Algebra problem - traces and norms

Question

This question:

Assume $F$ has $p$ distinct $p$th roots of $1$, $p$ a prime, and $E|F$ is cyclic of dimension $p^f$. Let $z$ be a primitie $p$th root of $1$. Show that if $E|F$ can be imbedded in a cyclic field $K|F$ of dimension $p^{f+1}$, then $z = N_{E|F}(u)$ for some u $\in E$.

is in the book Basic Algebra, Jacobson. $N_{E|F}(u)$ is the norm of u (If $E|F$ is Galois then the norm of $u$ is $\prod_{\phi \in Gal(E|F)} \phi_i (u) $). Can anyone give a hint on how to solve it? For example, what would be a good way to show that an element of $E$ is norm of some other element?

Answer 1

Thanks for the chance to dust out some cobwebs and refigure this out :-)

Trying to write this as some kind of a hint. Let's begin by working out the simple case, where $p=2$, $f=1$. Let $\sigma$ be a generator of the Galois group $Gal(K/F)\simeq C_4$. By Galois correspondence we know that $E$ is the fixed field of $\tau=\sigma^2$. Now we assume that $z=-1\in F$ ($z\neq1$, so we exclude the characteristic two case). We also know that we have root tower extensions all the way down. In particular we have $K=E(w)$, where $w^2=v\in E$. Because $\tau(w)\neq w$ and $$ v=\tau(v)=\tau(w^2)=(\tau(w))^2, $$ we can deduce that $\tau(w)=-w=zw$.

Consider the element $u=w/\sigma(w)$.

• We see that $u\in E$. This is because $\tau$ and $\sigma$ commute, and therefore $$\tau(u)=\tau(\frac w{\sigma(w)})=\frac{\tau(w)}{\sigma(\tau(w))}=\frac{-w}{-\sigma(w)}=u.$$
• We also know that the restriction of $\sigma\vert_E$ to $E$ generates the Galois group $Gal(E/F)$. Therefore $$N_{E/F}(u)=u\sigma(u)=\frac w{\sigma(w)}\cdot\frac{\sigma(w)}{\sigma^2(w)}=\frac{w}{\tau(w)}=-1=z. $$

Your task is to generalize this construct. In the general case you can similarly deduce that:

• The field $E$ is the fixed field of $\tau=\sigma^{p^f}$, where $\sigma$ is a generator of $Gal(K/F)$.
• You also have $K=E(w)$ with $w^p=v\in E$.
• By replacing $\sigma$ with its suitable power (exponent coprime to $p$) you can arrange that $\tau(w)=zw$.

Good luck!

Moral: The ratio of conjugates has always norm $1$. But with a ratio of conjugates coming from an extension field we get something nearly as good, and can use that here. This can be rewritten in cohomological language, but I'm not the right guy to describe all that.