I was wondering why ring of integers $\mathcal O_K$ for field $K$ is called ring of integers.

Definition says that elements in this ring will be a solution for monic equation with coefficients of rational integers. But wouldn't natural definition of ring of integers be ring of elements cannot be expressed as $\frac{a}{b}$ where $a$ is not divisible by $b$?

So what properties of $\mathbb{Z}$ does this ring inherit?

Edit: One can just answer about what properties get inherited from $\mathbb{Z}$.

  • 1
    $\begingroup$ Just a comment on your "natural definition"; how are you defining "not divisible by" in a field? The existence of inverses means that everything non-zero divides everything else, in the sense that for all $a,b\in K$ with $b\ne0$, there exists $c\in K$ (i.e. $c=a/b$) such that $a=bc$. $\endgroup$
    – mdp
    Sep 8, 2014 at 12:38
  • $\begingroup$ @MattPressland You're right. But then, I was thinking inside the ring, not field here. But then how I would be able to define this would be an issue though. $\endgroup$
    – FAM
    Sep 8, 2014 at 12:39
  • $\begingroup$ I see - but yes, then you have a logical loop. $\endgroup$
    – mdp
    Sep 8, 2014 at 13:19
  • 2
    $\begingroup$ This boils down to motivating the definition of an algebraic integer. $\endgroup$ Sep 8, 2014 at 13:43

1 Answer 1


Just like $\mathbb Z$ is integrally closed (due to rational root theorem), one can show that $\mathcal O _K$ is also integrally closed, that is, if $\beta \in K$ is the root of a monic polynomial $f$ with coefficients in $\mathcal O _K$, then $\beta \in \mathcal O _K$. And they both are finitely-generated $\mathbb Z$-modules (have integral basis).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .