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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 $\sigma$ be one of its generators. Then $\sigma(\zeta) = \zeta^r$, where $r$ is a primitive root mod $l$. Let $f$ be a positive divisor of $l - 1$. Let $e = (l - 1)/f$. There exists a unique subgroup $G_f$ of $G$ whose order is $f$. Let $K_f$ be the fixed subfield of $K$ by $G_f$. Then $e = [K_f : \mathbb{Q}]$. Let $A_f$ be the ring of algebraic integers in $K_f$.

We introduce the following algebraic integers.

\begin{align*} \eta_0 &= \zeta + \sigma^e(\zeta) + \sigma^{2e}(\zeta) + \dots + \sigma^{(f-1)e}(\zeta) \\ \eta_1 &= \sigma(\zeta) + \sigma^{e+1}(\zeta) + \dots + \sigma^{(f-1)e + 1}(\zeta) \\ &\vdots \\ \eta_{e-1} &= \sigma^{e-1}(\zeta) + \sigma^{2e-1}(\zeta) + \dots + \sigma^{fe - 1}(\zeta) \end{align*}

$\eta_0, \ldots, \eta_{e-1}$ are called periods of length $f$.

Applying $\sigma$ to $\eta_0$, we get $\eta_1$. Similarly,

$\eta_0 \to \eta_1 \to \dots \to \eta_{e-1} \to \eta_0$

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

Proposition The set $\{\eta_0, \ldots, \eta_{e-1}\}$ is an integral base of $A_f$.

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  • $\begingroup$ I played around with the $\TeX$ a bit. Take a look — there are some things in there that might save you time. Interesting series of questions, by the way. I wish I knew the answers offhand. $\endgroup$ – Dylan Moreland Jul 23 '12 at 22:25
  • $\begingroup$ @DylanMoreland Thanks, Dylan. $\endgroup$ – Makoto Kato Jul 23 '12 at 22:26
  • $\begingroup$ Doesn't Washington discuss this in the first or second chapter of his book on cyclotomic fields, at the place where he defines periods? $\endgroup$ – KCd Jul 24 '12 at 0:45
  • $\begingroup$ @KCd I don't have the book at hand. $\endgroup$ – Makoto Kato Jul 24 '12 at 3:17
  • $\begingroup$ What's the reason for the downvote? Unless you make it clear, I cannot improve my question. $\endgroup$ – Makoto Kato Aug 13 '12 at 2:24
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The $\eta_i$ form a basis for $A_f$, because the algebraic integers of $K_f$ are the algebraic integers of $K$ that are in $K_f$.

Since $l$ is an odd prime, the family $\{\sigma^i(\zeta)\}$ is the same as $\{\zeta^k,k=1 \ldots l-1\}$, and $1 = - (\zeta^1 + \ldots \zeta^{l-1})$, which should convince you that this family form a basis for the algebraic integers of $K$. Because it is also an orbit of the Galois group, we say it is an integral normal basis.
The elements of $K$ that are elements of $K_f$ are those fixed by $\sigma^e$. Thus, an element of $A_f$ is of the form $x = \sum_{i=0}^{l-1} a_i \sigma^i(\zeta)$ where $a_i \in \mathbb Z$, such that $\sigma^e(x) = x$, which means $a_{i+e} = a_i$, so regrouping the terms you get $x = a_0\eta_0 + a_1 \eta_1 + \ldots a_{e-1} \eta_{e-1}$

This works because we have an integral normal basis. Some number fields, such as $\mathbb Q(i)$, don't have such an integral basis for their ring of integers.

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