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Let $E= \mathbb{Q}(\sqrt{2} + \sqrt{7})$

For every $H < \mathrm{Gal}(E/\mathbb{Q})$ identify $\mathcal{F}(H)$

My attempt:

Notice that we need to adjoin four elements to reach the splitting field, namely $\pm \sqrt{2}, \pm \sqrt{7}$ so then the automorphisms can only permute these four elements. Naturally we would have the identity in this group

$\begin{align*} \sigma_0: \sqrt{2} \to \sqrt{2} & \hspace{10pt} \sqrt{7} \to \sqrt{7} \\ \sigma_1: \sqrt{2} \to \sqrt{2} & \hspace{10pt} \sqrt{7} \to -\sqrt{7} \\ \sigma_2: \sqrt{2} \to -\sqrt{2} & \hspace{10pt} \sqrt{7} \to \sqrt{7} \\ \sigma_3: \sqrt{2} \to -\sqrt{2} & \hspace{10pt} \sqrt{7} \to -\sqrt{7} \end{align*}$

Since $E/\mathbb{Q}$ is a Galois extension,

$$\therefore \mathrm{Gal}(E/\mathbb{Q}) = \mathrm{Aut}(E/\mathbb{Q}) = \{ \sigma_0, \sigma_1, \sigma_2,\sigma_3\}$$

Now we have that $\forall i = 0,...,3$,

$\begin{align*} \sigma_i(\sigma_i(\sqrt{2})) = \sqrt{2} = \sigma_0(\sqrt{2}),\hspace{10pt} \mbox{ and } & \hspace{10pt} \sigma_i(\sigma_i(\sqrt{7})) = \sqrt{7} = \sigma_0(\sqrt{7}) \end{align*}$

Hence, each element is of order two (it's own inverse) and

$$\therefore \mathrm{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$$

Now that we have that $\mathrm{Gal}(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$, we must describe $\mathcal{F}(H)$ in terms of the subgroups of the Klein group. All possible proper subgroups H of $\mathrm{Gal}(E/\mathbb{Q})$ are given by

$$ H_0 = \{\sigma_0\}, H_1 = \{\sigma_0,\sigma_1\}, H_2 = \{\sigma_0,\sigma_2\}, H_3 = \{\sigma_0,\sigma_3\}$$

Now we must describe for each $H_i$ $$\mathcal{F}(H) = \{ x \in E: \sigma(x) = x, \forall \sigma \in H \} $$

Now for $H_0$ $$\mathcal{F}(H_0) = \mathbb{Q}(\sqrt{2},\sqrt{7})$$

Similarly, $$\mathcal{F}(H_1) = \mathbb{Q}(\sqrt{2})$$ $$\mathcal{F}(H_2) = \mathbb{Q}(\sqrt{7})$$ However, my problem is this one:

$$\mathcal{F}(H_3) = ??$$

I am correct, or this are not the fixed fields ? Can anyone help me computing the one for $H_3$ ?

I guess it must be $\mathbb{Q}(\sqrt{2},\sqrt{7})$ since $\sigma_3 = -\sigma_0$, but I dont know how to prove it.

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    $\begingroup$ We don't have $\sigma_3=-\sigma_0$ on all the field, only if restricted to this subset $\{\sqrt2,\sqrt7\}$. $\endgroup$
    – Berci
    Commented Apr 22, 2018 at 19:33
  • $\begingroup$ @Berci thank you. $\endgroup$ Commented Apr 22, 2018 at 19:35

2 Answers 2

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Note that $\mathbb{Q}(\sqrt2,\sqrt7) = \mathbb{Q}(\sqrt2+\sqrt7)$, the whole field you started with, so this can't be the fixed field of $H_3$ -- in fact, we know $H_3$ doesn't fix $\sqrt2$ or $\sqrt7$.

Hint: We know the degree of $\mathbb{Q}(\sqrt2+\sqrt7)$ over $\mathbb{Q}$ is $4$, so we can find a basis of four elements over $\mathbb{Q}$. Three of those are $1$, $\sqrt2$, and $\sqrt7$. What could be fourth element of the basis? Then check if this element is fixed by $H_3$.

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It's not $\mathbb{Q}(\sqrt{2},\sqrt{7})$ since $\sqrt{2} \in \mathbb{Q}(\sqrt{2},\sqrt{7})$ is not fixed by $\sigma_3$.

You correctly identified the first two fixed fields by noting that $\sigma_1(\sqrt{2})=\sqrt{2}$ and $\sigma_2(\sqrt{7})=\sqrt{7}$.

Now $\sigma_1 \circ \sigma_2 = \sigma_3$. So $\sigma_3(\sqrt{2}\sqrt{7})=\sqrt{2}\sqrt{7}$.

I think you can guess $H_3$ now.

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