3
$\begingroup$

In the study of cyclic extensions we have the following theorem:

Theorem Let $K$ be a field containing an $n$-th primitive root of unity $\zeta$. Then the following claims hold:

  • If $E/K$ is a cyclic field extension of degree $n$, there exists $\alpha \in E$ such that $E=K(\alpha)$ and $\alpha$ has minimal polynomial $X^n-\alpha^n$ over $K[X]$.

  • If $\alpha\in\overline{K}$ is such that $\alpha^n \in K$, then $K(\alpha)/K$ is cyclic of degree $d$ dividing $n$ and the minimal polynomial of $\alpha$ over $K[X]$ is $X^d-\alpha^d$.

Observe that for the theorem to apply is necessary that $char(K)=0$ or $char(K)=p>0$ and $p\nmid n$. But this is not sufficient because we may still not have an $n$-th primitive root of unity in $K$.

My question is if we have a general approach to attack the problem of characterizing $E/K$ cyclic of degree $n$ in the cases where $char(K)=0$ or $char(K)=p>0$ and $p\nmid n$ but there is no $n$-th primitive root of unity in $K$. Note that this is not an Artin-Schreier case.

So, under these hypothesis, is it always true that $E=K(\alpha)$ for some $\alpha \in E$ with minimal polynomial $X^n-\alpha^n$ over $K[X]$? And if that is not the case, what can we say about $E/K$?

My attempt

Let $K$ be a field such that $char(K)=0$ or $char(K)=p>0$ and $p\nmid n$. Let $E/K$ be a cyclic extension of degree $n$. Now consider an algebraic closure $\overline{K}$ of $K$ containing $E$. We have an $n$-th primitive root of unity $\zeta$ in $\overline{K}$. We can consider the extensions $E(\zeta)/E\,$, $\,E(\zeta)/K(\zeta)$ and $K(\zeta)/K$. Let's summarize what we have so far:

  • $E/K\,$, $\,E(\zeta)/E\,$, $\,E(\zeta)/K(\zeta)$ and $K(\zeta)/K$ are finite Galois extensions.
  • $E/K$ is cyclic of degree $n\,$, $[E:K]=n$.
  • $K(\zeta)/K$ is a cylcotomic extension with Galois group $H \leq \mathbb{Z}_n^{\times}$. Put $[K(\zeta):K]=m$ so $m\mid \varphi(n)$.
  • Put $[E(\zeta):E]=e$ and $[E(\zeta):K(\zeta)]=d$. We have $n\mid d$ and, because $E/K$ is Galois, $d \mid n$ and thus $[E(\zeta):K(\zeta)]=d =n$.

If $E(\zeta)/K(\zeta)$ were cyclic we could use the quoted theorem. Is it always the case that $E(\zeta)/K(\zeta)$ is cyclic under these hypothesis? What do we need for $E(\zeta)/K(\zeta)$ to be cyclic?

If we were able to conclude something about $E(\zeta)/K(\zeta)$, is it possible to descend back to $E/K$ and conclude something?

One last comment. Because $E/K$ is Galois and $Gal(E/K)$ is finite, the extension $E/K$ has a finite number of subfields and the primitive element theorem applies. Thus, we know that $E=K(\alpha)$ for some $\alpha \in E$ but it might not happen that the minimal polynomial of $\alpha$ over $K$ is of the form $X^n-\alpha^n$. At this point we can consider $\beta \in K$ to achieve $m(\alpha+\beta,K) = X^n + a_{n-2}X^{n-2}+\cdots+a_{1}X + a_{0}$ and this would solve the problem for $n=2$ because $E=K(\alpha)=K(\alpha+\beta)$. The case $n=2$ can nevertheless be dealt with using no more than the definition of field extension.

Thanks!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.