Cyclic extensions Hi. Are there nice/simple examples of cyclic extensions $L/K$ (that is, Galois extensions with cyclic Galois group) for which $L$ cannot be written as $K(a)$ with $a^n\in K$?
Thanks.
 A: Dear Justin, your question is subtly ambiguous.
First interpretation:  Is there a cyclic extension $L/K$ of degree $n$ that cannot be written $L=K(a)$ with $a\in L$ and $a^n\in K ?$.
Answer Yes. For example, let $p$ be a prime number and $n$ an integer.The extension 
$\mathbb F_p\subset \mathbb F_{p^{p^n}}$ is cyclic and not of the required form since $a^{p^n} \in \mathbb F_p$  implies $a\in \mathbb F_p .\;$  So the simplest possible example for your question is  $\mathbb F_2 \subset \mathbb F_4$ ! [Franz and David give excellent examples in characteristic zero]
Second interpretation:  Is there a cyclic extension $L/K$ of degreee $n$ that cannot be written $L=K(a)$ with $a\in L$ and $a^N\in K$ for some N that might be different from n ?
Partial answer Some examples in the preceding interpretation disappear! For instance if you take
$\mathbb F_2 \subset \mathbb F_4$, you CAN write $\mathbb F_4= \mathbb F_2(a)$ with $a^\textbf{3}=1$ : just take for $a$ one of the two elements in $\mathbb F_4 \setminus \mathbb F_2$. Actually if $K\subset L$ are finite all examples disappar: all such extensions are cyclic and can be written $L=K(a)$ with $a^N\in K$. Indeed if $a\in L$ is a primitive element (which always exists: our extension is separable), we have  $L=K(a)$ and if $q=card(L)$ we can be sure that $a^{q-1}=1 \in K$.
A: Take $\zeta = e^{2\pi i/p}$ for a prime number $p\equiv1$ (mod 3), e.g. $p=7$.
Then  $Q(\zeta+\bar\zeta)$ is a totally real cyclic Galois extension of $\mathbf{Q}$ of degree a multiple of 3, hence contains a cubic extension $L$ that is Galois with cyclic Galois group.
Being totally real it cannot be the splitting field of a polynomial of the
form $X^3-a$. (or use David Loeffler's argument above).
Dirichlet's theorem on primes in arithmetic progressions assures us that  we have infinitely many such examples over the base $\mathbf{Q}$.
A: If $K$ has characteristic $p>0$, an
Artin-Schreier extension
$L/K$ is cyclic of degree $p$, and is generated by an element $a \in L$ such that $a^p \not \in K$. Such
an $L$  has the form $K(a)$ for a root $a$ of a polynomial of the 
form $X^p - X - \alpha$ for suitable $\alpha \in K$; the polynomial
then splits over $L$ since its roots are $a+i$ for $i \in \mathbf{F}_p$
(in this way you can see the action of the Galois group $\mathbf{Z}/p$ on the roots).
An example of such an extension is $\mathbf{F}_4/\mathbf{F}_2$ since $\mathbf{F}_4 = \mathbf{F}_2(a)$ for a root $a$ of $X^2 - X - 1 = X^2 + X +1$.
Edit(x2):
Looking at Georges Elencwajg's answer, I see that I made what he labels the "First Interpretation" of the question. Since his "Second Interpetation" seems interesting, here is what I hope is an example.
Let $k$ be a field of char $p>0$, and let
$K = k(t)$ for transcendental $t$. Let $L/K$ be the Artin-Schreier extension obtained by adjoining to $K$ a root $a$ of $X^p - X - t$.
I claim that $a^N \not \in K$ for any $N \ge 1$.
Well, note that $k[a,t]$ is an integral extension of $k[t]$. Hence if $a^N \in K$,
then in fact $a^N \in \langle t \rangle \subset k[t]$, so that $a$ satisfies a polynomial
$$X^N - tg(t) \quad \text{for some $g(t) \in k[t]$}.$$
But then $X^p - X - t$ divides $X^N - tg(t)$ and reducing mod $t$ 
gives a contradiction. (i.e. the condition $a^N \in K$ would mean that $L/K$ should be
ramified at the prime $\langle t \rangle$ of $k[t]$, which isn't so).
So: this cyclic extension is generated by an element $a$ for which $a^N \not \in K$
for any $N \ge 1$. I expect (?!) that this should give an example for Elencwajg's "Second Interpretation" of the question, but I didn't think about other possible generators
for the extension $L/K$.
A: An example is given by any cubic Galois extension of $\mathbb{Q}$, e.g. the extension $\mathbb{Q}(\alpha)$ where $\alpha = \cos \frac{2\pi}{9}$ is a root of $x^3 - 3x + 1$. 
This works because if the extension were of the form $\mathbb{Q}(\beta)$ with $\beta^3 \in \mathbb{Q}$, then since it is Galois it would have to contain a nontrivial cube root of unity, which it obviously doesn't.
