Set Aut F of all field automorphisms is a group This question is from my abstract algebra course.
Instructor told us to prove this by ourselves.

The set Aut F of all field automorphisms $F \to F$ forms a group under the operation of composition of functions.

I have problem in proving how inverse of a function $\sigma$ exists. Which result I should use?
Can you please tell how to prove that inverse exists.
 A: This is not always the case. For example the mapping
$$
\sigma:\frac{a(t)}{b(t)}\mapsto \frac{a(t^2)}{b(t^2)}
$$
is a non-surjective homomorphism from the field $F=\Bbb{R}(t)$ of rational functions to itself. Here $a(t),b(t)$ are arbitrary polynomials.

However, the following general facts do allow us to conclude this in many special cases:

*

*A homomorphism of fields necessarily has a trivial kernel, and is thus always injective.

*A homomorphism of fields is always linear over the prime field $k$. Therefore by rank-nullity an injective homomorphism is always surjective, if $[F:k]<\infty$.

So if $F$ is a finite extension of $\Bbb{Q}$ or $\Bbb{F}_p$ for some prime number $p$, the claim holds automatically.
It is true in some other cases as well. For example in the cases $F=\Bbb{R}$ and $F=\Bbb{Q}_p$ (= the field of $p$-adic numbers) simply because in those cases the identity mapping is the only automorphism.
A: If $ \sigma \in Aut F$ then $\exists \sigma^{-1} : F \rightarrow F$ which is a bijection.
We have
$$\forall (y_1, y_2) \in F^2 \quad y_1+y_2 = \sigma(\sigma^{-1}(y_1)) + \sigma(\sigma^{-1}(y_2)) = \sigma(\sigma^{-1}(y_1) + \sigma^{-1}(y_2)).$$
Hence
$$\sigma^{-1}(y_1+y_2) = \sigma^{-1}(y_1) + \sigma^{-1}(y_2).$$
And
$\forall \lambda \in \mathbb R \quad \forall y \in F$ we have
$$\lambda y= \lambda\sigma\sigma^{-1}(y) =\sigma(\lambda\sigma^{-1}(y)).$$
Then
$$\sigma^{-1}(\lambda y)=\lambda \sigma^{-1}(y).$$
