The rational algebraic integers are exactly the integers

Given: The hypotheses and objects in the Statement.

Let $A\to B$ be a homomorphism of commutative rings. An element $b\in B$ is integral over $A$ when it is a root of a monic polynomial in $A[X]$. The extension is integral when every element is integral. An algebraic integer is a complex number integral over $\mathbb{Z}$. (Integral elements over a commutative ring and algebraic integers).

Let $f=a_n{x}^n+\cdots +a_{1}x+a_0\in \mathbb{Z}[x]$ with $a_n\mathrm{\ne }0$. If a reduced rational number $r\mathrm{/}s$, where $r,s\in \mathbb{Z}$, $s>0$, and $\gcd (r,s)=1$, is a root of $f$, then $r\mid a_0\text{and}s\mid a_n\mathrm{.}$. (Rational root theorem).

$(\mathbb{Q},+,\cdot ,0,1)$ with the operations of def-rat-operations is a field: a commutative ring with $1\mathrm{\ne }0$ in which every nonzero element has a multiplicative inverse. (The rationals form a field).

The map $j(k)=[(k,1)]$ is injective and preserves addition, multiplication, and order. Composing with lem-nat-embeds-int embeds $\mathbb{N}$ in $\mathbb{Q}$; we write $k$ for $j(k)$ throughout. (The integers embed in the rationals).