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I'm looking for an elementary proof (one which does not use Galois theory). For the case $p = 3$, we have that $tr(A^3) = tr(A)^3 - 3e_1e_2 + 3e_3$ where the $e_i$ are coefficients of the characterstic polynomial of $A$, and are thus integers, so the result follows. I cannot see a way to generalize this for arbitrary $p$. Any help would be great!

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    $\begingroup$ It may help to consider the equivalent formulation $$ \operatorname{tr}(n \,A^p - \operatorname{tr}(A)^p I) = 0 $$ where $n$ is the size of the matrix $\endgroup$ Commented Dec 18, 2018 at 3:31
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    $\begingroup$ I'm not sure whether my answer below is "elementary enough" (I'm still using the language of finite fields, but not any "deep theory" of). And Galois theory seems to be an overkill to apply here ;) $\endgroup$
    – metamorphy
    Commented Dec 18, 2018 at 6:31

1 Answer 1

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If $A,B$ are commuting square matrices over $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$ of the same size, then $$(A+B)^p=A^p+B^p$$ (the binomial formula is valid here, and $p\mid\binom{p}{k}$ for $0<k<p$). (The same holds for $A,B$ over $\mathbb{F}_p[\lambda]$, for the same reason.) It follows, for a matrix $A$ over $\mathbb{F}_p$, denoting by $\chi_A(\lambda)=\det(\lambda I-A)$ its characteristic polynomial, that $(\lambda I-A)^p=\lambda^p I-A^p$ and therefore $\chi_{A^p}(\lambda^p)=\big(\chi_A(\lambda)\big)^p$. But $\big(f(x)\big)^p=f(x^p)$ for any polynomial $f\in\mathbb{F}_p[x]$ (this is usually proven by induction on the degree of $f$ using the same "binomial argument" as above, and the fact that $a^p=a$ for any $a\in\mathbb{F}_p$).

Thus actually we have $\chi_{A^p}(\lambda^p)=\chi_A(\lambda^p)$, i.e. just $\color{blue}{\chi_{A^p}\equiv\chi_A}$. As a corollary (comparing the coefficients), we get $\operatorname{tr}(A^p)=\operatorname{tr}(A)$, and we're done (again by $a^p=a$ for any $a\in\mathbb{F}_p$).

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  • $\begingroup$ Could you explain why the line about the characteristic polynomials being equal? $\endgroup$
    – John
    Commented Dec 18, 2018 at 6:47
  • $\begingroup$ I meant the equality between characteristic polynomials right after "and therefore ..." $\endgroup$
    – John
    Commented Dec 18, 2018 at 6:53
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    $\begingroup$ Yes, I've edited this right before. We take $\det$ of the both sides of $(\lambda I-A)^p=\lambda^p I-A^p$. $\endgroup$
    – metamorphy
    Commented Dec 18, 2018 at 6:55
  • $\begingroup$ Sorry, just saw that. It's clear now, thanks! $\endgroup$
    – John
    Commented Dec 18, 2018 at 6:56
  • $\begingroup$ @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold. $\endgroup$
    – metamorphy
    Commented Dec 18, 2018 at 8:01

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