Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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.

share|improve this answer

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.