Non-cyclic finite extensions of fixed fields of infinite order automorphisms of non-algebraically closed fields This is a problem from Galois theory:

Suppose that $F$ is algebraically closed, $\lambda$ is an automorphism of infinite order, and $f=F^\lambda$. Show that any finite extension of $f$ is cyclic.

What is an example of $F$ which is not algebraically closed so the conclusion fails?
Can these examples be characterize in some way?
 A: Filling in the details of the counterexample suggested by Jacob's comment:

Write $\mathbb{Q}\left(\{x_i\}_{i\in \mathbb{Z}}\right)$ as $\mathbb{Q}(x_\mathbb{Z})$. Note that $x^2-x_i=0$ has no root in $\mathbb{Q}(x_\mathbb{Z})$, so $\mathbb{Q}(x_\mathbb{Z})$ is not algebraically closed.  Let $\sigma:\mathbb{Q}(x_\mathbb{Z})\rightarrow\mathbb{Q}(x_\mathbb{Z})$ be the automorphism defined by $\sigma:x_i\rightarrow x_{i+1}$.  Then $\sigma^r:x_i\mapsto x_{i+r}$ is the identity map only when $r=0$, so $\sigma$ has infinite order. Take any rational polynomial $f\in \mathbb{Q}(x_\mathbb{Z})$.  If $f$ is a constant polynomial (that is, $f\in \mathbb{Q}$), then certainly $\sigma(f)=f$.  If $f$ is not a constant polynomial, then let $m_f$ denote the lowest index in $\mathbb{Z}$ of a variable appearing in $f$ with a nonzero coefficient.  Then, by assumption, the lowest index of a variable appearing in $\sigma(f)$ with a nonzero coefficient is $m_f+1$.  It follows that $\sigma(f)\ne f$, so $\mathbb{Q}(x_\mathbb{Z})^\sigma=\mathbb{Q}$.  Of course, not every finite extension of $\mathbb{Q}$ is cyclic.

