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I would like to determine the Galois structure of the field $K=\Bbb Q(\zeta_{27})$--the rationals adjoined a primitive $27^{th}$ root of unity. That is to say I would like to determine the intermediate fields corresponding to the various subgroups of the Galois group, but don't know how to go about doing so. Here are my thoughts.

I know that if an extension is Galois then it obeys the fundamental theorem of Galois theory and as such the closed subgroups of the Galois group correspond to the intermediate fields of the field extension (Galois correspondence). I also know that these intermediate fields $E/\Bbb Q$ have a corresponding subgroup $\operatorname{Aut}(K/E)$ which is the set of automorphisms in $G=\operatorname{Gal}(E/\Bbb Q)$ which fix every element of $E$. Now somehow the degrees of the extensions are related to the orders of the subgroups and so to find the Galois structure of the field I need to find the various groups of the Galois group. I am not sure how to go about doing this.

I know that the number of primitive roots of unity for the $27^{th}$ order cyclotomic polynomial, $\Phi_{27}(x)$, are given by Euler's totient function. The basis for this analysis is the set

$$\left\{e^{2i\pi\frac{k}{27}} : k \in \mathbb{N}, \gcd(k,27) = 1\right\}.$$

A minimal set of such $k$ is $\{1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26\}$ Plugging these in and computing the polynomial I get

$$\Phi_{27}(x) = (x-e^{2i\pi\frac{1}{27}})(x-e^{2i\pi\frac{2}{27}})(x-e^{2i\pi\frac{4}{27}})(x-e^{2i\pi\frac{5}{27}})(x-e^{2i\pi\frac{7}{27}})(x-e^{2i\pi\frac{8}{27}})(x-e^{2i\pi\frac{10}{27}})(x-e^{2i\pi\frac{11}{27}})(x-e^{2i\pi\frac{13}{27}})(x-e^{2i\pi\frac{14}{27}})(x-e^{2i\pi\frac{16}{27}})(x-e^{2i\pi\frac{17}{27}})(x-e^{2i\pi\frac{19}{27}})(x-e^{2i\pi\frac{20}{27}})(x-e^{2i\pi\frac{22}{27}})(x-e^{2i\pi\frac{23}{27}})(x-e^{2i\pi\frac{25}{27}})(x-e^{2i\pi\frac{26}{27}}) = x^{18} + x^9 + 1$$ as $\Phi_n(x) = \prod_{\stackrel{1 \le k \le n}{\gcd(k,n)=1}} \left(x-e^{2i\pi\frac{k}{n}}\right)$

Now How do I find the "intermediate fields" so as to find the Galois groups? Do I just adjoin element by element to the base field, i.e. if we take the extension field $K$ to be the splitting field for a polynomial with coefficients in the base field $\Bbb Q$ we have that $K$ is a Galois extension and the set of automorphisms is denoted $\operatorname{Gal}(K/\Bbb Q)$. We end up with the splitting field $K = F(\alpha_1,\alpha_2,...\alpha_n)$ where each $\alpha_i$ is one of the primitive $n^{th}$ roots of unity, hence the splitting field is the base field adjoined with roots of the polynomial. Do I just adjoin them one by one i.e. $F(\alpha_1)(\alpha_2)...(\alpha_n)$?

Any thoughts is very appreciated,

Thanks,

Brian

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    $\begingroup$ Do you know what the Galois group, call it $G$, looks like? That should be the first thing given that by fundamental results of Galois theory the intermediate fields are in 1-1 correspondence with the subgroups of $G$. IOW: First find the group, then the intermediate fields. $\endgroup$ – Jyrki Lahtonen Mar 12 '15 at 19:48
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    $\begingroup$ And to find $G$: 1) Observe that an automorphism $\sigma\in G$ is fully determined, if you know $\sigma(e^{2\pi i/27})$. 2) What choices are there for $\sigma(e^{2\pi i/27})$? 3) What does the composition of two automorphisms that you found look like? Can you parametrize them somehow naturally so that composition is translated into the operation of another group you are more familiar with? $\endgroup$ – Jyrki Lahtonen Mar 12 '15 at 19:52
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    $\begingroup$ I hope that you’ve used the responses of @JyrkiLahtonen to find the Galois group and the intermediate fields. If not, I give an additional hint: for primes $p$ other than $2$, the Galois group of the field of $p^m$-th roots, over $\Bbb Q$, is always cyclic. So there will be precisely one intermediate field of each possible degree. $\endgroup$ – Lubin Mar 13 '15 at 18:21
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The easiest way to do this is to note that

