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This question already has an answer here:

I faced the following question while prove the proposition "the group of automorphisms of a finite field is cyclic"-
Let $F$ be a finite field with $p^n$ elements where $\operatorname{char}(F)=p$. Then how to show that any automorphism of $F$ fixes $F_p$ pointwise i.e. if $f\in \operatorname{Aut}(F)$ then $f(x)=x \forall x\in F_p$ where $F_p\simeq \Bbb{Z}_p$.
Can anybody clear up query? Thanks for assistance in advance.

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marked as duplicate by Jyrki Lahtonen abstract-algebra Oct 28 '18 at 20:30

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  • $\begingroup$ And for field, you certainly have seen this question, with a proof. $\endgroup$ – Dietrich Burde Oct 28 '18 at 20:04
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Because $f(1)=1$, $f(1+1)=f(1)+f(1)=1+1$ and so on. And, of course, $f(0)=0$. Since$$F_p=\{0,0,\ldots,\overbrace{1+\cdots+1}^{p-1\text{ times}}\},$$this is enough.

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  • $\begingroup$ Why, $f(1)=1$? I know that an automorphism sends a generator to a generator, here $1$ is generator of $F_p$, but all nonzero elements of $F_p$ is generator here, so why 1 is mapped to 1? $\endgroup$ – Biswarup Saha Oct 28 '18 at 20:03
  • $\begingroup$ $f(1)=f(1\times1)=f(1)\times f(1)$. Since $f(1)\neq0$, $f(1)=1$. $\endgroup$ – José Carlos Santos Oct 28 '18 at 20:17
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Since $\Bbb{F}_p$ is the prime field of $F$, it is fixed by any automorphism of $F$.

Reference: how would you show that field automorphisms fix prime subfields?

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  • $\begingroup$ Yes, but why any automorphism will map 1 to 1? $\endgroup$ – Biswarup Saha Oct 28 '18 at 20:09
  • $\begingroup$ Because $f(1)=f(1\cdot 1)=f(1)\cdot f(1)$. $\endgroup$ – Dietrich Burde Oct 28 '18 at 20:21
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I think that the key here is to recognize that, for any non-trivial homomorphism $\phi$ 'twixt fields $K$ and $L$,

$\phi:K \to L, \tag 1$

we must have

$\phi(1_K) = 1_L, \tag 2$

where $1_K$ and $1_L$ are the multiplicative identity elements of $K$ and $L$, respectively.

Indeed, this fact is even more general; suppose $R$ and $S$ are commutative unital rings with in addition $S$ an integral domain, and again that $\phi$ is a non-trivial homomorphism

$\phi:R \to S; \tag 3$

then in $S$ we have the following equation:

$(\phi(1_R))^2 = \phi(1_R) \phi(1_R) = \phi(1_R^2) = \phi(1_R), \tag 4$

or

$\phi(1_R)(\phi(1_R) - 1_S) = (\phi(1_R))^2 - \phi(1_R) = 0; \tag 5$

now if $\phi$ is non-trivial, we have

$\phi(1_R) \ne 0_S, \tag 6$

lest, for every $r \in R$,

$\phi(r) = \phi(r 1_R) = \phi(r) \phi(1_R) = \phi(r)(0_S) = 0_S, \tag 7$

clearly not possible for non-trivial $\phi$; therefore, by (5)-(6) and the hypothesis that $S$ is an integral domain,

$\phi(1_R) - 1_S = 0, \tag 8$

or

$\phi(1_R) = 1_S. \tag 9$

Now $K$ and $L$ being fields manifestly satisfy our assumptions with regard to $R$ and $S$, whence (2) binds. Going back a step further, we return to the automorphism

$f:F \to F \tag{10}$

of the question as stated; we have shown that

$f(1_F) = 1_F, \tag{11}$

and now it is easy to see that $f$ fixes the prime subfield $F_p \simeq \Bbb Z_p$, since any $x \in F_p$ may be expressed as the sum of a certain number of $1_F$s:

$x = 1_F + 1_F + \ldots + 1_F, \; k \; \text{times}; \tag{12}$

then

$f(x) = f(1_F + 1_F + \ldots + 1_F)$ $= f(1_F) + f(1_F) + \ldots + f(1_F) = 1_F + 1_F + \ldots + 1_F, \; k \; \text{times}; \tag{12}$

(11) and (12) together imply

$f(x) = x, \; \forall x \in F_p \simeq \Bbb Z_p, \tag{13}$

which was to be seen.

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