Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$. Let $G$ be the Galois group of $\mathbb{Q}(\zeta)/\mathbb{Q}$. $G$ is isomorphic to $(\mathbb{Z}/l\mathbb{Z})^*$. Hence $G$ is a cyclic group of order $l - 1$. Let $f = (l - 1)/2$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. $K_f$ is a unique quadratic subfield of $K$. Let $A_f$ be the ring of algebraic integers in $K_f$.

Let $p$ be a prime number such that $p \neq l$. Let $pA_f = P_1\cdots P_r$, where $P_1, \dots, P_r$ are distinct prime ideals of $A_f$.

Since $p^{l - 1} \equiv 1$ (mod $l$), $p^f = p^{(l - 1)/2} \equiv \pm$1 (mod $l$).

My question: Is the following proposition true? If yes, how would you prove this?


(1) If $p^{(l - 1)/2} \equiv 1$ (mod $l$), then $r = 2$.

(2) If $p^{(l - 1)/2} \equiv -1$ (mod $l$), then $r = 1$.

This is a related question.

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    $\begingroup$ You seem to like cyclotomic stuff a lot $\endgroup$ – Jorge Fernández Hidalgo Jul 24 '12 at 23:24
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    $\begingroup$ Because I believe they are at the heart of algebraic number theory. $\endgroup$ – Makoto Kato Jul 24 '12 at 23:26
  • $\begingroup$ I think you should invest in a good reference for algebraic number theory like Neukirch. Many of your recent questions are standard cyclotomic ideas that are often addressed. $\endgroup$ – Brandon Carter Jul 24 '12 at 23:38
  • $\begingroup$ @BrandonCarter Does the Neukirch's book have the following result? math.stackexchange.com/questions/174703/… $\endgroup$ – Makoto Kato Jul 24 '12 at 23:53
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    $\begingroup$ Dear Makoto, It may not state precisely that result, but it will certainly establish (much more than) the tools necessary to prove it, which is just the theory of decomposition groups. Neukirch's book is not my own favourite, but any book (or online set of lecture notes) on algebraic number theory will develop these tools. Since they are applicable to many of your questions, you may well want to learn them. Regards, $\endgroup$ – Matt E Jul 26 '12 at 3:28

This is most easily established by decomposition group calculations; it is a special case of the more general result proved here (which is an answer to OP's linked question).


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