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What are the unique subfields $L$ of $\mathbb{Q}(\zeta_{13})$ such that $[L:\mathbb{Q}]=2, 3, 4$, and $6$? I get that the Euler's totient function of $13$ is $12$ and that must have some relation, as its factors are listed above. Let's take the first case, $[L:\mathbb{Q}]=2$ and I'll show you why I'm confused.

I would assume for $[L:\mathbb{Q}]=2$ we have $\mathbb{Q}$-basis $\{e,\sigma\}$ where $\sigma: \zeta^x\mapsto\zeta^{x+6}$, because this cycles round our $\zeta^1$ to $\zeta^{12}$ in $2$ steps, however, why can we just miss out $\zeta^{13}=1$? Surely when it cycles from, say, $\zeta^6$ it goes $\zeta^6\mapsto \zeta^{12}\mapsto \zeta^{18}=\zeta^5\mapsto \zeta^{11}$ etc and we generate the whole group? Then $[L:\mathbb{Q}]\neq2$?

Sorry I'm not phrasing myself very well, but I honestly don't know how to. Thanks to all in advance!

carmichael561
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DurhamManiac
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  • Then what is the mapping I should be looking for? – DurhamManiac May 29 '16 at 18:33
  • Our basis for Q(ζ13)/Q is {ζ^1,ζ^2,ζ^3,ζ^4,ζ^5,ζ^6,ζ^7,ζ^8,ζ^9,ζ^10,ζ^11,ζ^12}, so how can there be a mapping that takes an element to itself after 2 iterations? Or am I wrong in thinking that is what [L:Q]=2 means? – DurhamManiac May 29 '16 at 18:37
  • In @DietrichBurde's notation, if $u \in U(p)$, (i.e., $u$ a non-zero element of $\mathbb F_{p}$) then every $\sigma\in {\rm Gal}(\mathbb Q(\zeta_p)/\mathbb Q )$ is of the form $\sigma_u( \zeta^k) = \zeta^{ku}$, for all $k$. OK? And you forgot $\zeta^0=1$ in your list above – peter a g May 29 '16 at 18:39
  • the autormorphisms of $\mathbb{Q}(e^{2i \pi / 13}) / \mathbb{Q}$ are $\sigma(e^{2i \pi / 13}) = e^{2i k \pi / 13}$ for some integer $k$ such that $gcd(k,13) = 1$. so the Galois group of $\mathbb{Q}(e^{2i \pi / 13}) / \mathbb{Q}$ is $((\mathbb{Z}/\mathbb{Z}{13})^*,\times)$ which is $\simeq C{12}$ the cyclic group with $12$ elements. let $g$ being a generator of the Galois group, the subgroups are the ones generated by $g^2$, $g^3$ and $g^6$, – reuns May 29 '16 at 18:40
  • hence by thefundamental theorem of Galois theory, the subfields are the ones fixed by those subgroups : $\mathbb{Q}(e^{12 i \pi / 13}), \mathbb{Q}(e^{ 6i \pi / 13})$ and $ \mathbb{Q}(e^{4 i \pi / 13})$ . is all this correct ? – reuns May 29 '16 at 18:40
  • @user1952009 - what about $g^4$? and your last comment is incorrect... e.g., if $u\in \mathbb F_{13}^*$, then $\zeta^{4u} =\zeta^4$ implies $u=1$.. – peter a g May 29 '16 at 18:54
  • @peterag from my understanding I'm not missing 1 from that basis, as the standard basis is {1,ζ^1,ζ^2,ζ^3,ζ^4,ζ^5,ζ^6,ζ^7,ζ^8,ζ^9,ζ^10,ζ^11} which as ζ^12 can be written as -(1+ζ^1+ζ^2+ζ^3+ζ^4+ζ^5+ζ^6+ζ^7+ζ^8+ζ^9+ζ^10+ζ^11), gives us an equivalent basis of {ζ^1,ζ^2,ζ^3,ζ^4,ζ^5,ζ^6,ζ^7,ζ^8,ζ^9,ζ^10,ζ^11,ζ^12}, correct? – DurhamManiac May 29 '16 at 18:55
  • Ooops you're right. shame.... Sorry! – peter a g May 29 '16 at 18:55
  • @peterag : yes I forgot $g^4$ of course, but what I meant is not that the subfields are fixed pointwise but are fixed as a whole, i.e. $\sigma(x) \in K$ for every $x \in K$. so if an automorphism $\sigma$ of $\mathbb{Q}(e^{2i \pi/13})$ send for some $d | 12$ : $K = \mathbb{Q}(e^{2i d\pi/13})$ to itself then $\sigma(e^{2i \pi/13}) = e^{n 2 (12/d) i \pi/13}$ for some $n, gcd(n,13) = 1$ – reuns May 29 '16 at 19:03
  • @user1952009 The fields you listed are all the same field - $\mathbb Q(\zeta_{13)}$ - by your comment preceding the one that contains your list. For instance, $(\zeta^6)^{-2} =\zeta$. – peter a g May 29 '16 at 19:14
  • Okay, now I understand how the Galois groups are generated, but I'm still uncertain of what the unique subfields are, like what combination of ζ do we have in our subfield corresponding to the galois group {e, g^4, g^8}? – DurhamManiac May 29 '16 at 19:16
  • The Galois correspondence $H\mapsto\mathbb Q (\zeta)^H$ gives you the lattice of subfields. To see this explicitly, for $H$ a multiplicative subgroup of $\mathbb F^$, play with the sum $$\sum_{u\in H} \zeta^u$$ (and their Galois conjugates) . Try $H= \mathbb (F_p^)^2$, the squares, and $H= {\pm 1}$. For the former, one gets a quadratic extension of $\mathbb Q$ - and this is related to quadratic reciprocity... For the latter case, the fixed field is the maximal (totally) real subfield, and is $\mathbb Q(\cos 2\pi/13)$. You might have an easier time first playing with $p =5$ and $p=7$. – peter a g May 30 '16 at 02:19
  • BTW the word Gaussian period shows up in this context - but it's not a bad idea to play at least a bit with the sums first before looking it up... You'll get your own way of understanding them. – peter a g May 30 '16 at 02:22

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The subfields of $\mathbb{Q}(\zeta_p)$ for $p$ prime correspond to the subgroups of its Galois group $U(p)\cong C_{p-1}$. For $p=13$ we have to find the subgroups of the cyclic group $C_{12}$. We know that for each divisor $d\mid 12$ there exists exactly one subgroup of order $d$, because $C_{12}$ is cyclic - see here.

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Dietrich Burde
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