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Thread: Galois extension

  1. #1
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    Galois extension

    Explain why $\displaystyle \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. Find the smallest extension field $\displaystyle E/\mathbb{Q}(^3\sqrt2)$ so that $\displaystyle E/\mathbb{Q}$ is Galois. Determine the isomorphism class of $\displaystyle \text{Gal}(E/\mathbb{Q})$.

    I know how to prove that $\displaystyle \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. But I don't know how to do the next two parts of this question. Thanks in advance.
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  2. #2
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    Quote Originally Posted by vaevictis59 View Post
    Explain why $\displaystyle \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. Find the smallest extension field $\displaystyle E/\mathbb{Q}(^3\sqrt2)$ so that $\displaystyle E/\mathbb{Q}$ is Galois. Determine the isomorphism class of $\displaystyle \text{Gal}(E/\mathbb{Q})$.

    I know how to prove that $\displaystyle \mathbb{Q} \subset \mathbb{Q}(^3\sqrt2)$ is not a Galois extension. But I don't know how to do the next two parts of this question. Thanks in advance.
    Notice that the minimal polynomial of $\displaystyle \sqrt[3]{2}$ over $\displaystyle \mathbb{Q}$ is $\displaystyle x^3 - 2$. The roots of this polynomial is $\displaystyle \sqrt[3]{2},\zeta\sqrt[3]{2},\zeta^2\sqrt[3]{2}$. Thus, the splitting field of this polynomial is $\displaystyle E=\mathbb{Q}(\zeta,\sqrt[3]{2})$. Notice that $\displaystyle E/\mathbb{Q}$ is Galois with $\displaystyle \mathbb{Q}(\sqrt[3]{2})\subset E$ and it is the smallest extension because it is splitting field of the minimal polynomial. Now we see that $\displaystyle \text{Gal}(E/\mathbb{Q}) = S_3$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Notice that the minimal polynomial of $\displaystyle \sqrt[3]{2}$ over $\displaystyle \mathbb{Q}$ is $\displaystyle x^3 - 2$. The roots of this polynomial is $\displaystyle \sqrt[3]{2},\zeta\sqrt[3]{2},\zeta^2\sqrt[3]{2}$. Thus, the splitting field of this polynomial is $\displaystyle E=\mathbb{Q}(\zeta,\sqrt[3]{2})$. Notice that $\displaystyle E/\mathbb{Q}$ is Galois with $\displaystyle \mathbb{Q}(\sqrt[3]{2})\subset E$ and it is the smallest extension because it is splitting field of the minimal polynomial. Now we see that $\displaystyle \text{Gal}(E/\mathbb{Q}) = S_3$.
    How do we know this is isomorphic to $\displaystyle S_3$ and not $\displaystyle \mathbb{Z}_6$? I don't see how we know which group of order 6 it is isomorphic to.
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  4. #4
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    There are only two groups of order 6.

    A cyclic (abelian) one.

    And $\displaystyle D_6 \cong S_3$ which is non abelian.

    Write out the automorphisms it is pretty clear it is not cyclic nor abelian.
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