Q:

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:

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. 

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 $\mathbf{Q}$.

