3.9 Cyclic and Cyclotomic (分圆) Extension

In this section, we aim to classify all cyclic and cyclotomic extensions for $K$ satisfying certain condition.

Cyclic Extensions

Definition

We say $F/K$ is a cyclic (rep. abelian) extension if $F/K$ is Galois with $\mathrm{Gal}(F/K)$ is a cyclic (rep. abelian) group.

Definition

Let $F/K$ be a Galois extension, and let $G=\mathrm{Gal}(F/K)$. For any $u\in F$, define

• the norm of $u$ as $N_{F/K}(u)={\prod }_{g\in G}g(u)$;
• the trace of $u$ as ${\mathrm{tr}}_{F/K}(u)={\sum }_{g\in G}g(u)$.

Clearly, $N_{F/K}(u)$ and ${\mathrm{tr}}_{F/K}(u)$ are contained in $F^G=K$.

Theorem

Suppose $F/K$ is a cyclic extension of degree $n$, with $\mathrm{Gal}(F/K)=⟨\sigma ⟩\simeq \mathbb{Z}/n\mathbb{Z}$. Let $u\in F$, then:

• $\mathrm{tr}(u)=0$ iff $u=v-\sigma (v)$;
• (Hilbert 90) $N(u)=1$ iff $u=v\sigma (v{)}^{-1}$ for some $v\in F$.

\begin{proof} i) One direction is easy. Now we assume that $\mathrm{tr}(u)=0$, and set

$$v=\frac{1}{n}\left(\sum _{i=0}^{n-2}(n-1-i){\sigma }^i(u)\right).$$

Remark that $1/n\in K$ as $\mathrm{char}=0$. Then we can compute directly to show $v-\sigma (v)=u$.

ii) One direction is easy. Consider ${\sigma }^i:F^{\times }\to F^{\times }$ as characters. Define $h$ as a linear combination of characters

$$h=u{\sigma }^0+(u\sigma (u)){\sigma }^1+\cdots +(u\sigma (u)\cdots {\sigma }^{n-1}(u)){\sigma }^{n-1},$$

and there exists $y\in F^{\times }$ such that $h(y)\ne 0$. Let $v=h(y)$, then it is easy to verify $\sigma (v)=\frac{1}{u}v$ by $N(u)=1$ and so $u=v\sigma (v{)}^{-1}$. \end{proof}

Lemma

Let $K$ be a field, and let $n$ be a positive integer. Suppose $K$ contains a primitive $n$-th root of unity $\zeta$.

• If $d\mid n$, then $\eta ={\zeta }^{n/d}$ is a primitive $d$-th root of unity.
• If $d\mid n$ and $u$ is a root of $x^d-a\in K[x]$, then $x^d-a$ has $d$ distinct roots $u\cdot {\eta }^i$ with $0⩽i⩽d-1$. In addition, $K(u)$ is a splitting field of $x^d-a$ over $K$ and is Galois over $K$, with $\mathrm{Gal}(K(u)/K)<\mathbb{Z}/d\mathbb{Z}$, where the equality holds iff $x^d-a$ is irreducible over $K$.

\begin{proof} i) is easy. For ii), $g\in \mathrm{Gal}(K(u)/K)$ is uniquely determined by $g(u)=u\cdot {\zeta }^i$. Take $g_1(u)=u\cdot {\zeta }^i$ and $g_2(u)=u\cdot {\zeta }^j$, then $g_1{g}_2(u)=u\cdot {\zeta }^{i+j}$ yields that there is a group homomorphism $G\to \mathbb{Z}/d\mathbb{Z}$, which is obviously injective. Hence, $G⩽\mathbb{Z}/d\mathbb{Z}$. Note that $G\simeq \mathbb{Z}/d\mathbb{Z}$ iff $|G|=d$ iff $[K(u):K]=d$ iff $\mathrm{Irr}(u,K)=d$ iff $x^d-a$ is irreducible. \end{proof}

Theorem

Use $K,n,\zeta \in K$ as above and suppose $d\mid n$, $\eta ={\zeta }^{n/d}$. Assume that $\mathrm{char}K=0$. Let $F/K$ be an extension, then TFAE:

• $F/K$ is cyclic of degree $d$;
• there exists $a\in K$ such that $x^d-a\in K[x]$ is irreducible, and $F$ is a splitting field of it.

\begin{proof} ii)→i) by ^e267b2.

i)→ii). Write $G=\mathrm{Gal}(F/K)=⟨\sigma ⟩\simeq \mathbb{Z}/d\mathbb{Z}$. Since $\eta \in K$, we have $N_{F/K}(\eta )={\prod }_{g\in G}g(\eta )={\eta }^d=1$ and by ^21b820 $\eta =v/\sigma (v)$ for some $v\in F$. It deduces that $\sigma (v)=v{\eta }^{-1}$ and $\sigma (v^d)=v^d$. Thus, $v^d\in F^G=K$. Let $a=v^d$. We claim that $x^d-a$ is irreducible. Note that $x^d-a$ irreducible iff $x^d-a=\mathrm{Irr}(v,K)$ iff $[K(v):K]=d$ iff $F=K(v)$. Since $\mathrm{Gal}(F/K)$ is abelian, $\mathrm{Gal}(F/K(v))$ is normal and so $K(v)/K$ is a Galois extension. Note that restriction map induces a group homomorphism

$$\mathrm{Gal}(F/K)\to \mathrm{Gal}(K(v)/K),g\mapsto g{|}_{K(v)}$$

and it is injective by $\sigma (v)=v\cdot {\eta }^{-1}$ and ${\sigma }^i(v)=v\cdot {\eta }^{-i}$ for $0⩽i⩽d-1$. It deduces that $|\mathrm{Gal}(K(v)/K)|⩾d$. Hence $|\mathrm{Gal}(K(v)/K)|=d$ and now we finish the proof. \end{proof}

Cyclotomic Extension

Theorem

Assume that $\mathrm{char}K=0$. Let $F/K$ be a splitting field of $x^n-1$. We call $F/K$ a cyclostome extension. Then:

• $F=K(\zeta )$, where $\zeta$ is a primitive $n$-th root of unity;
• $G=\mathrm{Gal}(F/K)<(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$;
• $[F:K]\mid \phi (n)$ where $\phi$ is the Euler function. In particular, if $n=p$, then $\mathrm{Gal}(F/K)$ is cyclic of order dividing $p-1$.

\begin{proof} i) is obvious.

ii) Since $g\in G$ is uniquely determined by $g(\zeta )={\zeta }^i$, it induces a group homomorphism $G\stackrel{\theta }{\to }\mathbb{Z}/n\mathbb{Z}$. Then we can easily verify that $\theta (G)⩽(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ and $\theta$ is injective.

iii) Note that $|(\mathbb{Z}/n\mathbb{Z}{)}^{\times }|=\phi (n)$. For $n=p$, $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }=({\mathbb{F}}_p{)}^{\times }$ is a cyclic group of order $p-1$. \end{proof}