Step in Proof of Lemma in Narkiewicz _Elementary and Analytic Theory of Algebraic Numbers_

Question

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?

score 1 · Accepted Answer · answered Mar 24 '11 at 21:37

1

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 $\xi$.

answered Mar 24 '11 at 21:37

Arturo Magidin

398,050

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

score 0 · Answer 2 · edited Oct 01 '17 at 20:38

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.

Step in Proof of Lemma in Narkiewicz _Elementary and Analytic Theory of Algebraic Numbers_

2 Answers2