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Let $\zeta$ be a primitive 12th-root of unity. How many field extensions are there between $\mathbb{Q}(\zeta^3)$ and $\mathbb{Q}(\zeta)$?

Here is my solution:

$\mathbb{Q}(\zeta)$ is the splitting field of $X^{12}-1$, so $\mathbb{Q}(\zeta):\mathbb{Q}$ is a Galois extension. The minimal polynomial of $\zeta$ is the cyclotomic polynomial $\Phi_{12}$, whose degree is 4, therefore $G(\mathbb{Q}(\zeta):\mathbb{Q})=\mathbb{Z}_{12}^*$. Then, every non trivial subgroup has order 2, so every intermediate extension of $\mathbb{Q}(\zeta):\mathbb{Q}$ is quadratic, which shows that there is no intermediate field extensions between $\mathbb{Q}(\zeta^3)$ et $\mathbb{Q}(\zeta)$.

I would like to know if my reasoning is right.

Thanks in advance.

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    $\begingroup$ As long as you are confident in that $\Bbb{Q}(\zeta^3)$ is neither of the end points, then, yes, this is one way to conclude. $\endgroup$ Commented Dec 29, 2020 at 20:22
  • $\begingroup$ What do you mean by "end points"? $\endgroup$
    – QGM
    Commented Dec 29, 2020 at 20:41
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    $\begingroup$ The fields $L=\Bbb{Q}(\zeta)$ and $K=\Bbb{Q}(\zeta^3)$ themselves could be called the end points of $L/K$. If I were a teacher who assigned this as an exercise I would probably insist that a full solution include an argument as to why $L\neq K$, when the "interval" from $K$ to $L$ would have two end points rather than just one :-) Sorry about being unclear. $\endgroup$ Commented Dec 29, 2020 at 20:45

1 Answer 1

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Since $[Q(ζ):Q)]=|Z_{12}^*|=4$ and $[Q(ζ^3):Q)]=2$, then we get $[Q(ζ):Q(ζ^3))]=2$ and hence your reasons are right.

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