Smallest subring of $\mathbb{C}$ containing $\alpha$

Question

Let $\alpha\in\mathbb{C}$, and let $\mathbb{Z}[\alpha]$ be the smallest subring of $\mathbb{C}$ containing $\alpha$; that is, $\mathbb{Z}[\alpha]=\cap\ S$, where $S$ ranges over all those subrings of $\mathbb{C}$ containing $\alpha$. Prove that $$\mathbb{Z}[\alpha]=\{f(\alpha): f(x)\in\mathbb{Z}[x]\}$$

My attempt:

I think I know how to solve this problem but don't know how to present it as a "mathematical" proof.

Since $\mathbb{Z}[\alpha]$ is a subring of $\mathbb{C}$, it contains $0$ and $1$.

Since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all the natural numbers. It must also contain the additive inverses of the natural numbers, i.e, the negative integers. Thus $\mathbb{Z}[\alpha]$ contains $\mathbb{Z}$.

Since $\mathbb{Z}[\alpha]$ is closed under multiplication, it must contain all elements of the form $m\alpha^n$ where $m\in\mathbb{Z}$ and $n\in\mathbb{N}\cup\{0\}$.

Finally, since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all elements of the form $\sum m\alpha^n$ where $m$ and $n$ range over $\mathbb{Z}$ and $\mathbb{N}\cup\{0\}$ respectively.

How do I finish this proof and make it "mathematically presentable"?

Answer 1

It is easy to prove that the set you defined is a subring of $\mathbb{C}$, so $\mathbb{Z}[\alpha]$ must be contained in it by definition. On the other hand, we know that $\mathbb{Z}[\alpha]$ is closed under multiplication, sums, and contains $0$, $1$ and $\alpha$. Therefore, it must contain all polynomials in $\alpha$. Thus, $\mathbb{Z}[\alpha]$ contains the set you gave, concluding the proof.