$\mathbb{Z}[\alpha]$ is f.g. if $\alpha$ is an algebraic integer.

Question

I want to prove the following theorem, I came up to some point but I need some help to finish the proof:

Theorem.$\mathbb{Z}[\alpha]$ is finitely generated $\implies$ $\alpha$ is an algebraic integer.

Proof. Let $\mathbb{Z}[\alpha]$ is generated by $S=\{1=\beta_0,\beta_1, \dots, \beta_k\}$.

Now I define $\varphi:S \rightarrow \alpha S$. If $\alpha \beta_i = \alpha\beta_j$ for some $i\ne j \in \{0,\dots, k\}$ then $\alpha(\beta_i - \beta_j)=0 \implies (\beta_i -\beta_j)=0 $ since $\alpha,\beta_0, \dots, \beta_k \in \mathbb{Z}[\alpha]$ and $\mathbb{Z}[\alpha]$ is an integral domain since $\mathbb{Z}$ is an integral domain. Thus, $\beta_i = \beta_j$ so $\varphi$ is injective and $|S|<\infty \implies$ it is surjective.

What should I do next?

Answer 1

If $\mathbb Z [\alpha]$ is finitely generated, then there is some $n\in \mathbb N$, such that $1, \alpha, \alpha^2, \dots , \alpha^n$ generates $\mathbb{Z} [\alpha]$. (This follows as $(\alpha^i)_{i\in \mathbb{N}_0}$ generates $\mathbb Z [\alpha]$) Now write $\alpha^{n+1}= \sum_{i=0}^{n}a_i \alpha^i$, so $\alpha^{n+1}- \sum_{i=0}^{n}a_i \alpha^i=0$. Thus $\alpha$ is an algebraic integer.

Answer 2

Every $\beta_0,...,\beta_k$ is a polynomial with integer coefficients on $\alpha$. Take $n\in\mathbb{N}$ greater than all the dregrees of $\beta_1(\alpha),...,\beta_n(\alpha)$. Then $\alpha^n=\sum_{i=0}^{k}a_i\beta_i(\alpha)$ for some $a_i\in\mathbb{Z}$. The polynomial $p(x):=x^n-\sum_{i=0}^{k}a_i\beta_i(x)$ is monic with degree $n$ and such that $p(\alpha)=0$, so $\alpha$ is algebraic.