Let $K$ be a field, $G\leq\mbox{Aut}(K)$ a group of field automorphisms of $K$. When is the extension $K/K^G$ algebraic?
Recall that $K^G$ is the subfield of $K$ consisting of the elements of $K$ that are fixed by every element of $G$.
If the group is finite, then the extension is finite and Galois, but in the infinite case this is not always true. For example, consider the field $K=\mathbb C(t)$ of rational functions with complex coefficients, and consider the subgroup $G$ generated by the translation $t\mapsto t+1$, which is clearly an automorphism. It is not hard to see that $K^G =\mathbb C$ (the constant rational functions) and that therefore $K/K^G$ is transcendental.
Best Answer
I claim an element $a\in K$ is algebraic over $K^G$ if and only if the orbit of $a$ under $G$ is finite. The $\impliedby$ direction is as usual; let $\mathcal{O}$ be the orbit of $a$. If $\mathcal{O}$ is finite, then the polynomial $\prod_{\alpha\in\mathcal{O}}(x-\alpha)$ lies in $K^G[x]$, and is satisfied by $a$, whence $a$ is algebraic over $K^G$, as desired.
For the $\implies$ direction, suppose $a$ is algebraic over $K^G$, and let $p$ be its minimal polynomial over $K^G$. Then $G$ fixes $p$, so that $g(a)$ is a root of $p$ for every $g\in G$. But polynomials can have only finitely many roots in a field, so the orbit of $a$ under $G$ must be finite, as claimed.
In particular, the extension $K:K^G$ will be algebraic if and only if every element of $K$ has finite orbit under $G$. I'm not sure if this is an entirely satisfying answer for you; perhaps there is more to say in particular cases, but in full generality I don't know if it is possible to say anything more than this.