Lecture 17
Cyclotomic Extensions I
 


Definition 16.1  

A splitting field of $ x^{n} - 1$ over $ F$ is called a cyclotomic field of order $ n$ over $ F$.


Proposition 16.2  

Let $ ($char$ \;F,n)=1$ and $ f(x)=x^{n} - 1 \in F[x].$ Then $ G_f$ is isomorphic to a subgroup of $ U(n).$ In particular $ G_f$ is an abelian group and $ o(G_f) \mid \varphi(n).$

Proof. As $ f(x)$ is separable, it has $ n$ distinct roots. Let $ \{z_{1},z_{2}, \ldots ,z_{n}\} = Z$ be the set of roots of $ f(x)$ in $ F^{a}$ and $ E=F(z_{1},z_{2}, \ldots,z_{n}).$ Since $ Z \subseteq E^{\times}$ is a subgroup, it is cyclic. The map $ \psi : G(E/F) \rightarrow Aut(Z)$ such that $ \sigma \mapsto \sigma \vert _Z$ is an injective group homomorphism. Since $ Aut(Z) \simeq \{\bar{m} \mid (m,n) = 1\}:=U(n)$ is an abelian group, $ G(E/F)$ is also an abelian group whose order divides $ \varphi(n)$.

Example 16.3  

Let $ F=\mathbb{F}_2.$ Then $ x^3-1=(x-1)(x^2+x+1).$ Any root $ z$ of $ x^2+x+1$ is a primitive cube root of unity over $ F.$ Hence $ [F(z):F]=2.$ To find the degree of a primitive seventh root of unity over $ F,$ consider the factorization of $ x^7-1$ into irreducible polynomials over $ F:$
$\displaystyle x^7-1=(x-1)(x^3+x^2+1)(x^3+x+1).$

Therefore there are $ 6$ primitive $ 7th$ roots of unity over $ F$ with two minimal polynomials. In contrast to this, we shall see that all the primitive $ n^{th}$ roots of unity over $ \mathbb{Q}$ have the same irreducible polynomial called the $ n^{th}$ cyclotomic polynomial $ \Phi_n(x).$