Why is $\mathbb{Z}[\alpha ]$ not finitely generated as $\mathbb{Z}$-module? 
Assume that $\alpha \in \mathbb{C}$ is an algebraic number which is not an algebraic integer. My question is why $\mathbb{Z}[\alpha]$ is not finitely generated as  $\mathbb{Z}$-module.

Clearly there exists $ a_i \in \mathbb{Z}$, which are not all zero and $n>0$ such that $a_n\alpha^n+a_{n-1}\alpha^{n-1}+\dots + a_1\alpha + a_0=0$. Shouldn't this imply that $\mathbb{Z}[\alpha]$ is finitely generated? The motivation for this question is that it may be shown that $\alpha \in \mathbb{C}$ is an algebraic integer if and only if $\mathbb{Z}[\alpha ]$ is finitely generated as a $\mathbb{Z}$-module. Why does the coefficient $a_n$ has to be one for the module to be finitely generated? And, to check that I haven't misunderstood the terminology, $\mathbb{Z}[\alpha ]$ is equal to all polynomials in $\alpha$ over $\mathbb{Z}$, right?
 A: The standard proof that $\mathbb Z[\alpha]$ is finitely generated if $\alpha$ is an algebraic integer uses the fact that if
$\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_0=0$ then $\alpha^n=-(a_{n-1}\alpha^{n-1}+\cdots+a_0)$, so for any $m\geq n$ we have
$$\alpha^m=-(a_{n-1}\alpha^{m-1}+\cdots+a_0\alpha^{m-n})$$
i.e. $\alpha^m$ is in the module generated by $1,\alpha,\ldots,\alpha^{m-1}$. By induction this shows us that $\alpha^m$ is in the module generated by $1,\alpha,\cdots,\alpha^{n-1}$, hence $\mathbb Z[\alpha]$ is generated by these elements.
If $a_n$ is not a unit this all breaks down, because we no longer have a way to write $\alpha^n$ in terms of $a_{n-1}\alpha^{n-1}+\cdots+a_0$.
A: What you want is to be able to express $\alpha^n$ and all other higher powers of $\alpha$ as a $\mathbb{Z}$-linear combination of the powers from $0$ to $n-1$. But if $a_n$ is not $\pm 1$, you will not be able to do this.
A: Let me modify M Turgeon's answer slightly, so as not to assume that you want to use the powers $\alpha^0$ through $\alpha^{n-1}$ as your generators.  Suppose you had some more complicated or more clever finite set of generators $\{g_1,\dots,g_k\}$ for $\mathbb Z[\alpha]$.  Each $g_i$ is a polynomial in $\alpha$ with integer coefficients, but we know nothing more about these polynomials, not even how many of them there are.  Nevertheless, we can fix a natural number $N$ strictly greater than the degrees of all these $g_i$'s, and, because $\alpha^N$ is in $\mathbb Z[\alpha]$, we must be able to express $\alpha^N$ as a linear combination of the alleged generators $g_i$ with integer coefficients, say $\alpha^N=\sum_{i=1}^kc_ig_i$.  But now, transposing everything to the left side in this equation, you have a monic polynomial equation, with integer coefficients, satisfied by $\alpha$.  (It's monic because $\alpha^N$ has coefficient $1$, unaffected by any of the other terms, as these have lower degree.)  So $\alpha$ is an algebraic integer, contrary to your assumption.
