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.

$\endgroup$
3
  • $\begingroup$ 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 $\alpha$, $\endgroup$ Feb 21, 2018 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. $\endgroup$ Feb 21, 2018 at 13:21
  • 2
    $\begingroup$ Ok. Then he was wrong. $\endgroup$ Feb 21, 2018 at 13:47

1 Answer 1

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

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

$\endgroup$
2
  • $\begingroup$ 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)... $\endgroup$ Feb 21, 2018 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) $\endgroup$
    – user14972
    Feb 21, 2018 at 3:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .