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Let $p$ be an odd prime number.

Let $A\in M_{n\times n}(\mathbb{Z})$ be a matrix satisfiying $a_{ij}\equiv \delta_{ij}\pmod{p}$.

Prove that: if $|\det(A)|=1$ and $A^m=I$ for some $m\in\mathbb{N}^+$, then we have $A=I$.

I tried to write $A$ as $pX+I$ where $X\in M_{n\times n}(\mathbb{Z})$. Then there is an annihilation polynomial of $X$ by $(pX+I)^m=I$, and the problem turns to be proving $X=O$. But I don't know how to complete this proof.

Arturo Magidin
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dailycrazy
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  • My first thought is that while you can find matrices which satisfy $(px+1)^m - 1$, they won't be integer matrices. Because an integer matrix always has a characteristic polynomial with constant leading coefficient. – Sera Gunn Oct 10 '21 at 15:07
  • Also, over $\mathbb{Z}$, that polynomial factors as $px\left( \sum_{i = 0}^{m - 1} \binom{m}{i + 1} (px)^i \right)$. Those factors should be irreducible if $p$ is prime, but someone should check that. – Sera Gunn Oct 10 '21 at 15:11
  • @TrevorGunn However, we have a counterexample that $((3x+1)^4-1)=3x(3x+2)(9x^2+6x+2)$ : ( – dailycrazy Oct 10 '21 at 15:34
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    Right, $m$ would have to be prime for that conclusion. Still, we can make a substitution $y = px$ which shows that any factor is going to be an integer polynomial in $y$ and from that one wants to see that there can be no factor without a $p$ in the leading coefficient. (Some factors may be divisible by $p$ but hopefully not more than once or twice.) – Sera Gunn Oct 10 '21 at 15:45

1 Answers1

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Hints.

We can assume that $m$ is a prime (why?).

Now $$ A^m=(I+pX)^m=I+mpX+{m\choose2}p^2X^2+\ldots=I $$ If $m\neq p$ or $m=p$ in any case $X=pY$ for some matrix $Y$.

Continuing also we obtain in the next step that $X=p^2Z$ and so on.

kabenyuk
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