# Tuesday October 15th

## Cyclotomic Extensions

**Definition:**
Let $K$ denote the splitting field of $x^n-1$ over $F$.
Then $K$ is called the **$n$th cyclotomic extension of $F$**.

If we set $f(x) = x^n-1$, then $f'(x) = nx^{n-1}$.

So if $\ch F$ does not divide $n$, then the splitting field is separable.
So this splitting field is in fact normal.

Suppose that $\ch F$ doesn't divide $n$, then $f(x)$ has $n$ zeros, and let $\zeta_1, \zeta_2$ be two zeros.
Then $(\zeta_1 \zeta_2)^n = \zeta_1^n \zeta_2^n = 1$, so the product is a zero as well, and the roots of $f$ form a subgroup in $K\units$.

So let's specialize to $F = \QQ$.

The roots of $f$ are the $n$th roots of unity, i.e. $\zeta_n = e^{2\pi i / n}$, and are given by $\theset{\zeta_n, \zeta_n^2, \zeta_n^3, \cdots, \zeta_n^{n-1}}$.

The *primitive* roots of unity are given by $\theset{\zeta_n^m \suchthat \gcd(m, n) = 1}$.

**Definition:**
Let
$$
\Phi_n(x) = \prod_{i=1}^{\varphi(n)} (x-\alpha_i)
,$$
where this product runs over all of the primitive $n$th roots of unity.

Let $G$ be $\Gal(K/\QQ)$.
Then any $\sigma\in G$ will permute the primitive $n$th roots of unity.
Moreover, it *only* permutes primitive roots, so every $\sigma$ fixes $\Phi_n(x)$.
But this means that the coefficients must lie in $\QQ$.

Since $\zeta$ generates all of the roots of $\Phi_n$, we in fact have $K = \QQ(\zeta)$.
But what is the group structure of $G$?

Since any automorphism is determined by where it sends a generator, we have automorphisms $\tau_m(\zeta) = \zeta^m$ for each $m$ such that $\gcd(m, n) = 1$.

But then $\tau_{m_1} \circ \tau_{m_2} = \tau_{m_1 + m_2}$, and so $G \cong G_m \leq \ZZ_n$ as a ring, where

$$
G_m = \theset{[m] \suchthat \gcd(m, n) = 1}
$$
and $\abs G = \varphi(n)$.

> Note that as a *set*, there are the units $\ZZ_n\units$.

**Theorem:**
The Galois group of the $n$th cyclotomic extension over $\QQ$ has $\varphi(n)$ elements and is isomorphic to $G_m$.

**Special case**:
$n=p$ where $p$ is a prime.

Then $\phi(p) = p-1$, and
$$
\Phi_p(x) = \frac{x^p - 1}{x-1} = x^{p-1} + x^{p-2} + \cdots + x + 1
.$$

Note that $\ZZ_p\units$ is in fact cyclic, although this may not always happen.
In this case, we have $\Gal(K/\QQ) \cong \ZZ_p\units$.

## Construction of n-gons

To construct the vertices of an n-gon, we will need to construct the angle $2\pi/n$, or equivalently, $\zeta_n$.
Note that if $[\QQ(\zeta_n) : \QQ] \neq 2^\ell$ for some $\ell\in\NN$, then the $n\dash$gon is *not* constructible.

*Example:*
An 11-gon. Noting that $[\QQ(\zeta_{11}) : \QQ] = 10 \neq 2^\ell$, the 11-gon is not constructible.

Since this is only a sufficient condition, we'll refine this.

**Definition:**
A prime of the form $p = 2^{2^k}+1$ are called **Fermat primes**.

**Theorem:**
The regular $n\dash$gon is constructible $\iff$ all odd primes dividing $n$ are *Fermat primes* $p$ where $p^2$ does not divide $n$.

*Example:*
Consider
$$
\Phi_5(x) = x^4 + x^3 + x^2 + x + 1
.$$

Then take $\zeta = \zeta_5$; we then obtain the roots as $\theset{1, \zeta, \zeta^2, \zeta^3, \zeta^4}$ and $\QQ(\zeta)$ is the splitting field.

Any automorphism is of the form $\sigma_r: \zeta \mapsto \zeta^r$ for $r=1,2,3,4$.
So $\abs{\Gal(K/\QQ)} = 4$, and is cyclic and thus isomorphic to $\ZZ_4$.
Corresponding to $0 \to \ZZ_2 \to \ZZ_4$, we have the extensions
$$
\QQ \to \QQ(\zeta^2) \to \QQ(\zeta)
.$$

How can we get a basis for the degree 2 extension $\QQ(\zeta^2)/\QQ$?
Let
$$
\lambda(E) = \theset{\sigma \in \Gal(\QQ(\zeta)/\QQ) \suchthat \sigma(e) = e ~\forall e\in E }
,$$
$\lambda(K_H) = H$ where $H$ is a subgroup of $\Gal(\QQ(\zeta)/\QQ)$, and
$$
K_H = \theset{x\in K \suchthat \sigma(x) = x ~\forall \sigma\in H}
.$$

Note that if $\ZZ_4 = \generators{\psi}$, then $\ZZ_2 \leq \ZZ_4$ is given by $\ZZ_2 = \generators{\psi^2}$.

We can compute that if $\psi(\zeta) = \zeta^2$, then

\begin{align*}
\psi^2(\zeta) &= \zeta\inv \\
\psi^2(\zeta^2) &= \zeta^{-2}\\
\psi^2(\zeta^3) &= \zeta^{-3}
.\end{align*}

Noting that $\zeta_4$ is a linear combination of the other $\zeta$s, we have a basis $\theset{1, \zeta, \zeta^2, \zeta^3}$.

Then you can explicitly compute the fixed field by writing out
$$
\sigma(a + b\zeta + c\zeta^2 + d\zeta^3) = a + b\sigma(\zeta) + c\sigma(\zeta^2) + \cdots
,$$
gathering terms, and seeing how this restricts the coefficients.

In this case, it yields $\QQ(\zeta^2 + \zeta^3)$.

## The Frobenius Automorphism

**Definition:**
Let $p$ be a prime and $F$ be a field of characteristic $p>0$.
Then

\begin{align*}
\sigma_p: F &\to F \\
\sigma_p(x) &= x^p
\end{align*}

is denoted the *Frobenius map*.

**Theorem:**
Let $F$ be a finite field of characteristic $p > 0$.
Then

1. $\phi_p$ is an automorphism, and
2. $\phi_p$ fixes $F_{\sigma_p} = \ZZ_p$.

*Proof of part 1:*
Since $\sigma_p$ is a field homomorphism, we have
$$
\sigma_p(x+y) = (x+y)^p = x^p + y^p
\text{ and }
\sigma(xy) = (xy)^p = x^p y^p
$$

Note that $\sigma_p$ is injective, since $\sigma_p(x) =0 \implies x^p=0 \implies x=0$ since we are in a field.
Since $F$ is finite, $\sigma_p$ is also surjective, and is thus an automorphism.

*Proof of part 2:*
If $\sigma(x) = x$, then
$$
x^p = x \implies x^p-x = 0
,$$
which implies that $x$ is a root of $f(x) = x^p - x$.
But these are exactly the elements in the prime ring $\ZZ_p$.

$\qed$