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So basically the question is: Let $K$ be the splitting field of $t^5-2$ (over the rationals). Let $\theta=\sqrt[5]{2}$, and $\eta$ a primitive $5$-th root of unity. First we note that the splitting field of $t^5-2$ is given by $K=\mathbb{Q}(\eta,\theta)$. Since $\theta$ is an extension of degree $5$ and $\eta$ of degree $4$, then $K$ is an extension of degree $20$. Define $\tau$ to be the automorphism that fixes $\theta$ and it maps $\eta$ to $\eta^2$. Then the order of $\tau$ is $4$. Also, define $\sigma$ to be the automorphism fixing $\eta$ and mapping $\theta$ to $\eta\theta$. Then we see that $\sigma$ is of order $5$. I already constructed the lattice for the Galois group, but now I am trying to construct the lattice for the corresponding fixed fields.

By definition we have that $\langle\sigma\rangle$ has $\mathbb{Q}(\eta)$ as a fixed field. Note that $\langle \sigma,\tau^2\rangle$ is a subgroup of $\langle \sigma, \tau\rangle=G(K/\mathbb{Q})$ which contains $\langle \sigma\rangle$, so the fixed field of $\langle \sigma,\tau^2\rangle$ should be an intermediate field contained in $\mathbb{Q}(\eta)$ (since the correspondance betweeen galois subgroups and fixed fields is order reversing), but I cannot figure out what the fixed field of $\langle \sigma,\tau^2\rangle$ is. At first I thought it was $\mathbb{Q}(\eta^i)$ for some $i$, and we trivially have that $\sigma$ leaves $\eta^i$ unchanged and $\tau^2(\eta^i)=\eta^i$, but this does not work since $\tau^2(\eta)=\eta^4$, $\tau^2(\eta^2)=\eta^3$, $\tau^2(\eta^3)=\eta^2$ and $\tau^2(\eta^4)=\eta$.

So what would then the fixed field of $\langle \sigma,\tau^2\rangle$ be?

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1 Answer 1

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Note that $\tau^2$ maps $\eta$ to $\eta^4=\eta^{-1}$. So you might want to consider $\eta+\eta^{-1}$, which is contained in $\mathbb{Q}(\eta)$, and is fixed by both $\tau^2$ and $\sigma$. Now it's just a question of determining if it is the complete fixed field, or just a subfield thereof. But this should get you started.

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Thank you! this is the fixed field since I am looking for a field of degree $2$ over $\mathbb{Q}$ and $\mathbb{Q}(\eta+\eta^4)$ is indeed of degree $2$! –  Daniel Montealegre May 30 '12 at 3:30

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