$\Bbb{Z}_p$中等价于特定形式矩阵的中心化子及其阶求解

Alright, let's work through finding the centralizer of the matrix $M = \begin{bmatrix} a & b \ -b & a \end{bmatrix}$ over $\Bbb{Z}_p$ (the field with $p$ elements) and calculating its order. First, a quick reminder: the centralizer of $M$ is the set of all matrices $A$ (either from the ring of all 2x2 matrices $M_2(\Bbb{Z}_p)$, or the general linear group $GL_2(\Bbb{Z}_p)$ of invertible matrices) that commute with $M$—meaning $AM = MA$.

We'll split this into two key cases based on whether $b$ is zero or not.

Case 1: $b = 0$ (Scalar Matrix) #

When $b=0$, $M$ simplifies to $aI$, a scalar multiple of the identity matrix. Scalar matrices have a super straightforward centralizer: they commute with every 2x2 matrix over $\Bbb{Z}_p$. Multiplying by a scalar on the left or right does exactly the same thing, so no restrictions on $A$.

• If we're talking about the centralizer in $M_2(\Bbb{Z}_p)$:
  • $C(M) = M_2(\Bbb{Z}_p)$ (all 2x2 matrices over $\Bbb{Z}_p$)
  • Order: There are $p$ choices for each of the 4 entries, so total order is $p^4$.
• If we restrict to invertible matrices ($GL_2(\Bbb{Z}_p)$):
  • $C(M) = GL_2(\Bbb{Z}_p)$ (all invertible 2x2 matrices)
  • Order: The size of $GL_2(\Bbb{Z}_p)$ is $(p^2 - 1)(p^2 - p)$. This comes from counting valid invertible matrices: pick a non-zero first column ($p^2 - 1$ options), then pick a second column that's not a scalar multiple of the first ($p^2 - p$ options).

Case 2: $b \neq 0$ (Non-Scalar "Complex-like" Matrix) #

When $b \neq 0$, $M$ acts like a "complex number" $a + bi$ where $i^2 = -1$—matrix multiplication here matches the multiplication of elements in the ring $\Bbb{Z}_p[i] = \Bbb{Z}_p[x]/(x^2 + 1)$. To find the centralizer, let's let $A = \begin{bmatrix} x & y \ z & w \end{bmatrix}$ be a matrix that commutes with $M$, then solve $AM = MA$.

First, compute both products:

• $AM = \begin{bmatrix} xa - yb & xb + ya \ za - wb & zb + wa \end{bmatrix}$
• $MA = \begin{bmatrix} ax + bz & ay + bw \ -bx + az & -by + aw \end{bmatrix}$

Set corresponding entries equal and simplify:

1. From the (1,1) entry: $xa - yb = ax + bz$ → $-yb = bz$. Since $b \neq 0$, we can divide by $b$ to get $z = -y$.
2. From the (1,2) entry: $xb + ya = ay + bw$ → $xb = bw$. Again, $b \neq 0$, so $x = w$.

The remaining entries (2,1) and (2,2) will automatically hold if we substitute $z = -y$ and $w = x$, so we don't get any new constraints. That means the centralizer only includes matrices of the same form as $M$.

Centralizer in $M_2(\Bbb{Z}_p)$ #

• $C(M) = \left{ \begin{bmatrix} x & y \ -y & x \end{bmatrix} \mid x, y \in \Bbb{Z}_p \right}$
• Order: We have $p$ choices for $x$ and $p$ choices for $y$, so total order is $p \times p = p^2$.

Centralizer in $GL_2(\Bbb{Z}_p)$ #

We need matrices in the above form that are invertible, which requires their determinant to be non-zero. The determinant of $\begin{bmatrix} x & y \ -y & x \end{bmatrix}$ is $x^2 + y^2$, so we count all pairs $(x,y)$ where $x^2 + y^2 \neq 0$.

Let's break this down by $p$:

• If $p \equiv 3 \pmod{4}$: $-1$ is not a quadratic residue in $\Bbb{Z}_p$, so $x^2 + y^2 = 0$ if and only if $x = y = 0$. That means all non-zero matrices in the centralizer are invertible.
  • Order: $p^2 - 1$
• If $p \equiv 1 \pmod{4}$: $-1$ is a quadratic residue (there exists some $s \in \Bbb{Z}_p$ where $s^2 = -1$). Then $x^2 + y^2 = 0$ implies $x = \pm s y$ (for $y \neq 0$) or $x = y = 0$. The total number of such pairs is $2p - 1$, so the number of invertible matrices is $p^2 - (2p - 1) = (p - 1)^2$.
• If $p = 2$: In $\Bbb{Z}_2$, $x^2 = x$ for all $x$, so $x^2 + y^2 = x + y$. The pairs where $x + y = 0$ are $(0,0)$ and $(1,1)$, so the number of invertible matrices is $4 - 2 = 2$.

内容的提问来源于stack exchange，提问作者User432477438