutation of the roots when I do this, just choosing where a single root goes, namely $i$, which uniquely determines where $-i$ goes. 
Here's a more complicated but similar example. Consider now $\mathbb{Q}(z)$ where $z^5 = 1$ is a primitive fifth root of unity, which is a degree $4$ extension. This is equivalently $\mathbb{Q}[x]/(x^4 + x^3 + x^2 + x + 1)$ (this requires a little proof) which shows that homomorphisms $\mathbb{Q}(z) \to \mathbb{Q}(z)$ are uniquely determined by where they send $z$, which is subject to the constraint that $z^5 = 1$ but $z \neq 1$. Thus there are four such homomorphisms, sending $z$ to $z, z^2, z^3, z^4$, and we find that the Galois group is 
$$(\mathbb{Z}/5\mathbb{Z})^{\times} \cong \mathbb{Z}/4\mathbb{Z}.$$
In particular, because all the other fifth roots of unity are powers of $z$, a homomorphism is completely determined by where it sends $z$, and once that's known you have no choice about where the other roots go.