Prove $A^{p} = A^{p^{3}}$, where $A\in M_{2}(\mathbb Z_{p})$?

Question

What is the idea to prove $A^{p} = A^{p^{3}}$, where $A$ is a matrix in $M_{2}(\mathbb Z_{p})$?

Answer 1

Let $f$ denote the characteristic polynomial of $A$. We note that every irreducible monic polynomial with degree at most $2$ divides $x^{p^2} - x$. It follows that $f \mid (x^{p^2} - x)^p$, but by the "freshman's dream" we have $(x^{p^2} - x)^p = x^{p^3} - x^p$.

If we write $x^{p^3} - x^p = f(x)g(x)$, then we find that $A^{p^3} - A^p = f(A)g(A) = 0\cdot g(A) = 0$, as desired.

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Alternative approach: let $f$ denote the characteristic polynomial of $A$. We see that the splitting field of $f$ has degree at most $2$. However, $\Bbb F_{p^2}$ is the unique degree-$2$ extension of $\Bbb F_p$, which means that $A$ has all its eigenvalues in $\Bbb F_{p^2}$. Thus, we can write $A$ in its Jordan form over $\Bbb F_{p^2}$.

It now suffices to consider two cases:

Case 1: If $A$ is diagonalizable, then we note that by Lagrange's theorem, all elements of $\Bbb F_{p^2}$ satisfy $x^{p^2} = x$, and thus $x^{p^3} = x^p$. The result follows.

Case 2: In the remaining case, $A$ has a repeated eigenvalue $\lambda \in \Bbb F_p$ and can necessarily be written as $\lambda I + N$, where $N^2 = 0$. It follows that $$ A^p = (\lambda I)^p + N^p = \lambda^p I = \lambda I. $$ From there, it is easy to see that $A^{p^3} = (A^p)^{p^2} = \lambda I = A$, as desired.