Questions Regarding Proposition 9.3 of Schoof's Algebraic Number Theory Notes In the proof of Proposition 9.3 in Schoof (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.168.6261&rep=rep1&type=pdf), he writes the following: for $p \in \mathbb{Z}$ a prime, $O_F$ the algebraic integers of a number field $F$, and $\alpha$ a primitive element of $F$, we have that if $p \mid [O_F: \mathbb{Z}[\alpha]]$ yields that there exists an element $x \in O_F \setminus \mathbb{Z}[\alpha]$ such that $px \in \mathbb{Z}[\alpha]$.
I don't see how this is true. In particular, I don't see the implication of $p$ dividing this index yielding there is an algebraic integer of the form $\frac{1}{p}\mathbb{Z}[\alpha]$.
My second question concerns the phrase, "the ideal $I$ is a proper divisor of $T^n$", where $I$ is an ideal of $\mathbb{F}_p[T]$. What does it mean for an ideal to be a proper divisor in this context, and perhaps in general?
 A: The index $[O_F : \mathbb{Z}[\alpha]]$ is finite by Corollary 7.4, so the quotient $O_F/\mathbb{Z}[\alpha]$ is a finite abelian group.
Since $p$ divides the order of $O_F/\mathbb{Z}[\alpha]$, there exists an element of order $p$ in $O_F/\mathbb{Z}[\alpha]$ by Cauchy's theorem.
Let $x \in O_F$ be a representative of such an element, and denote by $\bar{x} \in O_F/\mathbb{Z}[\alpha]$ the residue class.


*

*Since $\bar{x} \neq \bar{0}$ we have $x \not\in \mathbb{Z}[\alpha]$.

*Since $p\bar{x} = \bar{0}$ we have $px \in \mathbb{Z}[\alpha]$.


Since $x \in O_F \setminus \mathbb{Z}[\alpha]$ satisfies  $px \in \mathbb{Z}[\alpha]$, we have $x = 1\cdot x=\frac{p}{p}\cdot x=\frac{px}{p} = \frac{1}{p}(px) \in \frac{1}{p}\mathbb{Z}[\alpha]$.

Regarding the proper divisor:


*

*$I$ contains $T^n$, so $I$ divides the ideal generated by $T^n$.

*$I$ contains a non-zero polynomial of degree $<n$, so $I$ is not equal to $(T^n)$.


In general, a divisor of an ideal $J$ which is not equal to $J$ is called proper.
