existence of an automorphism of $k^a$ whose fixed field is $k$ Let $k$ be a field such that every finite extension is cyclic. Show that there is an automorphism of $k^a$ over $k$ whose fixed field is $k$. Here $k^a$ is the algebraic closure of $k$.
P.S. It's a homework problem. The first thing that came to my mind was to use Zorn's lemma. But I couldn't see why the maximum must be $k^a$.
P.S.S. I've got an nice argument to completely prove the statement. The assumption that every finite extension of $k$ is cyclic indeed plays a key role in my argument. 
Can anyone give an example to show the statement is false if we only assume that every finite extension of $k$ is Galois?
 A: Here is my argument. 
Let $S=\{(E,\, \sigma_E):\,k\le E\le k^a,\,\sigma_E\in Gal(E/k),\,\textrm{the fixed field of }\sigma_E \textrm{ is } k\}$.
By routine arguments using Zorn's lemma, we got a maximal element, say, ($M$, $\sigma_M$). Suppose $M\ne k^a$.
Let $\alpha\in k^a-M$ be of minimal (greater than 1) degree over $M$. Extend $\sigma_M$ to $M(\alpha)$, say $\sigma_1$. By the maximality of $(M,\sigma_M)$, $\sigma_1$ fixes some $a\not\in k$. Hence $a\not\in M$. By the minimality of $\alpha$, we have $M(\alpha)=M(a)$.
Let $f(x)=irr(a,M,x)$, then $\sigma_1(a)$ is a root of $f^{\sigma_1}$. Since $\sigma_1$ fixes $a$, we have $f=f^{\sigma_1}$. This means $f(x)\in k[x]$ and therefore $[k(a):k]=[M(a):M]$.
Now consider another extension $\sigma_2$ of $\sigma_M$ to $M(\alpha)=M(a)$, that maps $a$ to another root of $f$, say $a_0$. Note that $a_0\ne a$. Similar to the previous argument we have:


*

*$\sigma_2$ fixes some $b\not\in k$, and hence $b\not\in M$.

*$M(\alpha)=M(b)$. Hence $[M(b):M]=[M(a):M]$.

*$[k(b):k]=[M(b):M]$. Hence $[k(b):k]=[k(a):k]$.
Note that $k(a)\ne k(b)$, since otherwise we would have $a\in k(b)$ which means $a$ was fixed by $\sigma_2$. But recall that $\sigma_2$ is defined by $\sigma_2(a)=a_0\ne a$.
Since every finite extension of $k$ is cyclic, $k$ has at most one extension of a given finite degree in $k^a$. But we have $[k(b):k]=[k(a):k]$ and $k(a)\ne k(b)$, a contradiction.
A: Presumably you are applying Zorn's lemma to the set of pairs $(\ell,\sigma)$ such that $\ell$ is an algebraic extension of $k$ and $\ell\in\operatorname{Gal}(\ell/k)$ is an automorphism such that $k$ is its fixed field. The set of such pairs is partially ordered by the relation $(\ell,\sigma)\prec (\ell',\sigma')$ defined to hold, whenever $\ell\subset\ell'$ and $\sigma'\vert_\ell=\sigma$.
Zorn's lemma then does promise the existence of a maximal element $(k_m,\sigma_m)$.
If $z\in k^a, z\notin k_m$, then $k[z]$ is an algebraic extension of $k$ hence cyclic. I would next try to prove that it is possible to lift the restriction of $\sigma_m$ to $\ell=k[z]\cap k_m$ to a $k$-automorphism $\sigma_z$ of $k[z]$ with the prescribed property. 
Now that $\sigma_m$ and $\sigma_z$ are compatible in the sense that they agree whenever both are defined, it should be possible to do the following. Let $\ell$ be a finite extension of $k$ that is contained in $k_m$. Then prove that there exists a $k$-automorphism of $\ell[z]$ such that its restriction to $\ell$ agrees with $\sigma_m$, and that its restriction to $k[z]$ agrees with $\sigma_z$. If you have covered the Galois theory of linearly disjoint field extensions, then this is easy.
As $k_m$ is the union of such fields $\ell$, you have just proven the existence of a suitable $k$-automorphism of $k_m[z]$ violating maximality of $k_m$.
A lot of details to check and verify, but I think this works.
