# Finding primitive element for $\mathbb{Q}(\omega_3, \omega_7)$ and calculating a minimal polynomial.

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?

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 $$[\Bbb Q(\omega_3):\Bbb Q][\Bbb Q(\omega_7):\Bbb Q]=2\cdot 6=12=[\Bbb Q(\beta):\Bbb Q]=[\Bbb Q(\omega_3,\omega_7):\Bbb Q]$$