# If p divides the index, then there exists $\beta \in O_K, \beta \notin \mathbb{Z}[\alpha]$ such that $p\beta \in \mathbb{Z}[\alpha]$

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)

-

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!].
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