Let $L/K$ be a finite Galois extension. If $\sigma \in \mathrm{Gal}(L/K)$ has order $d$, is it the case that $$L^\sigma := \{ \ell \in L : \sigma(\ell) = \ell\}$$
satisfies $[L^\sigma:K] = d$?
This came up as a would-be useful fact while I was solving a homework problem, although I ended up not using it.
Such a $\sigma$ gives rise to an order $d$ subgroup $\langle \sigma \rangle \leq \mathrm{Gal}(L/K)$ and so we know that $[L^{\langle \sigma \rangle}:K] = |\mathrm{Gal}(L/K)|/d$. Since $L^{\langle \sigma \rangle}$ intermediates $L^\sigma/K$, we have $[L^\sigma:K] = [L^\sigma:L^{\langle \sigma \rangle}]\cdot|\mathrm{Gal}(L/K)|/d$. So when is $[L^\sigma:L^{\langle \sigma \rangle}]$ not $d^2/|\mathrm{Gal}(L/K)|$? I can't come up with a quick non-example.
Let me think out loud through an example: $L:=\mathbb{Q}(\sqrt2, i)$. This is a degree $4$ extension of $\mathbb{Q}$. It is the splitting field of $x^2+2$. One automorphism of $L$ fixes $\sqrt 2$ and takes $i$ t $-i$; denote this $\sigma$. Clearly $\sigma^2 = \mathrm{id}$ and $L^\sigma = \mathbb{Q}(\sqrt 2)$, whose degree over $\mathbb{Q}$ is the order of $\sigma$. Perhaps choosing a field with only order $=2$ automorphisms was a bad idea.
Note that $L^\sigma = L^{\langle \sigma \rangle}$. Indeed, the inclusion $L^{\langle \sigma \rangle} \subseteq L^\sigma$ is clear, and if we have $\ell \in L^\sigma$, then, for all $n \in \mathbb{N}$: $$\sigma^{n}(\ell) = \sigma^{n - 1}(\sigma(\ell)) = \sigma^{n - 1}(\ell) = \cdots = \sigma(\ell) = \ell$$ So $\ell \in L^{\langle \sigma \rangle}$.
Therefore $[L^\sigma : K] = |\mathrm{Gal}(L/K)|/d$. So in fact $[L^\sigma : K] = d$ only if $|\mathrm{Gal}(L/K)| = d^2$. What is actually always true is that $[L: L^\sigma] = d$.