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I have two questions in algebraic number theory which I have difficulty understanding. I would be grateful if something could help me.

Let $K$ be an algebraic number field and let $O_K$ be its ring of integers. In addition, let $P$ be an (integral) prime ideal (in $O_K$) lying above the rational number $p$ and let $e>1$ be its ramification index. Now,

(1) Is there a connection between the fact that $K$ contains a primitive $p^{\rm th}$ root of unity (call it $\zeta_p$) and the divisibility of $e$ by $p-1$? I guess that if $\zeta_p\in K$ then $p-1\mid e$. What can we say about the ramification index in the case where $\zeta_p\notin K$ ?

(2) Suppose $p-1\mid e$ but $p\nmid e$. I am searching a criterion for the congruence $x^{p-1}\equiv-p \pmod{P^{e+1}}$ to be solvable (in $O_K$). Is there “simple” condition that is equivalent to the existence of a solution of such a congruence?

Thanks in advance to anyone who could help me.

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Why, in (2), adding that $p$ does not divide $e$? It is automatic from the assumption $p-1\mid e$. –  Andrea Mori Jul 22 '12 at 11:09
    
No, take for instance $e=6$, $p=3$, Here $3-1\mid6$ and also $3\mid6$, so $p-1\mid e$ does not imply $p\nmid e$. –  boaz Jul 22 '12 at 21:32
    
Sorry, my misunderstanding –  Andrea Mori Jul 22 '12 at 22:11

1 Answer 1

Let $K={\bf Q}(\sqrt3)$. Then $p=3$ ramifies with $e=2$, so $p-1$ divides $e$, but $K$ has no primitive 3rd root of unity. So it seems that in (1), the converse is not, in general, true.

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Thanks Gerry. Is there at all a connection between the presence of a $p^{th}$ primitive root of unity in a given number field and the ramification index of $p$ ? What we can say about the ramification index in a field that does not contain a primitive $p^{th}$ root of unity ? –  boaz Jul 22 '12 at 21:54
    
Sorry, I think I have already exhausted my knowledge of ramification indices. –  Gerry Myerson Jul 22 '12 at 23:45

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