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.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Comments

  • Mert Sevinis  On December 13, 2012 at 7:46 pm

    A similar question arises as follows
    (http://books.google.fr/books/about/Algebraic_Extensions_of_Fields.html?hl=fr&id=dC5HvUrfy44C; p.70, exercise 32):

    $K, F$ be cyclic extensions of $k$ with $[K:k]= [F:k]=p$. Let $k\subset L\subset KF$ with $[L:k]=p$. Let $a\in k$ and suppose that $a=N_{K/k}(b) = N_{F/k}(c)$ where $b\in K$ and $c\in F$. Show that there is an element $d\in L $ such that $a=N_{L/K}(d)$.

    I think I have an idea about the proof; I just write all elements explicitly and at the end I obtain $p$ equations with $p$ unknowns. However, I don’t use Hilbert 90, and the question seems to ask an answer using Hilbert 90. A proof using Hilbert 90, is it always tricky and does it come out by surprise, or is there a theory behind it?

    By the way thank you for this answer, I gained very much of time.

    • Mert Sevinis  On December 15, 2012 at 8:50 pm

      I think we may erase this comment (including this one:)), because it has flaws. Instead, if I’m not mistaken, the question has a similar answer: (sketch) Let $\sigma, \tau$ be extensions of the generators of $G(K/k)$ and $G(F/k)$ respectively. For some $\eta = \sigma^{i_0}\tau^{j_0}$ $F()=L$. $N_{KF/L}(b^{-1}c)=1$, by Hilbert 90 $\exists l\in KF :\ b^{-1}c = l^{-1}\eta(l)$. We also take $G(L/k) = $, and then set $d = c l/\sigma^{i_0}(l)$.

      • Sun's World  On December 15, 2012 at 10:22 pm

        Hehe. I almost forgot galois theory. This is the course I learned last semester.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: