# Why is a finitely generated $\mathbb Z$-module a finitely generated $\mathcal O_K$-module

There is something I don't understand in Neukirch's Algebraic number theorey. He said that:

"every ideal is a finitely generated $$\mathbb Z$$-module by (2.10) and therefore a fortiori a finitely generated $$\mathcal O_K$$-module."

I know that every ideal is a finitely generated $$\mathbb Z$$-module but I fail to understand why then it is a finitely generated $$\mathcal O_K$$-module. Could someone tell me something? Appreciate that.

• Because an $\mathcal O_K$-module is also a $\mathbf Z$-module (by restriction of scalars), and ‘he who can do the more can do the less’. Oct 13, 2018 at 13:12

If $$R \subset S$$ are rings and $$M$$ is an $$S$$-module which is finitely generated as an $$R$$-module, then it is finitely generated as an $$S$$-module too. Indeed, if $$m_1, \ldots, m_n$$ is a list of generators, then everything in $$M$$ is a finite $$R$$-linear combination $$\sum r_i m_i$$, so it's also a finite $$S$$-linear combination because the coefficients $$r_i$$ are also in $$S$$.
• (not needed for your particular question, but this generalizes to the case that $f: R \to S$ is some not necessarily injective map and we view $M$ as an $R$-module via $f$.) Oct 13, 2018 at 13:11