Tag Archives: cyclic extension

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 pth 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.