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I'd like to compute the intermediate fields between $\mathbb{Q}$ and a splitting field for $f=X^4-10X^2+4$.

Here's what I've done:

The roots of $f$ are $\pm \sqrt{5\pm \sqrt{21}}$. Multiplying them we see that a splitting field for $f$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{5+\sqrt{21}})$ which has degree 4 over $\mathbb{Q}$.

Hence the order of the Galois group is 4. Its elements are determined by the action on $\sqrt{5+\sqrt{21}}$, which can be sent to either of the four roots of $f$. It's easily checked that the three non-identity elements have order $2$, hence the Galois group is isomorphic to $C_2\times C_2$.

By the Galois correspondence, there are three intermediate fields. One of them is obviously $\mathbb{Q}(\sqrt{21})$, but how do I get the other ones? I tried to manually determine the fixed fields for all three subgroups, but the calculations got too nasty for me.

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  • $\begingroup$ How do you know $\sqrt{5-\sqrt{21}}\in\mathbb{Q}(\sqrt{5+\sqrt{21}})$ ? $\endgroup$
    – Belgi
    Nov 14, 2012 at 13:43
  • $\begingroup$ Their product is $2$. $\endgroup$
    – Berci
    Nov 14, 2012 at 13:51
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    $\begingroup$ Now that you know that the roots lie in $\mathbb Q(\sqrt 6, \sqrt {14}, \sqrt {21})$, you can compute $((\sqrt 6 + \sqrt{14})/2)^2$ and act surprised. $\endgroup$
    – mercio
    Nov 14, 2012 at 14:43

1 Answer 1

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  1. $\Bbb Q\left(\sqrt{5+\sqrt{21}} + \sqrt{5-\sqrt{21}}\right)$
  2. $\Bbb Q\left(\sqrt{5+\sqrt{21}} - \sqrt{5-\sqrt{21}}\right)$

For an involutive automorphism $\phi$, for all elements $s$, we have that $s+\phi(s)$ and $s\cdot \phi(s)$ are always fixed, so they have a chance to generate the subfield..

Update: interestingly, $$ \begin{align} \left(\sqrt{5+\sqrt{21}} + \sqrt{5-\sqrt{21}}\right)^2 &= 14 \\ \left(\sqrt{5+\sqrt{21}} - \sqrt{5-\sqrt{21}}\right)^2 &= 6. \end{align} $$ Hmm.. In other words, the following 3 intermediate fields we have: $$\Bbb Q(\sqrt 6),\ \Bbb Q(\sqrt{14}),\ \Bbb Q(\sqrt{21})$$

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  • $\begingroup$ Did you agree with the OP that the degree of the extension is $4$ ? $\endgroup$
    – Belgi
    Nov 14, 2012 at 13:51
  • $\begingroup$ Yes, it seems so. $\endgroup$
    – Berci
    Nov 14, 2012 at 13:51
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    $\begingroup$ Very clear. Thanks a lot. $\endgroup$
    – user46225
    Nov 14, 2012 at 14:37

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