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In a nutshell, my question is: what is degree of the field extension $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) $ over $\mathbb{Q}$?

As to why I'm asking this, I was trying to find the subfields of $\mathbb{Q}\, (\zeta_{14})$, the cyclotomic field of order $14$, and I found that the subfield $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) \subseteq \mathbb{Q}\, (\zeta_{14})$ should have degree $2$ over $\mathbb{Q}$.

This is because it is the fixed field of the following subgroup of automorphisms $$\{1,9,11 \} \subset (\mathbb{Z}/14\mathbb{Z})^{\times} \cong \text{Gal} \, \mathbb{Q}\, (\zeta_{14}) / \mathbb{Q}, $$ and since this subgroup has index $2$, the corresponding fixed field must have degree $2$ over $\mathbb{Q}$ by Galois Theory.

However, when I put the sum $\zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}$ in Wolfram alpha, I get the following expansion: $$ \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11} = \sqrt[-7]{-1} + (-1)^{2/7} - (-1)^{4/7},$$ which doesn't look like the root of a quadratic polynomial! Futhermore, I also get that this number is a root of the sixth degree polynomial $x^6 - x^5 + x^4 - 15x^3 + 22x^2 - 8x + 8$. This suggests that this number isn't the root of a quadratic polynomial, which goes against the fact that the extension $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) $ should be quadratic by the Galois theory reasoning I gave above.

My question is: am I right to assume that the extension above is a degree $2$ extension of $\mathbb{Q}$, by Galois theory? And if I am right, why does this element of a quadratic extension seem to have degree $>2$ over the rationals?

Adithya Chakravarthy
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  • That number is a root of $x^2-x+2$ – J. W. Tanner Apr 21 '21 at 16:51
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    Did you try computing the minimal polynomial of $\zeta_{14}+\zeta_{14}^{9} + \zeta_{14}^{11} $ directly? – Dionel Jaime Apr 21 '21 at 17:12
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    In case this simplifies matters: we have $\mathbb Q(\zeta_{14}) = \mathbb Q(\zeta_7)$ by identifying $\zeta_7 = \zeta_{14}^2$, $-1 = \zeta_{14}^7$, and thus $\zeta_{14} = -\zeta_7^4$. Your element can be rewritten as $-(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. – Ravi Fernando Apr 21 '21 at 17:47
  • @RaviFernando That simplifies things significantly. Since the degree of the minimal polynomial is $2$, and complex roots come in complex conjugate pairs, all we have to do is find $(x-\alpha)(x-\bar{\alpha})$. – Dionel Jaime Aug 11 '21 at 03:41

2 Answers2

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I'm not sure what went wrong when you plugged it into Wolfram Alpha, but you could try and compute the minimal polynomial (which is indeed quadratic by your own reasoning) of $\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11}$ directly. Upon squaring, you should find that

$$(\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11} )^2 = (\zeta_{14}^2+ \zeta_{14}^4+\zeta_{14}^{6} + \zeta_{14}^{8} + \zeta_{14}^{10}+\zeta_{14}^{12}) + (\zeta_{14}^{6} +\zeta_{14}^{10}+\zeta_{14}^{12}) $$

Here I grouped the terms that are the $7^{th}$ roots of unity. I can now exploit the fact that the sum of the $n^{th}$ roots of unity is $0$ and obtain that the above sum is equal to $(-1) + \zeta_{14}^{6} + \zeta_{14}^{10} + \zeta_{14}^{12}$. You can draw these roots of unity on the unit circle to convince yourself that

$$\zeta_{14}^{6} + \zeta_{14}^{10} + \zeta_{14}^{12} = -(\zeta_{14}^{13} + \zeta_{14}^{3} + \zeta_{14}^{5} ) . $$

Notice that the $3$ primitive $14^{th}$ roots of unity that are NOT the ones you've given appear. We can use this to exploit the fact that the sum of $14^{th}$ roots of unity is $0$. If we set $\alpha = \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}$, we obtain that

$$ \underbrace{(-1) - (\zeta_{14}^{13} + \zeta_{14}^{3} + \zeta_{14}^{5})}_{\alpha^2} -\underbrace{(\zeta_{14}+\zeta_{14}^9 + \zeta_{14}^{11})}_{\alpha} = (\zeta_{14}^2+ \zeta_{14}^4+\zeta_{14}^{6} + \zeta_{14}^{8} + \zeta_{14}^{10}+\zeta_{14}^{12}) + \zeta_{14}^7 = -2$$

