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오리톢 [902596] · MS 2019 · 쪽지

2021-10-20 00:23:53
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게시글 주소: https://orbi.kr/00040120085


(a) Recall that $x^{q^2}-x\in \Bbb F_q[x]$ is the product of all irreducible degree 1 and 2 polynomials over $\Bbb F_q$. Since any quadratic irreducible polynomial is the minimal polynomial of its two roots, $q^2 = \phi(1)+2\phi(2)$ where $\phi(n)$ counts the number of irreducible polynomials of degree $n$ in $\Bbb F_q[x]$. Since $\phi(1) = q$, $\phi(2) = \frac{1}{2}(q^2-q)$.\\

Similar argument shows that $\phi(3) = \frac{1}{3}(q^3-q)$.\\

\\

(b) Recall that any conjugacy class in $GL_3(\Bbb F_q)$ can be represented by a unique matrix in rational canonical form. We can view any elements in $GL_3(\Bbb F_q)$ as a nonsingular endomorphism of $3$-dimensional vector space $V$ over $\Bbb F_q$. Once we view $V$ as $\Bbb F_q[x]$-module, there are total $3$ decomposition of $V$;

$$\Bbb F_q[x]/(x^3+b_2x^2+b_1x+b_0),\ \Bbb F_q[x]/(x+a_0)\oplus\Bbb F_q[x]/(x+a_0)(x+b_0)$$

$$\Bbb F_q[x]/(x+b_0)\oplus\Bbb F_q[x]/(x+b_0)\oplus\Bbb F_q[x]/(x+b_0)$$

which correspond to matrices;

$A_1 =\begin{pmatrix}0 & 0 & -b_0\\ 1 & 0 & -b_1\\ 0 & 1 & -b_2\\ \end{pmatrix},\ A_2 =\begin{pmatrix}0 & -a_0b_0 & 0\\ 1 & -(a_0+b_0) & 0\\ 0 & 0 & -a_0\\ \end{pmatrix},\ A_3 = \begin{pmatrix}-b_0 & 0 & 0\\ 0 & -b_0 & 0\\ 0 & 0 & -b_0\\ \end{pmatrix}$.\\

Now, $\det(A_1) = -b_0$ so there are total $q^2(q-1)$ many invertible matrices of $A_1$. $\det(A_2) = -a_0^2b_0$ so there are total $(q-1)^2$ many invertible matrix of $A_2$. $\det(A_3) = -b_0^3$ so there are total $(q-1)$ many invertible matrices of $A_3$.\\

Summing up, the number of conjugacy classes of $GL_3(\Bbb F_q) = q(q+1)(q-1)$.\\

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