Let F subset of $\mathbb{C}$ be the splitting field of $X^7-2$ over $\mathbb{Q}$, and $z=e^{\frac{2πi}{7}}$, a primitive seventh root of unity.

Question

Let $F$ subset of $\mathbb{C}$ be the splitting field of $X^7-2$ over $\mathbb{Q}$, and $z=e^{\frac{2πi}{7}}$, a primitive seventh root of unity.

Let $[F:\mathbb{Q}(z)]=a$ and $[F:\mathbb{Q}(2^{1/7})]=b$.Then

$a=b=7$.
$a=b=6$.
$a>b$.
$a< b$.

I got $b=7$ but how to calculate $a$ ?

More to the point: do you know what $;\Bbb Q(z);$ is within your question's context? — DonAntonio, Jun 04 '14 at 12:48

score 4 · Answer 1 · answered Jun 04 '14 at 13:48

Let us define $\zeta_7 := e^{\frac{2\pi i}{7}}$, to make the notation simpler.

First we claim that $F = \mathbb{Q}(\zeta_7, \sqrt[7]{2})$. To see why this is true, we first have that the roots of $x^7-2$ are $\sqrt[7]{2}, \zeta_7 \sqrt[7]{2}, \zeta_7^2 \sqrt[7]{2}, \ldots , \zeta_7^6 \sqrt[7]{2}$; certainly these satisfy the polynomial and there are 7 of them, so this must be all of them. Then we can see that $\mathbb{Q}(\zeta_7, \sqrt[7]{2}) \subset F$ because $\sqrt[7]{2}\in F$ and $\zeta_7 = \frac{\zeta_7 \sqrt[7]{2}}{\sqrt[7]{2}} \in F$. Also $F\subset \mathbb{Q}(\zeta_7, \sqrt[7]{2})$ because all of the roots of $x^7-2$ are contained in $\mathbb{Q}(\zeta_7, \sqrt[7]{2})$. Hence we have $F = \mathbb{Q}(\zeta_7, \sqrt[7]{2})$.

Now we consider the degree $[\mathbb{Q}(\zeta_7, \sqrt[7]{2}):\mathbb{Q}]$. There are two obvious intermediate subfields:

$\mathbb{Q}\subset \mathbb{Q}(\sqrt[7]{2})\subset \mathbb{Q}(\zeta_7, \sqrt[7]{2})$, the degree of the first extension is $[\mathbb{Q}(\sqrt[7]{2}):\mathbb{Q}]=7$ because the minimal polynomial of $\sqrt[7]{2}$ is $x^7-2$.

$\mathbb{Q}\subset \mathbb{Q}(\zeta_7)\subset \mathbb{Q}(\zeta_7, \sqrt[7]{2})$, the degree of the first extension is $[\mathbb{Q}(\zeta_7):\mathbb{Q}] = 6$ because the minimal polynomial of $\zeta_7$ is the 7-th cyclotomic polynomial, which has degree 6.

This tells us that $[\mathbb{Q}(\zeta_7, \sqrt[7]{2}):\mathbb{Q}]$ is divisible by both 6 and 7, which are coprime, hence it is divisible by 42, hence we have $[\mathbb{Q}(\zeta_7, \sqrt[7]{2}):\mathbb{Q}] = 42$. Now we can answer the original question: $$[F:\mathbb{Q}(\zeta_7)] = \frac{[\mathbb{Q}(\zeta_7, \sqrt[7]{2}):\mathbb{Q}]}{[\mathbb{Q}(\zeta_7):\mathbb{Q}]} = \frac{42}{6} = 7$$ And $$[F:\mathbb{Q}(\sqrt[7]{2})] = \frac{[\mathbb{Q}(\zeta_7, \sqrt[7]{2}):\mathbb{Q}]}{[\mathbb{Q}(\sqrt[7]{2}):\mathbb{Q}]} = \frac{42}{7} = 6$$

Let F subset of $\mathbb{C}$ be the splitting field of $X^7-2$ over $\mathbb{Q}$, and $z=e^{\frac{2πi}{7}}$, a primitive seventh root of unity.

1 Answers1