I am trying to understand a proof of the following theorem:
Theorem: Let $\alpha$ be an algebraic integer wich has degree $n$ over $\mathbb{Q}$. Then the field extension $\mathbb{Q}\left (\alpha\right )/\mathbb{Q}$ has an integral basis of the form $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{n-1}\left (\alpha\right )}{d_{n-1}}\right \}$ where the $f_i$ are monic polynomials over $\mathbb{Z}$ which have degree $i$, and the $d_i$ are in $\mathbb{N}$ and satisfy $d_i\mid d_{i+1}$. The $d_i$ are uniquely determined.
We call $R$ the ring of algebraic integers over the number field definen above. The first part of the proof that Marcus gives on his book (page 37) goes as follows:
For each $k$, $1\leq k\leq n$, let $F_k$ be the free abelian group of rank $k$ generated by $1/d,\alpha /d,\cdots ,\alpha ^{k-1}/d$, where $d=\text{disc}\left (\alpha\right )$, and set $R_k=R\cap F_k$. Thus we have $R_1=\mathbb{Z}$ and $R_n=R$ (why?). We will define the $d_i$ and the $f_i$ so that for each $k$, $1\leq k\leq n$, $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{k-1}\left (\alpha\right )}{d_{k-1}}\right \}$ is a basis over $\mathbb{Z}$ for $R_k$.
This is certainly true for $k=1$. Thus fix $k<n$ and assume that $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{k-1}\left (\alpha\right )}{d_{k-1}}\right \}$ is a basis over $\mathbb{Z}$ for $R_k$, whith the $f_i$ and $d_i$ as in the theorem. We have to define $f_k$ and $d_k$ and show that we get a basis for $R_{k+1}$ by throwing in $f_k\left (\alpha\right )/d_k$.
Let $\pi :F_{k+1}=\bigoplus_{i=0}^k\mathbb{Z}\frac{\alpha^i}{d}\to \mathbb{Z}\frac{\alpha^k}{d}$ be the canonical projection on the last factor. That is, $\pi$ selects the term of degree $k$. Then $\pi \left (R_{k+1}\right )$ is a subgroup of the infinite cyclic group $\mathbb{Z}\frac{\alpha^k}{d}$, which implies that $\pi \left (R_{k+1}\right )$ is itself cyclic. Fixing any $\beta \in R_{k+1}$ such that $\pi \left (\beta\right )$ generates $\pi \left (R_{k+1}\right )$, we leave to the reader to show that $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{k-1}\left (\alpha\right )}{d_{k-1}},\beta\right \}$ is a basis over $\mathbb{Z}$ for $R_{k+1}$.
At this point I got stuck, since I am having troubles to show this last assertion.
It is clear that $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{k-1}\left (\alpha\right )}{d_{k-1}},\beta\right \}\subset R_{k+1}$ and it is easy to show that they are linearly independent over $\mathbb{Z}$, therefore the only thing we have to prove is that the above set generates $R_{k+1}$ over $\mathbb{Z}$.
This was my attempt to prove it:
We fix $x\in R_{k+1}$ and then we have that $x-\pi \left (x\right )=x-m\pi \left (\beta\right )$ (for some $m\in \mathbb{Z}$) is an element with zero last factor, then $x-m\pi \left (\beta\right )\in F_k\cap R=R_{k}$, therefore $x-m\pi \left (\beta\right )$ is a $\mathbb{Z}$-linear combination of $\left \{1,\frac{f_1\left (\alpha \right )}{d_1},\cdots ,\frac{f_{k-1}\left (\alpha\right )}{d_{k-1}}\right \}$ and this proves what we want.
The problem with the above solution is that when I assert that $x-m\pi \left (\beta\right )\in F_k\cap R$, I am implicitly saying that $\pi \left (\beta\right )$ is an algebraic integer. By definition $\beta$ is an algebraic integer, but why is the same true for $\pi \left (\beta\right )$? The projection is not a ring-homomorphism, thus it is not that clear for me that $\pi \left (\beta\right )$ is an algebraic integer, but of course it must be in order to be able to show what we want.
So, my question is: why is $\pi \left (\beta\right )$ an algebraic integer?