Integral closure of $\mathbb{Z}$ in $\mathbb{C}$ is not finitely generated as a $\mathbb{Z}$-module?

Question

Let

$$ \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} $$

be the integral closure of $ \mathbb{Z} $ in $ \mathbb{C} $. Prove that $ \mathbb{Z}^{'}_{\mathbb{C}} $ is not finitely generated as a $ \mathbb{Z} $-module.

My attempt: Suppose $ \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} $ for some $ \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} $. I know that given $ \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} $ with minimal polynomials of degrees $ m $ and $ n $ respectively, then we can construct a monic polynomial with integral coefficients $ P(z)=\prod(z-(\alpha_{i}+\beta_{j})) $ where $ \alpha_{i},\beta_{j} $ are the conjugates of $\alpha $ , $\beta $ and therefore the minimal polynomial of $ \alpha + \beta $ has degree $ \leq mn $.

I was thinking that then the degrees of the minimal polynomials of the elements of $ \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} $ could be somehow bounded by a certain constant, whereas $ \mathbb{Z}^{'}_{\mathbb{C}}$ contains elements whose degrees of the minimal polynomials are arbitrarily large, take $ 2^{\frac{1}{n}} $ for example,$ n\geq 2$. I think this works when we are asked to prove that the set of all algebraic numbers $ \overline{\mathbb{Q}} $ is not a finitely dimensional $\mathbb{Q}$-vector space, but I don't know how to proceed in my case.

I would really appreciate any ideas, hints or solutions. Thank you very much!

Answer 1

Consider any candidate set of generators $\alpha_1,\ldots, \alpha_n$. Then we know that $\Bbb Q(\alpha_1,\ldots, \alpha_n)$ has degree bounded by

$$M=1+\prod_{i=1}^n [\Bbb Q(\alpha(i):\Bbb Q]$$

However, take any prime $p>M+1$. Then the primitive $p^{th}$ root of $1$, $\zeta_p$ has degree $\varphi(p)=p-1\ge M$, and we know that $\Bbb Z[\zeta_p]$ is a finitely generated $\Bbb Z$ module of rank $p-1$ contained in $\Bbb Z[\alpha_1,\ldots, \alpha_n]$, which implies that the rank of $\Bbb Z[\alpha_1,\ldots, \alpha_n]$ is simultaneously less than and greater than $M$, a contrdiction.

Answer 2

Here is an easy solution: Assume that $\mathbb{Z}_\mathbb{C}^\prime$ is finitely generated as a $\mathbb{Z}$-module. Assume its rank is $m$, and let $n>m$. The polynomial $f(x)=x^n-2$ is irreducible by Eisenstein's criterion. Let $\alpha$ be a root of $f$ (so $\alpha,\alpha^2,\dotsc$ are in $\mathbb{Z}_\mathbb{C}^\prime$.) We know from basic field theory that $1,\alpha,\alpha^2,\dotsc,\alpha^{n-1}$ are a $\mathbb{Q}$-basis for the field $K(\alpha)$, viewed as a $K$-vector space, so certainly these elements are linearly independent over $\mathbb{Z}$ (any nontrivial $\mathbb{Z}$-linear combination of them is also a $\mathbb{Q}$-linear combination.) It follows that the rank of $\mathbb{Z}_\mathbb{C}^\prime$ is at least $n>m$, a contradiction.