We've concluded that $\alpha^2 - \alpha = -2 \ $ so we have the minimal polynomial being $x^2 -x +2.$

Dionel Jaime
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    I think it's $x^2-x+\color{red}2$ – J. W. Tanner Apr 21 '21 at 17:47
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    The error is in the last long equation: if the $-1$ on the left is interpreted as $-\zeta_{14}^0$, then you're missing a $\zeta_{14}^7 = -1$ on the right-hand side. – Ravi Fernando Apr 21 '21 at 18:05
  • It might be a slightly more robust path to compute trace and norm, since we do know a-priori that the thing is quadratic (at worst) over $\mathbb Q$. – paul garrett Apr 21 '21 at 19:11
  • @Paul There are days where I just want to be a caveman. But yes, I agree that might be swifter. – Dionel Jaime Apr 21 '21 at 19:21
  • @DionelJaime, :) On another hand, sometimes it's better to just be slightly inefficient ... but finished... rather than still fretting about optimization. :) – paul garrett Apr 21 '21 at 19:28
  • Thanks for the post! This was very helpful. I didn't really have the guts to do the algebra with my bare hands, but I guess that's the cleanest way to go about finding the minimal polynomial. Appreciate the help :) – Adithya Chakravarthy Apr 21 '21 at 19:54
  • @paulgarrett I'm curious about your comment about trace and norm. How can computing the trace and norm of $\zeta + \zeta^9 + \zeta^{11}$ tell us the degree of its minimal polynomial? And also, to find the trace / norm, wouldn't we have to know all the Galois conjugates of $\zeta + \zeta^9 + \zeta^{11}$, which we could only find if we know the minimal polynomial to begin with? – Adithya Chakravarthy Apr 21 '21 at 19:58
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    @chaad, if you already know that the thing $\alpha$ is at most quadratic over \Q, and in fact know the only non-trivial Galois element $\tau$ that might move it, then you know its Galois trace from that quadratic extension is $\alpha+\tau\alpha$, and its Galois norm is $\alpha\cdot \tau\alpha$. My point was that there's no necessity of looking around for relations between $\alpha$ and $\alpha^2$. – paul garrett Apr 21 '21 at 20:30
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It is not generally true that if you add up the values of a subgroup $H$ of the Galois group on a generating element for the whole field that you'll get an element generating the subfield fixed by $H$. For example, ${\rm Gal}(\mathbf Q(\zeta_8)/\mathbf Q) \cong (\mathbf Z/8\mathbf Z)^\times$ and the subfield fixed by $H = \{1,5 \bmod 8\}$ is $\mathbf Q(i)$, but if you apply $H$ to $\zeta_8$ and add the results you get $\zeta_8 + \zeta_8^5$, which is $0$, so $\zeta_8 + \zeta_8^5$ is contained in $\mathbf Q(i)$ but that sum does not generate $\mathbf Q(i)$ over $\mathbf Q$. The fact that you did not run into such a problem using $\zeta_{14}$ depends on some accidental choices in your "random" example.

When $H$ is the subgroup of order $3$ in ${\rm Gal}(\mathbf Q(\zeta_{14})/\mathbf Q)$, summing the elements in the $H$-orbit of $\zeta_{14}$ gives you a generator of the subfield of $\mathbf Q(\zeta_{14})$ with codimension $3$ (equivalently, of dimension $6/3 = 2$ over the base field $\mathbf Q$) because (i) a sum over an $H$-orbit generates the subfield fixed by $H$ when you sum over the $H$-orbit of an element in a normal basis, as explained in Theorem 3.8 here, and (ii) the primitive $n$th roots of unity form a normal basis for $\mathbf Q(\zeta_n)/\mathbf Q$ when $n$ is squarefree (e.g., $n = 14$), as explained in my answer on the MSE page here.

KCd
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    It’s certainly true that the sum of the conjugates isn’t always a generator of the appropriate fixed field, but it’s surprising how often this strategy works. Next, try the product. If that doesn’t work either, then flounder and thrash about. – Lubin May 21 '21 at 02:04
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    @Lubin I agree that very often the sum will work in practice. Pete Clark's answer to the question https://mathoverflow.net/questions/26832/degree-of-sum-of-algebraic-numbers gives a setting where a sum of two elements generates a field extension without needing to use a normal basis or even a Galois extension. – KCd May 21 '21 at 02:09