Consider the field $\mathbb{Q}(\omega_3, \omega_7)$ with $\omega_n = e^{2 \pi i/n} = cos(\frac{2 \pi}{n}) + i sin(\frac{2 \pi}{n})$ the cyclotomic root of unity.

According to the theorem of the primitive element, if $\mathbb{F}$ is a field with characteristic 0 and there are 2 elements $a,b$ that are algebraic over the field $\mathbb{F}$, then there exists an $\alpha$ such that: $\mathbb{F}(a,b) = \mathbb{F}(\alpha)$.

a) Find a primitive element $\alpha$ for $\mathbb{Q}(\omega_3, \omega_7)$.

b) Calculate $[\mathbb{Q}(\omega_3, \omega_7): \mathbb{Q}]$.

c) What is the minimal polynomial of $\omega_7$ over $\mathbb{Q}(\omega_3)$.

For a) i tried to use a similar method as Primitive element of the extension $\mathbb Q(\sqrt{2},\sqrt{3})$ over $\mathbb Q$ . In other words I am trying to find an element $\alpha$ so that with simple calculations with $\alpha$, I can get $\omega_3$ and $\omega_7$ seperately.

$\alpha = \omega_3 + \omega_7$ hasn't worked for me.

$\alpha = \omega_3 \cdot \omega_7 = e^{\frac{10}{21}(2 \pi i)}$ looks like a good idea since then $\alpha^{21/70} = \omega_7$ and $\alpha^{21/30} = \omega_3$ but I'm not a 100% sure if this is correct. Since using this alpha, for any $n \in \mathbb{N}_0$, it holds that $\alpha^{21/n*10} = \omega_n$ so my instinct says that the field $\mathbb{Q}(\omega_3, \omega_7)$ is contained in $\mathbb{Q}(\alpha)$ but not that they are equal.

So now I think maybe the 'simple calculation' has to be a linear combination for them to be equal, but I don't know what would work.

Can someone help me further finding $\alpha$.

For b) I want to use the fact that 3 and 7 are prime, I can use the product rule and the fact that $X^n-1$ is the minimal polynomial of $\mathbb{Q}$, and get that both 3 and 7 are divisors. Which gives me that it has to be 21.

For c) I think that $X^7-1$ will hold since that it is the minimal polynomial of $\omega_7$ over $\mathbb{Q}$ and since gcd(7,3) = 1, that means that the basis $\{ 1, \omega_3, \omega_3^2\}$ doesn't contain $\omega_7$.

Is my reasoning in b and c correct?


1 Answer 1


For a): $\alpha=\omega_3\omega_7$ works as a primitive element but not for the reason you mentioned as we cannot take fractional powers in fields in general. We can argue as follows: Let $\beta=\alpha^{-2}=\exp(2\pi i/21)$. Then have $\omega_3=\beta^7,\omega_7=\beta^3$ and we get $\Bbb Q(\omega_3,\omega_7)\subseteq\Bbb Q(\alpha)$ and hence equality. Basically the reason here is that $10$ and $21$ are coprime, so $\alpha$ is a primitive $21$-th root of unity.
For b): This is not correct. The minimal polynomial of $\omega_3$ over $\Bbb Q$ is $x^2+x+1$, so $[\Bbb Q(\omega_3):\Bbb Q]=2$ and similarly for $\omega_7$. Since $\alpha$ is a primitive $21$-th root of unity we get $[\Bbb Q(\alpha):\Bbb Q]=\varphi(21)=12$ (this is not that trivial, see for example Lang's Algebra, chapter VI, 3 on 'Roots of unity').
For c): It is correct that the minimal polynomial of $\omega_7$ over $\Bbb Q$ (which is $x^6+x^5+\dots+x+1$) is the same as the minimal polynomial over $\Bbb Q(\omega_3)$. This follows from the fact that $\Bbb Q(\omega_3)$ and $\Bbb Q(\omega_7)$ are linear disjoint over $\Bbb Q$ (since $3$ and $7$ are coprime). We can see this for example by looking at the degree $$[\Bbb Q(\omega_3):\Bbb Q][\Bbb Q(\omega_7):\Bbb Q]=2\cdot 6=12=[\Bbb Q(\beta):\Bbb Q]=[\Bbb Q(\omega_3,\omega_7):\Bbb Q]$$

  • $\begingroup$ Wow, thank you for your detailed answer. It was very clear and now I understand it properly :) $\endgroup$
    – fieke_2000
    Jan 23, 2021 at 14:34

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