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The equation $x^3-2=0$ is irreducible over $\mathbb{Q}$. If $\alpha$ is a root and $\omega$ is a primitive 3rd root of 1, then $\mathbb{Q}(\alpha,\omega)$ is the minimal spliting field of $x^3-2=0$ over $\mathbb{Q}$. Since $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}]=6$, the Galois group of this extension (or the Galois group of $x^3-2=0$ over $\mathbb{Q}$) is $S_3$.

Since $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\omega)]=3$, the Galois group of the equation over $\mathbb{Q}(\omega)$ is a cyclic subgroup of $S_3$ of order 3.

Question: Which of the 3-cycles of $S_3$ is the generator of this cyclic group?

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The Galois group $Gal(\mathbb{Q}(\alpha,\omega)/\mathbb{Q})$ contains the six automorphisms below (where each $\sigma_i$ fixes $\mathbb{Q}$):

$$ \begin{matrix} \sigma_1(\alpha)=\alpha & \sigma_1(\omega)=\omega \\ \sigma_2(\alpha)=\alpha\omega & \sigma_2(\omega)=\omega \\ \sigma_3(\alpha)=\alpha\omega^2 & \sigma_3(\omega)=\omega \\ \sigma_4(\alpha)=\alpha & \sigma_4(\omega)=\omega^2 \\ \sigma_5(\alpha)=\alpha\omega & \sigma_5(\omega)=\omega^2 \\ \sigma_6(\alpha)=\alpha\omega^2 & \sigma_6(\omega)=\omega^2 \end{matrix} $$

These automorphisms are obtained by noting that any $\sigma\in Gal(\mathbb{Q}(\alpha,\omega)/\mathbb{Q})$ must send $\alpha$ to another root of $x^3-2$ and $\omega$ to another root of $x^2+x+1$ (these polynomials are the minimum polynomials of $\alpha$ and $\omega$ respectively). To see the isomorphism between this group and $S_3$ we look at how the roots of $x^3-2$ are permuted by the automorphisms. Let $(\alpha,\alpha\omega,\alpha\omega^2)$ correspond to $(1,2,3)$. Then calculating $\sigma_i(\alpha),\sigma_i(\alpha\omega),$ and $\sigma_i(\alpha\omega^2)$ we get the following map between the $\sigma_i$'s and $S_3$:

$$ \sigma_1 \mapsto (1) \\ \sigma_2\mapsto (1,2,3) \\ \sigma_3\mapsto (1,3,2) \\ \sigma_4\mapsto (2,3) \\ \sigma_5 \mapsto (1,2) \\ \sigma_6\mapsto (1,3) $$

The subgroup of the Galois group that corresponds to the subfield $\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\omega)$ is then the subgroup of order three containing $\sigma_1, \sigma_2, \sigma_3$. This subgroup is generated by either $\sigma_2$ or $\sigma_3$ (i.e. $\langle \sigma_2 \rangle = \langle \sigma_3 \rangle$).

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