Let be a cyclic extension of dimension over and let be a generator of . Let , and suppose is a non-zero element of such that for some . Show that there exists a in the (unique) subfield of of dimensionality such that .

is a cyclic group, then it has a unique subgroup of index . By the Galois corresponding theorem, there exists a unique subfield such that . . has dimensionality over .

Consider , then . We also have

,

by Hilbert’s Satz 90, such that .

Let , then , because

Surprisingly we have

.

Jacabson p300. This problem puzzled me for three weeks. Finally it turns out to be very easy.

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## Comments

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.

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 $d = c l/\sigma^{i_0}(l)$.

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