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Sep 4, 2020 at 6:10 history edited Mare CC BY-SA 4.0
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Sep 4, 2020 at 5:50 vote accept Mare
Sep 4, 2020 at 4:32 history became hot network question
Sep 3, 2020 at 22:32 comment added YCor If $\Phi(t)$ is a cyclotomic polynomial other than $t\pm 1$ then $\Phi$ has no real root, hence has a constant sign on the reals. Since it's monic, it's then positive on reals, and hence $\ge 1$ on integers.
Sep 3, 2020 at 22:27 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 22:20 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 22:19 answer added Qiaochu Yuan timeline score: 10
Sep 3, 2020 at 22:19 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 22:17 comment added Mare @JoeSilverman That answers Q1 together with the wikipedia link I guess. If you want you can turn it into an answer.
Sep 3, 2020 at 22:14 comment added Joe Silverman Usually people say that $M$ has finite order, rather than calling it periodic. In any case, the fact that $M^n=I$ means that the eigenvalues are roots of unity, which in turn implies that its characteristic polynomial is a product of cyclotomic polynomials (since the characteristic polynomial is in $\mathbb Z[x]$). Hence as noted earlier, your question is whether cyclotomic polynomials are non-negative when evaluated at 1, it has nothing ot do with matirces.
Sep 3, 2020 at 22:12 comment added Mare @AbdelmalekAbdesselam I forgot whether this holds in general, but it holds for acyclic quiver algebra and thus your comment answers Q3!
Sep 3, 2020 at 22:08 comment added Abdelmalek Abdesselam For Q1 one could perhaps show that $p_A$ is a product of cyclotomic polynomials and then the result would follow from en.wikipedia.org/wiki/Cyclotomic_polynomial#Polynomial_values
Sep 3, 2020 at 22:02 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 21:25 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 20:48 history edited Mare CC BY-SA 4.0
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Sep 3, 2020 at 20:29 history asked Mare CC BY-SA 4.0