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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?

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Have you tried checking this yourself for small n, say n = 1,2,3? How are eigenvalues related to the factorization of the characteristic polynomial? –  KCd Jun 19 '11 at 15:36

2 Answers 2

I can't comment at the moment - insufficient points. You need to go back to what the p-adic valuation means and basic facts about divisibility of integers by primes. I think it is a lot simpler than you realise.

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The eigenvalues are the roots of the characteristic polynomial. Now you just need to recall the relation between the roots of a (monic) polynomial and its coefficients.

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Let $r_{i}$ be the roots of $f(x)$. Then by the relations between roots and coefficients, I know that $|r_{1} + \cdots + r_{n}|_{p} = |a_{1}|_{p}$. As the $p$-adic valuation is nonarchimedean, we know that $|a_{1}|_{p} \leq \max(|r_{1}|_{p}, \ldots, |r_{n}|_{p})$. How can I prove that $|a_{1}|_{p} \leq 1/p$? –  120948 Jun 19 '11 at 15:49

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