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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.

  • This result is not true for all fields $F$. There has to be some additional quantifications. For example, the Frobenius endomorphism will not be invertible in a non-perfect field. – EuYu Nov 01 '20 at 07:03
  • Usually one denotes all automorphism with $\operatorname{Aut}$, are you sure it's supposed to be just homomorphisms? (One would rather write $\operatorname{End}{F}$.) – Qi Zhu Nov 01 '20 at 07:08

2 Answers2

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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.

Jyrki Lahtonen
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  • It was supposed to be automorphism instead of homomorphism but please don't delete your answer . it is quite helpful and informative . I am ashamed of my mistake and I am sorry!! –  Nov 01 '20 at 07:29
  • I think the homomorphism you mentioned is onto. Can you give a proof of it how its not onto –  Nov 01 '20 at 07:42
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    @User The rational function $t\in F$ is not in the image. It is impossible to find polynomials $a(t),b(t)$, $b(t)\neq0$, such that $t=a(t^2)/b(t^2)$. For that to happen we need the polynomial identity $$t b(t^2)= a(t^2),$$ but on the left we have only odd degree terms and on the right we only have even degree terms, so all the term must vanish contradicting $b(t)\neq0$. – Jyrki Lahtonen Nov 01 '20 at 08:38
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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 $$\sigma^{-1}(\lambda y)=\lambda \sigma^{-1}(y).$$