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.

Definition Let $p$ be a prime number. Let $a$ be an integer not divisible by $p$. Let $m > 1$ be an integer. Suppose that there exists an integer $n_0 \geq 1$ with the following property. $x^m \equiv a$ (mod $p^n$) has an integer solution whenever $n \geq n_0$. Then we say $a$ is an $m$-th power residue with respect to $p$.

Can we prove the following theorem without using $p$-adic numbers?

Theorem Let $p$ be a prime number. Let $m > 1$ be an integer. There exists an iteger $n \geq 1$ with the following property. If $a \equiv 1$ (mod $p^n$), then $a$ is an $m$-th power residue with respect to $p$.

Motivation This theorem and its generalization to algebraic number fields can be used in class field theory. I'm interested in proving CFT without $p$-adic numbers(see here).

share|improve this question
2  
Yes. All you have to do is prove that the solution mod p^k (k \ge n) lifts to a solution mod p^{k+1} etc. This is known as (a special case of) Hensel's lemma. If p is odd and does not divide a, then you can take n = 1 in the statement above. –  franz lemmermeyer Jul 18 '12 at 6:31
    
@franzlemmermeyer Thanks! –  Makoto Kato Jul 18 '12 at 7:04
    
@franzlemmermeyer Dear Franz Lemmermeyer, If m is not divisibe by p, Hensel lift can be applied as you wrote. However, I don't know how to do it when m is divisible by p. Regards. –  Makoto Kato Jul 18 '12 at 8:23
1  
Sanity check: you have seen the form of Hensel's lemma that doesn't require $f'(a)$ to be invertible? e.g. en.wikipedia.org/wiki/Hensel%27s_lemma#Generalizations –  Hurkyl Jul 20 '12 at 14:59
3  
As a philosophical aside, I'm not sure you've really thought out your goal. You remind me of the people who say "We shouldn't use real numbers, we should use Cauchy sequences instead" who, in the end, are using the real numbers anyways, they've just obscured things and made them more complicated by always thinking in terms of Cauchy sequences. It looks like you're making the p-adic analog of this mistake: you haven't really stopped using the $p$-adic numbers, you're just obscuring them by phrasing results in terms of the sequences that converge to them. –  Hurkyl Jul 20 '12 at 15:13

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.