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I'm to look at $\mathbb{Q}(\zeta_7)$ and $\mathbb{Q}(\zeta_{10})$, and they both have a common thing I can't see how to work around.

For $\mathbb{Q}(\zeta_7)$, skipping some details I doubt matter too much, I get that its Galois group has two subgroups, one of which corresponds to $\mathbb{Q}(i\sqrt{7})$ (which I got to calculating a trace), but the other subgroup, using the trace, gives me the element $\zeta + \zeta^6$; the problem is it turned out that $\mathbb{Q}(\zeta +\zeta^6)= \mathbb{Q}(\zeta_7)$. I tried the norm but that was worse; came out as 1.

Same issue with $\mathbb{Q}(\zeta_{10})$. This has a Galois group with only one subgroup, which using a trace to get a fixed element gave $\zeta+\zeta^9$ which has identical problems to the case above.

Given a group of automorphisms, are there any other methods of finding fixed elements to get the required subfield? I only know of the trace and norm, if there isn't some candidate suspected by inspection.

FireGarden
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    $\mathbb Q(\zeta+\zeta^{-1})$ is certainly not equal to $\mathbb Q(\zeta)$, for either of $\zeta = \zeta_7$ or $\zeta_{10}$. – Dustan Levenstein Mar 23 '14 at 20:55
  • Have a look at this: http://math.stackexchange.com/questions/132169/a-complete-picture-of-the-lattice-of-subfields-for-a-cyclotomic-extension-over?rq=1 – M Turgeon Mar 23 '14 at 20:57
  • @DustanLevenstein If we call $\zeta+\zeta^6 = \beta$, then I get that $\zeta^3 = \beta^3-\beta^2-2\beta+1$, and the powers of $\zeta^3$ generate the others.. Is this really not the case? I don't see mistakes in my working (i think ill re-do it to be sure..) – FireGarden Mar 23 '14 at 20:59
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    well that can't be right, because $\zeta+\zeta^6 \in \mathbb R$, and $\zeta^3 \notin \mathbb R$. – Dustan Levenstein Mar 23 '14 at 21:02
  • @DustanLevenstein I did make a mistake, Thank's for pointing that out, I probably wouldn't have questioned that stuff..

    Secondly, given it won't be the whole field, is there a simpler way of finding an explicit element to adjoin? I really had to mess about to get that $i\sqrt{7}$, making a polynomial the trace satisfied. Not a lot of fun.

     – FireGarden Mar 23 '14 at 21:09

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One way of finding an element fixed by automorphisms is to take the sum of an orbit: say $\sigma$ is an automorphism with $\sigma^4 = \mathrm{id}$, then $x + \sigma(x) + \sigma^2(x) + \sigma^3(x)$ is always fixed, regardless of $x$. Of course, whether or not this is an interesting example depends on what you choose for $x$. But, for example, you can immediately find $\zeta + \zeta^6$ this way, because that's just the sum of the orbit of $\zeta$ under complex conjugation.

Ben Millwood
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