Find the degree of extension $\Bbb Q(ζ_9 + ζ^{−1}_ 9 )$ over $\Bbb Q$

Question

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!

Answer 1

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 $2\pi/9$.

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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 $K$. $\qquad \blacksquare$

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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 $[\mathbb{Q}(\zeta_9 + \zeta_9^{-1}): \mathbb{Q}]$.

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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.

Answer 2

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.

Answer 3

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.