1
$\begingroup$

This is a statement made by my professor, but I suspect it to be false.

He said that if $k(\alpha)$ contains a primitive nth root of unity and $\alpha^n = a \in k$ then $k(\alpha)$ is a cyclic extension

I don't think this is true, consider the case $\alpha$ as a primitive $nth$ root, and $k = \mathbb{Q}$, then we know that this field extension is not always cyclic. right?

The reason he wanted this statement is because we are trying to prove that $k$ adjoinin a bunch of $ith$ roots of unity where $i$ is in a finite subset of $\mathbb{N}$ is a root extension with each successive intermediate field extension $K_{i+1}/K_i$ is cyclic.

$\endgroup$
  • $\begingroup$ Check the statement, please. The correct version is that when $k$ contains a primitive $n$th root of unity and $\alpha^n=a\in k$. then $k(\alpha)/k$ is cyclic. When $k=\Bbb{Q}$ this result applies only to $n=2$ because $\Bbb{Q}$ doesn't contain any higher roots of unity. For the claim to hold it is not sufficient that those roots of unity come along only as a by product of adjoining $\alpha$, $\endgroup$ – Jyrki Lahtonen Feb 21 '18 at 5:52
  • $\begingroup$ I am aware of your statement. I talked to him afterward and verified that he meant precisely what was stated in the original post which is why I wanted to verify. $\endgroup$ – Ecotistician Feb 21 '18 at 13:21
  • 2
    $\begingroup$ Ok. Then he was wrong. $\endgroup$ – Jyrki Lahtonen Feb 21 '18 at 13:47
3
$\begingroup$

Some counterexamples to the statement you posed:

$$\mathrm{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $$ $$\mathrm{Gal}(\mathbb{Q}(\zeta_{12}) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $$ $$\mathrm{Gal}(\mathbb{Q}(\zeta_{15}) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 4 \mathbb{Z} $$

More generally, it turns out that there is an isomorphism

$$ \mathrm{Gal}(\mathbb{Q}(\zeta_n) / \mathbb{Q}) \cong (\mathbb{Z} / n \mathbb{Z})^\times $$

where by $R^\times$ I mean the unit group of the ring $R$.

However, adjoining roots of unity always gives abelian extensions, and so they can always be decomposed into a tower of cyclic extensions; e.g. these examples can be decomposed into

$$\mathbb{Q}(\zeta_8) / \mathbb{Q}(\zeta_4) / \mathbb{Q} $$ $$\mathbb{Q}(\zeta_{12}) / \mathbb{Q}(\zeta_4) / \mathbb{Q} $$ $$\mathbb{Q}(\zeta_{15}) / \mathbb{Q}(\zeta_3) / \mathbb{Q} $$

$\endgroup$
  • $\begingroup$ 1. can you give me a hint to prove your last statement? It is my final piece to proving dummit and foote lemma 38 (every element in a root extension is in a galois root extension s.t. it's a tower of cyclic extensions. (I am trying to use this to prove k(zeta_n1, zeta_n2........zeta_nk) is a root extension s.t. the chains are cyclic) by expressing this field as k(zeta_n1).....(zeta_n(k-1) (zeta_nk)... $\endgroup$ – Ecotistician Feb 21 '18 at 2:22
  • $\begingroup$ @Ecotistician: I suppose the simplest explanation is the group theoretic statement: any finitely generated abelian group is a product of cyclic groups. The towers I built simply work one factor at a time. (this is a lot stronger than what you actually need here; but using this is a lot simpler) $\endgroup$ – Hurkyl Feb 21 '18 at 3:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.