# Proving that a Galois group is cyclic

Let $K$ be a ﬁeld containing a primitive $n$th root of unity and let $F = K(t)$ be the ﬁeld of rational functions over $K$. I'm having trouble proving that for each $n > 1$ the ﬁeld $F$ is Galois over $L = K(t^n) ⊂ F$ and that the Galois group $Aut_LF$ is cyclic of order $n$.

I know how to prove that if $K$ contains a primitive $n$th root of unity and $α$ is a root of $X^n − a$ where $a \ne 0$ is in $K$, then $G(K(α)/K)$ is cyclic (this is something I went over with my professor a couple of weeks ago), but I'm honestly not sure if/how I can apply that to this problem.

• The extension $K(t)/K(t^n)$ is just $F(\alpha)/F$ where $F=K(t^n)$ and $\alpha$ is a root of $X^n-t^n$. – anon Apr 14 '16 at 0:13
• This is very confusing till you get the hang of it. @anon has given you the key. – Lubin Apr 14 '16 at 0:49