Is the order of a field automorphism equal to the degree over the fixed-point subfield? Given any field $L$ and any automorphism $f:L \to L$, one could define the fixed-point subfield $K := \{x \in L \mid f(x)=x\}$ in the obvious way.
Now, suppose that $f$ has finite order $n$. Does this mean that the field extension $L/K$ has degree $n$ (i.e., $L$ is an $n$-dimensional $K$-vector space)?
The statement is true in a few examples:

*

*The identity has order $1$, and clearly, the fixed-point subfield is all of $L$ and $L$ is $1$-dimensional over $L$.

*Complex conjugation has order $2$, and the complex numbers are $2$-dimensional over the real numbers, which form the fixed-point subfield.

*The same is true for $\mathbb{Q}(\sqrt{x})$ where $x$ is any rational number that is not the square of another rational number.

*If $L$ is a finite field of order $p^n$ and $f$ is the Frobenius automorphism (i.e., $f(x)=x^p$), then the fixed-point subfield is the prime subfield of $L$, and clearly, $L$ is $n$-dimensional over its prime subfield.

Is the statement true in general, or are additional assumptions needed (e.g., $L$ is perfect, or $L$ has characteristic $0$)?
 A: Yes, this is true and as Jyrki says it's a corollary of a standard result in Galois theory. This special case can probably be proven by one's "bare hands" but the argument eludes me for now; here's an even specialer case that can be proven using more or less just linear algebra.

Specialer case: The result is true if $K$ contains the $n^{th}$ roots of unity (which in particular requires that it have characteristic not dividing $n$).

Proof. Because $f^n = 1$, the eigenvalues of $f$ are the $n^{th}$ roots of unity $1, \zeta, \dots \zeta^{n-1}$, and moreover because $f$ satisfies a polynomial with distinct roots over $K$, it is diagonalizable over $K$. If we write $L$ as the direct sum of the eigenspaces
$$K_i = \{ x \in L : f(x) = \zeta^i x \}$$
of $f$, so that $K_0 = K$, then we see that to prove the desired claim it suffices to prove that each $K_i$ is $1$-dimensional as a $K$-vector space. A quick calculation shows that $K_i \cdot K_j \subseteq K_{i+j}$; furthermore, because $f$ is assumed to have order exactly $n$, some $K_i$ must have dimension at least $1$ for $\zeta^i$ a primitive $n^{th}$ root of unity. By repeatedly taking powers we conclude that every $K_i$ has dimension at least $1$ (so far we use only that $L$ is an integral domain). Another quick calculation shows that if $x, y \in K_i$ are nonzero then $\frac{x}{y} \in K$ (here we need that $L$ is a field), from which it follows that every $K_i$ has dimension at most, hence exactly, $1$. $\Box$
So we've proven something stronger than what was asked for: not only is $\dim_K L = n$, but we can find a basis of $L$ consisting of eigenvectors of $f$, one for each $n^{th}$ root of unity. In the language of representation theory this shows that $L$, considered as a representation of the cyclic group $C_n$ over $K$, is the regular representation. This is a special case of the normal basis theorem. Also from here we're getting close to Kummer theory.
