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Nov 22, 2021 at 1:43 history edited David Roberts CC BY-SA 4.0
fixed arxiv front-end link and title; added MR link; formatting
Nov 4, 2009 at 7:02 history edited Ian Agol CC BY-SA 2.5
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Nov 4, 2009 at 1:11 comment added Qiaochu Yuan 1. For quadratic polynomials a necessary condition is that the linear term is non-positive, since the trace is non-negative. It's not hard to get all polynomials of the form x^2 - 2nx + 1, but admittedly I haven't thought too hard about the quadratic case. 2. No. For example, P(x) = x^n - 1 works.
Nov 4, 2009 at 0:45 history answered Ian Agol CC BY-SA 2.5