$$G=\operatorname{Gal}\left(\Bbb Q(\zeta_{27})/\Bbb Q\right)\cong \left(\Bbb Z/27\Bbb Z\right)^*$$

is a cyclic group of order $\phi(27)=27-9=18$. Then cyclic groups have unique subgroups of every order dividing the group order, that means there are exactly $6$ subgroups of $G=\langle x\rangle $, one for each divisor of $18$. They are $\{1\}, G, \langle x^2\rangle, \langle x^3\rangle, \langle x^6\rangle, \langle x^9\rangle$.

Here we can exploit the uniqueness to classify the fields. I give something of a more ad-hoc approach for each, because this one can be done with a quick-and-dirty approach, for the most part, and I think it's sometimes good to get your hands a bit dirty to explore the techniques you can employ. For $\{1\}$ and $G$, this is easy, as always we get $K^{\{1\}}=K$ and $K^G=\Bbb Q$. So let us turn to the others. $\Bbb Q(\zeta_3)$ is clearly a subfield, and has degree $\phi(3)=2$, so it corresponds to $\langle x^9\rangle$ which is the unique index $2$ subgroup of $G$. To see $\Bbb Q(\zeta_9)=K^{\langle x^3\rangle}$ (which is another obvious subfield) we check that it has degree $\phi(9)=6$ immediately giving the result.

For the others, we don't have the same low-hanging fruit, but we can note that $K\cap\Bbb R$ is the maximal real sub-field and is exactly the field fixed by complex conjugation--this is always how you can determine the maximal real subfield--which we know to have order $2$, hence index $9$. Ergo this must correspond to the group $\langle x^2\rangle$.

Finally we have $K^{\langle x^3\rangle}$. For this one I give an argument which is a flavor of the general case and in the spirit of the adjunction approach from your original question. We recall that the identification $G\cong \left(\Bbb Z/27\Bbb Z\right)^*$ comes from the fact that we can define the map $\sigma_r:\zeta_{27}\mapsto \zeta_{27}^r$. You can check that $\sigma_{2}$ generates $G$ by noting that $2^9=512=(19)(27)-1\equiv -1\mod{27}$. How can we use this to write out our field explicitly? Well, since $\sigma_2$ is a generator, we can just say our $x=\sigma_2$ without any loss of generality. We note that $x^3=\sigma_{2^3}=\sigma_8$. So to make our field fixed by this, we should just adjoin all points in the orbit of $\langle x^3\rangle$ somehow. We compute that

$$\begin{cases} 8^1\equiv 8 \\ 8^2\equiv 10 \\ 8^3\equiv -1\equiv 26 \\ 8^3\equiv -1\cdot 10 \equiv 17 \\ 8^5\equiv -1\cdot 8\equiv 15 \end{cases}\mod 27.$$

to get all the possibilities. We then guess that perhaps $K^{\langle x^3\rangle}=\Bbb Q(\alpha)$ with

$$\alpha=\zeta_{27}^8+\zeta_{27}^{10}+\zeta_{27}^{15}+\zeta_{27}^{17}+\zeta_{27}^{26}=\zeta_{27}^8(1+\zeta_{27}^2+\zeta_{27}^7+\underbrace{\zeta_{27}^{9}+\zeta_{27}^{18}}_{\zeta_3^2+\zeta_3=-1})=\zeta_{27}^{10}+\zeta_{27}^{15}.$$

