Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Again, this is a homework question, so (for the sake of learning) I'd be happy I get subtle hints only.

Anyhow, the setup is as follows: "$K=\mathbb{Q}(\alpha)$ is of degree n over $\mathbb{Q}$, where $\alpha$ is in the ring of integers $O_K$ of $K$, let the index of $\alpha$ be the index of $[O_K:\mathbb{Z}[\alpha]]$. Suppose that the minimal polynomial of $\alpha$ is Eistenstein at $p$. Then $p$ does not divide the index of $\alpha$.

Proceed as follows: If p divides the index, then there exists $\beta \in O_K, \beta \notin \mathbb{Z}[\alpha]$ such that $p\beta \in \mathbb{Z}[\alpha]$. [...]"

The rest of the exercise is clear. It is only the last sentence I don't follow. That is, why should such a $\beta$ exist?

(We say that a monic polynomial $x^n+a_{n-1}+x^{n-1}+\ldots+a_0$ is Eistenstein at p if p divides all the $a_i$ but $p^2$ does not divide $a_0$.)

(my creativity failed finding a good title for this question)

share|cite|improve this question
up vote 4 down vote accepted

Consider the quotient group (under addition) $\mathcal O_K/\mathbb Z[\alpha].$ You are assuming that $p$ divides its order. Therefore ... [apply a basic theorem in group theory here!].

share|cite|improve this answer
Else, apply the theorem of elementary divisors to the free abelian group ${\cal O}_K$ – Andrea Mori Feb 3 '11 at 21:28
Therefore there exists a subgroup of order $p$, hence a non-zero element $\beta$ such that $p\beta = 0 \in O_K/\mathbb{Z}[\alpha]$, that is, $p\beta \in \mathbb{Z}[\alpha]$. This is correct, right? – Fredrik Meyer Feb 3 '11 at 22:02
@Fredrik: Dear Fredrik, Yes, that's right. Regards, – Matt E Feb 3 '11 at 23:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.