# If $\mathbb{Z}[\alpha]$ is finitely generated, can we write $\mathbb{Z}[\alpha]=\mathbb{Z}+\mathbb{Z}\alpha+ \cdots + \mathbb{Z}\alpha^k$?

This is a very basic question in algebraic integers. I was trying to prove an implication of algebraic integers in my own way.

(i) $$\alpha\in\mathbb{C}$$ is an algebraic integer.

(ii) There is a finitely generated $$\mathbb{Z}$$-sub-module of $$\mathbb{C}$$ which contains the ring $$\mathbb{Z}[\alpha]$$.

Arguments in proving $$(ii)\Rightarrow (i)$$ Any notes on algebraic integers give proof of this which I can see but my question is about following arguments:

(1) Let $$S\subset \mathbb{C}$$ be a finitely generated $$\mathbb{Z}$$-sub-module which contains $$\mathbb{Z}[\alpha]$$.

(2) Since submodule of finitely generated module over PID is finitely generated, so $$\mathbb{Z}[\alpha]$$ is such!

(3) Then $$\mathbb{Z}[\alpha]=\mathbb{Z}+\mathbb{Z}\alpha+ \cdots + \mathbb{Z}\alpha^k$$ for some $$k\ge 1$$.

(4) Hence $$\alpha^{k+1}=c_0 + c_1\alpha+\cdots + c_k\alpha^k$$.

(5) Hence $$\alpha$$ satisfies a monic polynomial $$x^{k+1}-(c_kx^k+\cdots + c_0)$$ over $$\mathbb{Z}$$.

Is this proof correct?

• (3) $\Bbb Z[\alpha]$ is finitely generated, but is it necessary that it has a set of generators in the form of $\{1, \alpha, \alpha^2, \dots\}$? At least for me it is not that obvious :(
– xbh
Commented Nov 3, 2022 at 7:09
• Notice that the body of your question does not include a question, or rather, it has a question different from the one in the title. Commented Nov 3, 2022 at 7:17

If $$\mathbb{Z}[ \alpha ]$$ is finitely generated,then there are polynomials $$f_1,f_2,\cdots,f_m \in \mathbb{Z}[x]$$,such that $$\mathbb{Z}[ \alpha ]=\mathbb{Z}f_1(\alpha)+\mathbb{Z}f_2(\alpha)+\cdots+\mathbb{Z}f_m(\alpha)$$.
Let $$n$$=max{ deg $$f_1,\cdots,$$deg $$f_m$$},it's easily see that $$\mathbb{Z}[ \alpha ]=\mathbb{Z}+\mathbb{Z}\alpha+\cdots+\mathbb{Z}\alpha^n$$.