# Thread: Cyclotomic Fields over Q

1. ## Cyclotomic Fields over Q

Prove that E=Q(2^(1/3)) is not a subfield of any cyclotomic field over Q.

First I note that the extension E/Q is not Galois. It has only trivial automorphism group and is an extension of degree 3.

The Galois closure is F = Q(2^(1/3),z_3), where z_3 is a third root of unity.

I try to proceed by contradiction.
Suppose E is contained in Q(z_n) some cyclotomic field.

The Galois closure of E, F, must be contained in Q(z_n). So z_3 must be in z_n.

At this point I am stuck. I tried to show that 3|n. I know that n does not equal 3. Also if 3|n then n is not prime.

I can't see how to proceed (or if this is even a good method)

2. Originally Posted by robeuler
Prove that E=Q(2^(1/3)) is not a subfield of any cyclotomic field over Q.
We say that $\displaystyle K/\mathbb{Q}$ is an abelian extension iff it is a Galois extension with $\displaystyle \text{Gal}(K/\mathbb{Q})$ an abelian group. Now if $\displaystyle \mathbb{Q}\subseteq L \subseteq K$, where $\displaystyle L$ is a field, then $\displaystyle \text{Gal}(K/L)$ is a normal subgroup of $\displaystyle \text{Gal}(K/\mathbb{Q})$ therefore by the fundamental theorem of Galois theory it follows that $\displaystyle L/\mathbb{Q}$ must be a normal extension.

Cyclotomic extensions are a special case of abelian extensions. Thus for $\displaystyle E/\mathbb{Q}$ to be a subfield of a cyclotomic field we require for $\displaystyle E/\mathbb{Q}$ to be a Galois extension with an abelian Galois group. This is obviously not true because the minimal polynomial of $\displaystyle \sqrt[3]{2}$ is $\displaystyle x^3 - 2$, and only one root of the three is contained in $\displaystyle E$.

By the way there is an incredible theorem in algebraic number theory. Above we said that for $\displaystyle E/\mathbb{Q}$ to be contained in a cyclotomic field we require for $\displaystyle E/\mathbb{Q}$ to be an (finite) abelian extension. However, the converse is true!!! In other words, if $\displaystyle E/\mathbb{Q}$ is a (finite) abelian extension then it is a subfield of some cyclotomic extension. This deep result is known as the Kronecker-Weber theorem.

3. Originally Posted by ThePerfectHacker
We say that $\displaystyle K/\mathbb{Q}$ is an abelian extension iff it is a Galois extension with $\displaystyle \text{Gal}(K/\mathbb{Q})$ an abelian group. Now if $\displaystyle \mathbb{Q}\subseteq L \subseteq K$, where $\displaystyle L$ is a field, then $\displaystyle \text{Gal}(K/L)$ is a normal subgroup of $\displaystyle \text{Gal}(K/\mathbb{Q})$ therefore by the fundamental theorem of Galois theory it follows that $\displaystyle L/\mathbb{Q}$ must be a normal extension.

Cyclotomic extensions are a special case of abelian extensions. Thus for $\displaystyle E/\mathbb{Q}$ to be a subfield of a cyclotomic field we require for $\displaystyle E/\mathbb{Q}$ to be a Galois extension with an abelian Galois group. This is obviously not true because the minimal polynomial of $\displaystyle \sqrt[3]{2}$ is $\displaystyle x^3 - 2$, and only one root of the three is contained in $\displaystyle E$.

By the way there is an incredible theorem in algebraic number theory. Above we said that for $\displaystyle E/\mathbb{Q}$ to be contained in a cyclotomic field we require for $\displaystyle E/\mathbb{Q}$ to be an (finite) abelian extension. However, the converse is true!!! In other words, if $\displaystyle E/\mathbb{Q}$ is a (finite) abelian extension then it is a subfield of some cyclotomic extension. This deep result is known as the Kronecker-Weber theorem.

Thanks a lot! I read your explanation and knew exactly what was going on. I even know of the K-W theorem. And I agree that it is astonishing. It is one of those results that I am surprised about.

4. nevermind

5. I have one follow up question. Are subfields of abelian extensions normal extensions over the base field?

6. Originally Posted by robeuler
I have one follow up question. Are subfields of abelian extensions normal extensions over the base field?
I already answered that question. Say you have $\displaystyle F\subseteq E\subseteq K$ with $\displaystyle K/F$ to be an abelian extension. Then $\displaystyle E/F$ is normal if and only if $\displaystyle \text{Gal}(K/E)\triangleleft \text{Gal}(K/F)$. Of course this is true since all subgroups of an abelian group are normal.