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.

I first provide a background, or the context, where this question arises. Skip it if one wants so.

Background: In the book The Genus fields of algebraic number fields by Ishida, one finds the assertion: Let $K$ be an algebraic number field with $[K:Q]=n$ and $(p)$ totally ramified in $K$, $p\equiv 1\pmod{2n}$, $d=(n,p-1)$. Let $q$ be a prime with $\operatorname{ord}_p(q)=d$. If $(q)$ is a norm of an ideal in $K$, then $n\mid h_K$.

So I think that,

Condition: If $p, q$ are primes with $\begin{cases}\left(\dfrac{p}{q}\right)_n=1, \text{ i.e. } (x^n-p) \text{ has a solution modulo $q$}\\\left(n,\dfrac{p-1}{f}\right)=1, \text{ where } f=\operatorname{ord}_p(q) \end{cases}\tag{I}$, then the conditions of the theorem are satisfied, so that $n\mid h_K$.

And the question is

Question: Can the conditions $(I)$ be fulfilled? And how to construct explicitly such pairs? Or is there a name for such pairs of primes?

For example, if $n$ is also a prime, and $p\not\equiv 1\pmod n$, then the condition reduced to that $p$ is congruent to an $n$-th power modulo $q$, which by reciprocity, can of course fulfilled by infinitely many $q$, IIRC. And I know little, if anything, of the general situation. Thanks in advance in any case.

share|improve this question
1  
The notation $(\dfrac{p}{q})_n$ is non-standard, but I think it is more brief. Sorry if this is not appropriate. It means that $p$ is an $n$-th power modulo $q$. Moreover, I tagged this as algebraic-number-theory, chiefly because the proof of the assertion is involved with the theory; else the question is not so concerned with algebraic-numbers, I surmise. –  awllower May 6 '13 at 11:27
add comment

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.