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I am trying to find the degree of extension $\Bbb Q(ζ_9 + ζ^{−1}_ 9 )$ over $\Bbb Q$. I know that for a prime number $p$, we always have $[\Bbb Q(ζ_p):\Bbb Q]=p-1$, so we can use the tower of extension to find the degree $\Bbb Q(ζ_p + ζ^{−1}_ p )$ over $\Bbb Q$. But now I am wondering how to deal with a composite number? Could someone please help? Thanks in advance!

NOTE:I have never learnt about Galois theory so far. So please avoid the usage of the Galois theory. But I have learnt about the tower of extension and the minimal polynomial. Thanks!

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  • $\begingroup$ It's a shame you can't use Galois theory yet, since it would make this problem rather simple. For your future benefit, I think it's worth noting that it gives you a tool with which you can find the degree of the minimal polynomial of that element (and thus the degree of the extension in question). In particular, if you find the total number of distinct elements in the orbit of that element when acted upon by the Galois group for $\mathbb{Q}(\zeta_9)$ over $\mathbb{Q}$, that number is equal to the degree of the minimal polynomial (since the elements in the orbit are its roots). $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 4:54

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In greater generality to what you've written, we have $[ \mathbb{Q}( \zeta_n) : \mathbb{Q}] = \varphi(n)$, where $\varphi$ is the totient function. This is because the minimal polynomial is going to be $\Phi_n(x) = \displaystyle \prod_{ \ \ \ 1 \leq k \leq n \\ \gcd(k, n) = 1} \left( x - \zeta_n^k \right)$. This is a difficult thing to prove in general, but we are in luck: the special case where $n$ is a power of a prime is proven in Problem 3 here.

To begin, we can consider a tower of fields $\mathbb{Q} \subset \mathbb{Q}( \zeta_9 + \zeta_9^{-1} ) \subset \mathbb{Q}( \zeta_9)$, and since $[ \mathbb{Q}(\zeta_9): \mathbb{Q}] = \varphi(9) = 6$, the multiplicativity formula implies that $[\mathbb{Q}( \zeta_9 + \zeta_9^{-1}):\mathbb{Q}] = 2, 3$, or $6$.

You can check that we have $\zeta_9^{-1} = \zeta_9^8$. Computing this or looking at it in the complex plane, you'll find that $\zeta_9 + \zeta_9^{-1}$ has no imaginary part, implying that $\mathbb{Q}(\zeta_9 + \zeta_9^{-1}) \subset \mathbb{Q}(\zeta_9)$ is a strict inclusion. Thus, we can eliminate $6$.

From Euler's formula, you'll notice further that $\zeta_9 + \zeta_9^{-1} = 2\cos(2 \pi/9)$, which reveals that $\mathbb{Q}( \zeta_9 + \zeta_9^{-1}) = \mathbb{Q}( \cos(2 \pi/9))$. Using our knowledge of trig identities and the fact that cosine evaluates to rational numbers at some integer multiples of $\pi/9$, maybe we can find the minimal polynomial of this thing? Indeed! After some thought, I've found that one can use De Moivre's theorem to prove that $\cos(3 \theta) = 4 \cos^3(\theta) - 3 \cos(\theta)$, which yields a rational number when evaluated at $2\pi/9$.




Alternate method for future reference once you've been exposed to some Galois theory:

Lemma: If $K/F$ is a Galois extension, then the minimal polynomial for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$. That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.

Proof: This is a consequence of the fact that a separable polynomial is irreducible $\iff$ its Galois group acts transitively on its roots. For a proof of this fact, see Theorem 2.9(b) here. Notice that $\displaystyle \min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$. This latter fact is because, if $L \subseteq K$ is the splitting field of $\displaystyle \min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$. $\qquad \blacksquare$


Note that $\mathbb{Q}( \zeta_9)/\mathbb{Q}$ is a Galois extension (finite extensions of $\mathbb{Q}$ are separable and this is the splitting field for $f(x) = x^9 - 1$). First, compute the Galois group of this field. Next, find the number of elements in the orbit of $\zeta_9 + \zeta_9^{-1}$ under the action of $ \text{Gal}( \mathbb{Q}(\zeta_9)/ \mathbb{Q})$; this will be the degree of the minimal polynomial of $\zeta_9 + \zeta_9^{-1}$ per the above lemma. This, of course, will be equal to $[\mathbb{Q}(\zeta_9 + \zeta_9^{-1}): \mathbb{Q}]$.


If you have access to Artin's Algebra (second edition), there is (if I recall correctly) a paragraph or two in the Galois theory chapter discussing the above technique of using Galois groups to find minimal polynomials. It should fill in any missing details or context.

