MHB Field extensions and degree of odd prime numbers

[mathmari]

Hey!

Let $p$ be an odd prime number. We set $a=Re \left (e^{\frac{2\pi i}{p}}\right )$.
Show that:

1) $$\mathbb{Q}(a)\leq \mathbb{Q}(e^{\frac{2\pi i}{p}}) \text{ and } \left [\mathbb{Q} \left (e^{\frac{2\pi i}{p}} \right ):\mathbb{Q}(a)\right ]=2$$
2) $$[\mathbb{Q}(a):\mathbb{Q}]=\frac{p-1}{2}$$

1) To show that $ \mathbb{Q}(a)\leq \mathbb{Q} \left ( e^{\frac{2\pi i}{p}}\right ) $ we have to show that $a \in \mathbb{Q}(a) \Rightarrow a \in \mathbb{Q} \left (e^{\frac{2\pi i}{p}}\right ) $, right ??

$a = Re \left (e^{\frac{2\pi i}{p}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+\overline{e^{\frac{2\pi i}{p}}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+e^{\frac{-2\pi i}{p}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+\left( e^{\frac{2\pi i}{p}}\right )^{-1} \right ) \in \mathbb{Q} \left (e^{\frac{2\pi i}{p}}\right )$

Is this correct?? (Wondering)To find $ \left [\mathbb{Q}\left (e^{\frac{2\pi i}{p}}\right ):\mathbb{Q}(a)\right ]$ we have to find the degree of $Irr\left (e^{\frac{2 \pi I}{p}},\mathbb{Q}(a)\right )$, right?
But how could we do that?? (Wondering)2) How could we find $Irr(a,\mathbb{Q})$ ?? (Wondering)

 

[Euge]

1) To show that $ \mathbb{Q}(a)\leq \mathbb{Q} \left ( e^{\frac{2\pi i}{p}}\right ) $ we have to show that $a \in \mathbb{Q}(a) \Rightarrow a \in \mathbb{Q} \left (e^{\frac{2\pi i}{p}}\right ) $, right ??

$a = Re \left (e^{\frac{2\pi i}{p}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+\overline{e^{\frac{2\pi i}{p}}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+e^{\frac{-2\pi i}{p}}\right )=\frac{1}{2} \left ( e^{\frac{2\pi i}{p}}+\left( e^{\frac{2\pi i}{p}}\right )^{-1} \right ) \in \mathbb{Q} \left (e^{\frac{2\pi i}{p}}\right )$

Is this correct?? (Wondering)

It's right, but you don't need to say $a\in \Bbb Q(a) \Rightarrow a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$. You're simply trying to show that $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$.

To find $ \left [\mathbb{Q}\left (e^{\frac{2\pi i}{p}}\right ):\mathbb{Q}(a)\right ]$ we have to find the degree of $Irr\left (e^{\frac{2 \pi I}{p}},\mathbb{Q}(a)\right )$, right?
But how could we do that?? (Wondering)

Yes. Let $z = e^{\frac{2\pi i}{p}}$. Since $a = \frac{z + z^{-1}}{2}$, multiplying both sides by $2z$ yields $2az = z^2 + 1$, or $z^2 - 2az + 1 = 0$. Show that $x^2 - 2ax + 1 \in \Bbb Q(a)[x]$ is $\text{Irr}(z: \Bbb Q(a))$.

2) How could we find $Irr(a,\mathbb{Q})$ ?? (Wondering)

There's no need. Since $\Bbb Q \le \Bbb Q(a) \le Q(z)$, we have

$$[\Bbb Q(a) : \Bbb Q] = \frac{[\Bbb Q(z) : \Bbb Q]}{[\Bbb Q(z) : \Bbb Q(a)]}.$$

Since $[\Bbb Q(z) : \Bbb Q] = p - 1$ and $[\Bbb Q(z) : \Bbb Q(a)] = 2$, it follows that $[\Bbb Q(a) : \Bbb Q] = \frac{p-1}{2}$.

 

[mathmari]

It's right, but you don't need to say $a\in \Bbb Q(a) \Rightarrow a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$. You're simply trying to show that $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$.

Does it stand that $\mathbb{Q}(a) \leq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ when we show that $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$, because $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$ means that $\mathbb{Q}(a) \subseteq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ and since $\mathbb{Q}(a)$ and $\mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ are fields, we have that $\mathbb{Q}(a) \leq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ ?? (Wondering)

 

[Euge]

Does it stand that $\mathbb{Q}(a) \leq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ when we show that $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$, because $a\in \Bbb Q\left(e^{\frac{2\pi i}{p}}\right)$ means that $\mathbb{Q}(a) \subseteq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ and since $\mathbb{Q}(a)$ and $\mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ are fields, we have that $\mathbb{Q}(a) \leq \mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ ?? (Wondering)

Yes :D

 

[mathmari]

Yes :D

Ok! Do we have to prove that $\mathbb{Q}(a)$ and $\mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ are fields?? (Wondering)

 

[Euge]

Ok! Do we have to prove that $\mathbb{Q}(a)$ and $\mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ are fields?? (Wondering)

I would hope not, since by definition, $\Bbb Q(a)$ is the smallest field containing $\Bbb Q$ and $a$, and similarly for $\Bbb Q(e^{\frac{2\pi i}{p}})$.

 

[mathbalarka]

Ok! Do we have to prove that $\mathbb{Q}(a)$ and $\mathbb{Q}\left(e^{\frac{2\pi i}{p}}\right)$ are fields?? (Wondering)

$a$ and $\exp(2\pi i/p)$ are algebraics over $\Bbb Q$ and field while adjoined with algebraic are...?