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Math Help - Cyclotomic Fields over Q

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

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

    By the way there is an incredible theorem in algebraic number theory. Above we said that for E/\mathbb{Q} to be contained in a cyclotomic field we require for E/\mathbb{Q} to be an (finite) abelian extension. However, the converse is true!!! In other words, if 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.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    We say that K/\mathbb{Q} is an abelian extension iff it is a Galois extension with \text{Gal}(K/\mathbb{Q}) an abelian group. Now if \mathbb{Q}\subseteq L \subseteq K, where L is a field, then \text{Gal}(K/L) is a normal subgroup of \text{Gal}(K/\mathbb{Q}) therefore by the fundamental theorem of Galois theory it follows that L/\mathbb{Q} must be a normal extension.

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

    By the way there is an incredible theorem in algebraic number theory. Above we said that for E/\mathbb{Q} to be contained in a cyclotomic field we require for E/\mathbb{Q} to be an (finite) abelian extension. However, the converse is true!!! In other words, if 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.
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    nevermind
    Last edited by robeuler; April 16th 2009 at 11:18 PM.
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    I have one follow up question. Are subfields of abelian extensions normal extensions over the base field?
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  6. #6
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    Quote Originally Posted by robeuler View Post
    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 F\subseteq E\subseteq K with K/F to be an abelian extension. Then E/F is normal if and only if \text{Gal}(K/E)\triangleleft \text{Gal}(K/F). Of course this is true since all subgroups of an abelian group are normal.
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