## Tag Archives: trace and norm

### Albert radicals of norm and Hilbert’s Satz 90 $\mathbf{Problem(Albert):}$ Let $E$ be a cyclic extension of dimension $n$ over $F$ and let $\eta$ be a generator of $\text{Gal }E/F$. Let $r|n$, $n=rm$ and suppose $c$ is a non-zero element of $F$ such that $c^r=N_{E/F}(u)$ for some $u\in E$. Show that there exists a $v$ in the (unique) subfield $K$ of $E/F$ of dimensionality $m$ such that $c=N_{K/F}(v)$. $\mathbf{Proof:}$ $G=\text{Gal }E/F$ is a cyclic group, then it  has a unique subgroup $H=\{\eta^m,\eta^{2m},\cdots,\eta^{rm}=1\}$ of index $r$. By the Galois corresponding theorem, there exists a unique subfield $K=\text{Inv }H$ such that $\text{Gal K/F}\cong G/H$. $\text{Gal }K/F=\{\eta^1|_K,\eta^2|_K\cdots,\eta^m|_K\}$. $K$ has dimensionality $m$ over $F$.

Consider $w=c^{-1}u\eta(u)\eta^2(u)\cdots \eta^{m-1}(u)$, then $\displaystyle\eta(w)=\frac{\eta^m(u)}{u}w$. We also have $\displaystyle N_{E/K}(w)=\eta^m(w)\eta^{2m}(w)\cdots \eta^{rm}(w)=c^{-r}\eta(u)\eta^2(u)\cdots\eta^n(u)=1$,

by Hilbert’s Satz 90, $\exists \, l\in E$ such that $\displaystyle w=\frac{\eta^m(l)}{l}$.

Let $\displaystyle v=\frac{ul}{\eta(l)}$, then $v\in K$, because $\displaystyle \eta^m(v)=\frac{\eta^m(u)\eta^m(l)}{\eta^{m+1}(l)}=\frac{\eta^m(u)\eta(l)}{\eta^{m+1}(l)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{\eta^m(l)}{\eta(l)}=\frac{\eta^m(u)}{\eta(w)}\cdot\frac{wl}{\eta(l)}=u\frac{l}{\eta(l)}=v$

Surprisingly we have $\displaystyle N_{K/F}(v)=v\eta(v)\cdots \eta^{m-1}(v)=u\eta(u)\cdots\eta^{m-1}(u)\frac{l}{\eta^m(l)}=cw\frac{l}{\eta^m(l)}=c$. $\text{Q.E.D}\hfill \square$ $\mathbf{Remark:}$ Jacabson p300. This problem puzzled me for three weeks. Finally it turns out to be very easy.

### Criterion for embedding in cyclic field $\mathbf{Problem:}$ Assume $F$ has $p$ distinct pth roots of 1. $p$ a prime, and $E/F$ is cyclic of dimensional $p^f$. Let $z$ be a primitive $p$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 $u\in E$. $\mathbf{Proof:}$ Suppose $\sigma$ is the generating isomorphism of cyclic Galois group $K/F$. Then $\displaystyle E=\text{Inv }\sigma^{p^f}$. $N_{K/E}(z)=z^p=1$, so by Hilbert satz 90, there exists $a\in K$ such that $\displaystyle z=\sigma^{p^f}(a)a^{-1}$ . $\displaystyle \sigma^{p^f}(a^{-1}\sigma(a))=\sigma^{p^f}(a^{-1})\sigma^{p^f+1}(a)=(z a)^{-1}\sigma(z a)=a^{-1}\sigma(a)$.
So $a^{-1}\sigma(a)=a_0\in E$, then $N_{E/F}(a_0)=a^{-1}\sigma(a)\sigma(a^{-1})\sigma^2(a)\cdots \sigma^{p^f-1}(a)\sigma^{p^f}(a)=a^{-1}\sigma^{p^f}(a)=z$. $\text{Q.E.D}\hfill \square$ $\mathbf{Remark:}$ Jacobson p300.