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Let $\mathbb{F}_q$ be a finite field with $q=p^f$ elements. I know that for each $0\leqslant i\leqslant f-1$, the maps $\sigma_i(x)=x^{p^i}$ are $\mathbb{F}_p$-automorphsim of $\mathbb{F}_q$ (so $\sigma_i(x)=x$ for every $x\in \mathbb{F}_p$). Are there any other $\mathbb{F}_p$-automorphsim ?

My Attempt: suppose that $g$ is a generator of $\mathbb{F}_q$. If $\sigma$ is an $\mathbb{F}_p$-automorphsim, then there exist a positive integer $k$ such that $\gcd(k,q-1)=1$ and $\sigma(g)=g^k$. Does $k$ must be a power of $p$ ?

Teddy38
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boaz
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1 Answers1

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There's a problem with your attempt. it need little modification:

When we want to find automorphism of $\mathbb{F}_q$ over $\mathbb{F}_p$, we require that automorphism should preserves $\mathbb{F}_p$, i.e., restricting to $\mathbb{F}_p$, it should give identity.

Now, if we take $\sigma \in \text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q)$, $\sigma(x) = x$ for all $x\in \mathbb{F}_p$. Now, $\text{Frob}_p: x \mapsto x^p$ and $\text{Frob}_p \in \text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q)$.

Order of $\text{Frob}_p$ in $\text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q)$ is $f$. Now from degree calculation we know $|\text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q)| = f$. So, $\text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q) = \langle \text{Frob}_p \rangle$.

Remarks

(1) As $\mathbb{F}_p$-vector space, dimension of $\mathbb{F}_q$ is $f$ and hence $|\text{Aut}_{\mathbb{F}_p}(\mathbb{F}_q)| = f$.

(2) Here $\text{Frob}_p$ is known as Frobenius element.

Offlaw
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