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  • $\begingroup$ Thanks! May I please ask how can we know that the degree of $\Bbb Q(\zeta_9)$ is $6$? Is it possible to explicitly find its minimal polynomial? $\endgroup$
    – Y.X.
    Apr 12, 2017 at 5:41
  • $\begingroup$ @PropositionX, I am actually in the process of thinking about that right now with the intention of editing this post. You can check out the wiki article on "cyclotomic polynomials" for a starting point. Unfortunately, the only way I know how to do this at present is via Galois theory, but I'm sure there are other routes. $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 5:42
  • $\begingroup$ @PropositionX, "The fact that $\Phi_n(x)$ is irreducible of degree $\phi(n)$ is a nontrivial result due to Gauss". Quite discouraging; it might be worth making a new post to discuss this. In particular, there might be tricks for cracking the special case of $n=p^2$ for a prime $p$ if the general case is as difficult as it sounds. $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 5:48
  • $\begingroup$ Ah-ha! @PropositionX, some zealous Googling did actually reveal a trick for finding $\Phi_k(x)$ when $k = p^n$ for $p$ prime. Check out problem $3$ here: math.ucla.edu/~iacoley/hw/alghwwinter/HW%206.pdf $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 6:29
  • $\begingroup$ Thanks for patience! I would have a look. $\endgroup$
    – Y.X.
    Apr 12, 2017 at 6:41
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Let me simply write $\zeta$. Consider $K=\mathbf{Q}[\zeta +\bar\zeta]$. I'll prove that $\zeta$ is a root of quadratic polynomial over $K$ hence of degree 2 over $K$. Then by tower theorem the degree of $K$ over the rationals follows.

$f(X) = X^2 -(\zeta +\bar \zeta)X + 1$ (note that $\zeta\bar\zeta=1$).

This polynomial has coefficients in $K$ and has $\zeta$ and $\bar\zeta$ as its roots. Note that elements of $K$ are all real and so $\zeta$ is not in $K$. SO its degree over $K$ is 2.

No Galois theory is needed.

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    $\begingroup$ It also does not exactly answer the question, which is to compute the degree of $K$ over over $\mathbb{Q}$, not the degree of $\mathbb{Q}(\zeta_{9})$ over $K$. Of course, one can obtain the former if one knows the degree of $\mathbb{Q}(\zeta_{9})$ over $\mathbb{Q}$, but then one is back to the conundrum in your post @KajHansen. $\endgroup$ Apr 12, 2017 at 6:01
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    $\begingroup$ @KajHansen: of course this solution is useful to calculate degree of $\mathbf{Q}[\zeta+\bar\zeta]$ only when the degree of $\mathbf{Q}[\zeta]$ is known. That has to go through Gauss. $\endgroup$ Apr 12, 2017 at 6:02
  • $\begingroup$ Oh crap, you are very right @Alex. I got ahead of myself and was taking that little piece of information for granted, not thinking before commenting. :S At the very least, this avoids some of the messy search for trig identities from the middle of my post. $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 6:03
  • $\begingroup$ Ah-ha! @AlexWertheim, some zealous Googling did actually reveal a trick for finding $\Phi_k(x)$ when $k = p^n$ for $p$ prime. Check out problem $3$ here: math.ucla.edu/~iacoley/hw/alghwwinter/HW%206.pdf $\endgroup$
    – Kaj Hansen
    Apr 12, 2017 at 6:28
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    $\begingroup$ @KajHansen: heh, nice! Amusingly enough, I also solved this problem (in a very similar way to Ian), but had since forgotten: math.ucla.edu/~awertheim/Math%20210B/Homework/Math210BHW6.pdf $\endgroup$ Apr 12, 2017 at 6:44
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This is a very easy problem, if you just eschew fancy techniques and pull out a pencil. First, a primitive ninth root of unity is just a ninth root of $1$ that isn’t a cube root of $1$. Therefore it is a root of $$ \frac{X^9-1}{X^3-1}=X^6+X^3+1\,. $$ This is the ninth cyclotomic polynomial, $\Phi_9(X)$. You show it’s irreducible by calculating $\Phi_9(X+1)=X^6+6X^5+15X^5+21X^3+18X^2+9X+3$, and invoking Eisenstein.

Finding the minimal polynomial of $\xi=\zeta+\zeta^{-1}$ is not hard by any method, but I like writing \begin{align} 0&=\zeta^3+1+\zeta^{-3}\,\quad\text{and then using this in:}\\ \xi^3&=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}\\ &=3\zeta-1+3\zeta^{-1}\,\quad\text{giving}:\\ \xi^3-3\xi+1&=0\,. \end{align} Now you show that $\Psi(X)=X^3-3X+1$ is irreducible by calculating $\Psi(X+2)=X^3+6X^2+9X+3$, and again invoke Eisenstein, so that the minimal polynomial for $\xi=\zeta_9+\zeta_9^{-1}$ is $X^3-3X+1$, which says that $[\Bbb Q(\xi):\Bbb Q]=3$.

Please notice that there was no fancy mathematics and no machine-aided computation used, just pencil lead.

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