Suppose I had a linear operator $L$ whose characteristic polynomial was $f(x) = x^{n} + a_{1}x^{n-1} + \cdots + a_{n-1}x + a_{n}$. Furthermore, I also know that the eigenvalues of $L$ have $p$-adic valuation $\geq 1$. Why does it follow that $a_{i} \in p^{i}\mathbb{Z}_{p}$ for all $i$? What if all the eigenvalues have $p$-adic valuation equal to 1?
Characteristic polynomial and $p$-adic valuation
1
$\begingroup$
linear-algebra
p-adic-number-theory
-
2Have you tried checking this yourself for small n, say n = 1,2,3? How are eigenvalues related to the factorization of the characteristic polynomial? – 2011-06-19