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Let $K=\Bbb Q[3^{1/5},\zeta_5]$ where $\zeta_5=e^{i2\pi/5}$ is a fifth root of unity. Let $\sigma$, $\tau$ be Gal$(K/Q)$ be defined by $$\sigma(3^{1/5})=\zeta_5 3^{1/5},\qquad \sigma(\zeta_5)=\zeta_5,$$ $$\tau(3^{1/5})=3^{1/5},\qquad \tau(\zeta_5)=\zeta_5^2.$$ Prove that $K$ is Galois Over $\Bbb Q$ and that Galois$(K/\Bbb Q)=\langle\sigma,\tau\rangle$. Find all Subfields of $K$ and their Corresponding Subgroups of Gal$(K/\Bbb Q$). Which of the Subfields are Gal$(K/\Bbb Q)$?

I have found that in order to prove $K$ being Galois, we show it is a finite normal seperable extension of $\Bbb Q$, which then makes proving it being cyclic not difficult. I am struggling with the subfield and subgroups part however, and in showing which of the subgroups are galois.

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$K$ is Galois because it is the splitting field of a separable polynomial (one of the many characterizations of a Galois extension). Classifying Galois sub-fields is a simple matter of classifying normal subgroups. The Sylow $5$ subgroup must be normal since $\text{#}P_5\equiv 1\mod 5$ and $\text{#}P_5 | 4$ (the total degree is $20$). So the fixed field of $\sigma$ which is $\Bbb Q(\zeta_5)$ is a Galois sub-field.

We also see that the Sylow-$2$ subgroup cannot be normal as $\langle\tau\rangle$ is one such and conjugating by $\sigma$ produces a group which fixes $\zeta_5\sqrt[5]{3}$, so none of these are normal subfields (they are all conjugate by Sylow).

The subfield $K\cap\Bbb R = K^{\tau^2}$ fixed by complex conjugation is clearly normal--you can also realize it as the intersection of all the Sylow $2$ subgroups--and indeed this is the only subgroup of order $2$ using Sylow's theorems.

Finally we note that $H=\langle\sigma,\tau^2\rangle$ is normal by the preceding discussion, and as $\tau\not\in H$ we have that this subgroup has order $10$ and $K^H$ is also a normal subgroup, it is in-fact $\Bbb Q(\sqrt{5})$, which is the unique quadratic subfield of $K$.

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