How would you show that field automorphisms fix prime subfields?

abstract-algebra field-theory ring-homomorphism

Suppose $K$ is a prime subfield of $E$, then if $\phi$ is an automorphism from $E$ to $E$, we have for all $x \in K$, $\phi(x) = x$.

I feel like this is just the definition of a field automorphism, but my book says this should be proven as an exercise.


Solution 1:

This is because the prime subfield is generated as a field by $1$. Since you have no choice but to send $1$ to itself, the prime subfield remains fixed as well.

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