# Let $F$ be a finite field with $p^n$ elements where char$(F)=p$ [duplicate]

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.

## marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 28 '18 at 20:30

• And for field, you certainly have seen this question, with a proof. – Dietrich Burde Oct 28 '18 at 20:04

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.

• 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? – Biswarup Saha Oct 28 '18 at 20:03
• $f(1)=f(1\times1)=f(1)\times f(1)$. Since $f(1)\neq0$, $f(1)=1$. – José Carlos Santos Oct 28 '18 at 20:17

Since $$\Bbb{F}_p$$ is the prime field of $$F$$, it is fixed by any automorphism of $$F$$.

• Yes, but why any automorphism will map 1 to 1? – Biswarup Saha Oct 28 '18 at 20:09
• Because $f(1)=f(1\cdot 1)=f(1)\cdot f(1)$. – Dietrich Burde Oct 28 '18 at 20:21

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.