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I was looking at the proof of Lemma 2.17 in Narkiewicz Elementary and Analytic Theory of Algebraic Numbers but don't understand a step.

Let $p$ be a rational prime, $a$ be an algebraic integer of degree $n$, $K = \mathbb{Q}(a)$, and $R_K$ be the ring of integers of $K$. Assume the minimal polynomial of $a$ is Eisenstein with respect to $p$.

It is asserted that if $p$ divides $[R_K:\mathbb{Z}[a]]$ (the index of $a$ in $R_K$), then there exists $\xi$ in $R_K$ of the form $$ \xi = (b_0 + b_1 a + \cdots + b_{n-1}a^{n-1})/p $$ with the $b_i$'s integers not all divisble by $p$.

I don't see why this assertion is true. Can someone please help?

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2 Answers

up vote 1 down vote accepted

Consider the additive group $R_K/\mathbb{Z}[a]$. This is a group of order divisible by $p$; let $\xi$ be an element of $R_K$ that maps to an element of order $p$ of this quotient. In particular, $\xi\notin \mathbb{Z}[a]$.

But $p\xi\in\mathbb{Z}[a]$, since the image of $\xi$ has order $p$, so there exist $b_0,\ldots,b_{n-1}\in\mathbb{Z}$ such that $p\xi = b_0 + b_1a + \cdots + b_{n-1}a^{n-1}$ (since $1$, $a,\ldots,a^{n-1}$ are a $\mathbb{Z}$-basis for $\mathbb{Z}[a]$). Thus, we can write $\xi$ in the given form. If every $b_i$ were divisible by $p$, say $b_i = pc_i$, then $p\xi = p^2(c_0 + c_1a + \cdots + c_{n-1}a^{n-1})$, hence $\xi\in\mathbb{Z}[a]$, contradicting the choice of $\xi$.

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The group $R_k/\mathbb{Z}[a]$ is perhaps not of order $p$, but since $p$ divides its order, there is still an element of order $p$. –  user8268 Mar 24 '11 at 21:40
@user8268:Thanks! I misread that as "$p$ equals" for some strange reason. Fixed. –  Arturo Magidin Mar 24 '11 at 21:41
Perfect. I was thinking it might be something like this, but got stuck. Thanks. –  admchrch Mar 24 '11 at 21:54
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HINT $\ $ There is an element $\rm\:w\ = v/(p\:m),\ v\in \mathbb Z[a]\:$ with least denominator $\rm p\:m\ = [R_K:\mathbb Z[a]]\:,$ hence there is an element with least denominator $\rm\:p\ $, namely $\rm\:m\ w\:,\:$ i.e. $$\rm\displaystyle \xi\ :=\ m\ w\ =\ m\ \frac{v}{p\:m}\ =\ \frac{v}{p}\:$$

Note that this is just the trivial denominator ideal special-case of Cauchy's theorem that a finite group of order $\rm\:p\:m\:$ has an element of order $\rm\:p\:.$ But one need not know anything about Cauchy's theorem, since this fractional form is so simple that it is (implicitly) known since grade school.

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