Congruent zeta functions. No.2

Yoshifumi Tsuchimoto

In this lecture we define and observe some properties of conguent zeta functions.

For any prime $p$, $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$. To construct $\mathbb{F}_{p^r}$ for $r$,

1. We find an irreducible polynomial $u(X)\in \mathbb{F}_p[X]$ of degree $r$. (Such a thing exists always.)
2. $K=\mathbb{F}_p[X]/(u(X))$ is a field with $p^r$ elements. It is an extension field of $\mathbb{F}_p$ generated by the class $a=\bar X$ of $X$ in $K$.
3. In other words, $K=\mathbb{F}_p[a]$ where $a$ is a root of $u$.
4. The isomorphism class of $K$ is independent of the choice of $u$.

Proof of Lemma 1.3 (5). We prove the following more general result

LEMMA 2.1   Let $K$ be a field. Let $G$ be a finite subgroup of $K^\times$(=multiplicative group of $K$). Then $G$ is cyclic.

PROOF.. We first prove the lemma when $\vert G\vert=\ell^k$ for some prime number $\ell$. In such a case Euler-Lagrange theorem implies that any element $g$ of $G$ has an order $\ell^s$ for some $s\in \mathbb{N}$, $s \leq k$. Let $g_0\in G$ be an element which has the largest order $m$. Then we see that any element of $G$ satisfies the equation

$$x^m=1 .$$

Since $K$ is a field, there is at most $m$ solutions to the equation. Thus $\vert G\vert\leq m$. So we conclude that the order $m$ of $g_0$ is equal to $\vert G\vert$ and that $G$ is generated by $g_0$.

Let us proceed now to the general case. Let us factorize the order $\vert G\vert$:

$$\vert G\vert=\ell_1^{k_1} \ell_2 ^{k_2} \dots \ell_t ^{k_t} \qquad(\ell_1,\ell_2,\dots, \ell_t:$$prime, $$k_1,k_2,\dots,k_t \in \mathbb{Z}_{>0}).$$

Then $G$ may be decomposed into product of $p$-subgroups

$$G=G_1\times G_2 \times \dots \times G_t \qquad (\vert G_j\vert=\ell_j^{k_j} (j=1,2,3,\dots,t)).$$

By using the first step of this proof we see that each $G_j$ is cyclic. Thus we conclude that $G$ is also a cyclic group. $\qedsymbol$

EXERCISE 2.1   Let $G$ be a finite abelian group. Assume we have a decomposition $\vert G\vert=m_1 m_2$ of the order of $G$ such that $m_1$ and $m_2$ are coprime. Then show the following:

1. Let us put
   $$H_j=\{g\in G; g^{m_j}=e_G\} \qquad(j=1,2)$$
   Then $H_1,H_2$ are subgroups of $G$.
2. $\vert H_j\vert=m_j$ ($j=1,2$).
3. We have
   $$G=H_1 H_2.$$

EXERCISE 2.2   Let $G_1,G_2$ be finite cyclic groups. Assume $\vert G_1\vert$ and $\vert G_2\vert$ are coprime. Show that $G_1\times G_2$ is also cyclic.