And how can we see this is the case? We just check that this number, $\alpha$ is not fixed by a subgroup of $\langle x\rangle$ containing $\langle x^3\rangle$. Of course, this is only possible if $\alpha$ is fixed by all of $G$, which we can tell if we can demonstrate that some automorphism, say $\sigma_2$ for simplicity, doesn't fix $\alpha$. But this is simple: recall any $18$ consecutive powers of $\zeta_{27}$ will serve as a $\Bbb Q$-basis for $K$ treated as a vector space. Using the defining relationship $\zeta_{27}^{18}=-\zeta_{27}^9-1$ we note that

$$\begin{cases} \alpha = \zeta_{27}^{10}+\zeta_{27}^{18-3}=\zeta_{27}^{10}-\zeta_{27}^6-\zeta_{27}^{-3} \\ \sigma_2(\alpha)=\zeta_{27}^3+\zeta_{27}^{18+2}=\zeta_{27}^3-\zeta_{27}^{11}-\zeta_{27}^2 \end{cases}$$

Choose the basis to be $\zeta_{27}^{-3},\ldots , \zeta_{27}^{14}$. Then $\alpha$ and $\sigma_2(\alpha)$ are easily seen to be linearly independent over $\Bbb Q$, hence unequal, finishing the problem.


Addendum If you cannot see how any $18$ consecutive powers work, it's not hard to see: by degree considerations $1,\zeta_{27},\ldots, \zeta_{27}^{17}$ are a basis, but then multiplication by $\zeta_{27}^k$ is a linear transformation on the field with inverse give by multiplication by $\zeta_{27}^{-k}$, hence it sends a basis to a basis.

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  • $\begingroup$ Hello Adam, first off, I sincerely appreciate the in depth exposition! This helps a lot. First question though: you said "note that $K \cap \mathbb{R}$ is the maximal real sub-field and is exactly the field fixed by complex conjugation. I don't see how this works, mostly how this is fixed by complex conjugation. I think I understand that the order is 2 as $a+bi \cdot a - bi = a^2 + b^2 \in \mathbb{R}$ if this is indeed why it is order 2. But don't understand how this plays a role in why $\langle x^2 \rangle$ is the corresponding group (although I do see that 18/2 = 9). $\endgroup$ – Relative0 Apr 13 '15 at 21:22
  • $\begingroup$ @Relative0 The group has order $18$, so that any subgroup of order $2$ (in this case there is only one since the group is cyclic) has index $9$ by Lagrange's theorem. By the FTGT, this corresponds to the unique sub-extension of degree $9$. Now, complex conjugation is trivially of order $2$, because when you apply it to a complex number twice, you get the identity. $\endgroup$ – Adam Hughes Apr 13 '15 at 22:09
  • $\begingroup$ concerning the complex conjugate being order of 2, $(a+bi)^{**} = 1 \cdot a+bi$ ( where $^*$ is the complex conjugate)? Also, I am curious to as why you chose, for example, $2^9=512=(19)(27)-1\equiv -1\mod{27}$ instead of $2^{18} \equiv 1 \mod{27}$ as the root of unity would be 1 mod 27 (although I am guessing two things, that this has something to do with order 2, but also we could do this and get $(8^3)^2, (8^4)^2, \text{ and } (8^5)^2$ and in the we get a similar cancellation that leaves us with the correct answer, is this right? $\endgroup$ – Relative0 Apr 13 '15 at 22:30
  • $\begingroup$ @Relative0 re: complex conjugation: yes. As for the $2$ thing, remember the order is the minimal number so that I get $1$, what if $2^9$ were $1$? Then writing $2^{18}$ would tell me only that the order divided $18$. Furthermore, I did it by-hand, so I stopped searching for the order at $9$ because it was then apparent the order of $2$ was $18$, there's no need to actually compute $2^{18}$ when I'd already computed $2^9$ and gotten $-1$. $\endgroup$ – Adam Hughes Apr 14 '15 at 1:57

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