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Let $F$ be a field and f an automorphism on K. Is $f$ the the identity map on the prime field of $F$? I feel it should follow from the fact that the prime field is either $\mathbb{Q}$ or $F_p$, but I cannot find a proof.

Can someone help me out?

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    $\begingroup$ The prime field is generated by $1$. Since $f$ is a homomorphism and fixes $1$, it fixes what $1$ generates. $\endgroup$ – Daniel Fischer Feb 5 '14 at 22:31
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More generally let $f,g : K \to L$ be two field homomorphisms and let $P$ be the prime field of $K$, i.e. the unique smallest subfield of $K$. Since $\{a \in K : f(a)=g(a)\}$ is a subfield of $K$ (check it!), it has to contain $P$. This means $f|_P = g|_P$.

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Yes, the prime field is either $\mathbf Q$ or $\mathbf F_p$ for some $p$. The proof is very straightforward (think about it... what can you say about the image of the canonical map $\mathbf Z \to F$?).

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  • $\begingroup$ I think I got it: We have $f(1)=1$, for if $f(0)=0$ then f must be the 0-map. We have $f(1)=f(1)^2$ and this means that $f(1)$ must be the non-zero root of the eqn $x^2-x$, which means $f(1)=1$. So for $n\in\mathbb{Z}$ we have $f(n\cdot 1)=n\cdot 1$. So if $F$ has prime charistic $p$, then $f$ is the identity on $\mathbb{F}_p$. But how do I finish the proof for the subfield $\mathbb{Q}$? $\endgroup$ – david Feb 5 '14 at 23:03
  • $\begingroup$ @david: $1=f(1)=f(1/2+1/2)=f(1/2)+f(1/2)=2f(1/2)$, so... Then generalize $2$ to $n$. $\endgroup$ – user452 Feb 6 '14 at 0:24
  • $\begingroup$ @david: In this context, a ring homomorphism always takes $1$ to $1$ by definition so you don't need to check that. There is a unique map from $\mathbf Z$ to any ring $R$, which is completely determined by the fact that it takes $1$ to $1$. So you have two possibilities: either $\mathbf Z \to F$ is injective, in which case $F$ contains a copy of $\mathbf Z$ and hence of $\mathbf Q$ since it's a field, or else the map $\mathbf Z \to F$ is not injective, in which case its kernel must be a prime $p$, and then you get an injection $\mathbf Z/p\mathbf Z \to F$. $\endgroup$ – Bruno Joyal Feb 6 '14 at 0:48
  • $\begingroup$ @trb456 $f(1/2)$ does not make sense. There is no map $f: \mathbf Q \to \mathbf F_p$. $\endgroup$ – Bruno Joyal Feb 6 '14 at 0:49
  • $\begingroup$ @Bruno Joyal: I was only considering an automorphism on $\mathbb{Q}$, which was what the first comment asked for, or so I thought. Your answer is more general, of course. But the comment seemed to understand the answer for finite fields, but not $\mathbb{Q}$. $\endgroup$ – user452 Feb 6 '14 at 0:58

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