The Ring $ \mathbf {Z}$ is noetherian

The ring $ \mathbf {Z}$ of integers is noetherian because every ideal of $ \mathbf {Z}$ is generated by one element.

Proposition 2.2.10   Every ideal of the ring $ \mathbf {Z}$ of integers is principal.

Proof. Suppose $ I$ is a nonzero ideal in  $ \mathbf {Z}$. Let $ d$ the least positive element of $ I$. Suppose that $ a\in I$ is any nonzero element of $ I$. Using the division algorithm, write $ a=dq + r$, where $ q$ is an integer and $ 0\leq r < d$. We have $ r=a-dq\in I$ and $ r<d$, so our assumption that $ d$ is minimal implies that $ r=0$, so $ a=dq$ is in the ideal generated by $ d$. Thus $ I$ is the principal ideal generated by $ d$. $ \qedsymbol$

Example 2.2.11   Let $ I=(12,18)$ be the ideal of $ \mathbf {Z}$ generated by $ 12$ and $ 18$. If $ n=12a+18b\in I$, with $ a,b\in\mathbf{Z}$, then $ 6\mid n$, since $ 6\mid 12$ and $ 6\mid 18$. Also, $ 6=18-12\in I$, so $ I=(6)$.

The ring $ \mathbf {Z}$ in SAGE is ZZ, which is Noetherian.

sage: ZZ.is_noetherian()
We create the ideal $ I$ in SAGE as follows, and note that it is principal:
sage: I = ideal(12,18); I
Principal ideal (6) of Integer Ring
sage: I.is_principal()
We could also create $ I$ as follows:
sage: ZZ.ideal(12,18)
Principal ideal (6) of Integer Ring

Proposition 2.2.7 and 2.2.10 together imply that any finitely generated abelian group is noetherian. This means that subgroups of finitely generated abelian groups are finitely generated, which provides the missing step in our proof of the structure theorem for finitely generated abelian groups.

William Stein 2012-09-24