showing isomorphism

Question

Let { $\omega_1 ,\omega_2,...\omega_n$ } be an integral basis of $\mathcal{O_k}$ , so that $$\mathcal{O}_k=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2+...+\mathbb{Z}\omega_n$$ and $$(m)=m\mathbb{Z}\omega_1+m\mathbb{Z}\omega_2+...+m\mathbb{Z}\omega_n $$

Now I do not understand why it follows that $$\mathcal{O}_k/(m)\cong\mathbb{Z}/m+...+\mathbb{Z}/m$$

Thanks for the help .

Answer 1

This has nothing to do with the special setting of number fields, but is instead just an algebraic fact.

Let $A$ be a commutative ring and $M_1, …, M_n$ be $A$-modules and $M = M_1 × … × M_n$. Then for any Ideal $\mathfrak a ⊆ A$, $$M/(\mathfrak a M) \cong M_1/(\mathfrak a M_1) × … × M_n/(\mathfrak a M_n).$$ This is a simple application of the isomorphism theorem for the canonical projection $$M → M_1/(\mathfrak a M_1) × … × M_n/(\mathfrak a M_n).$$

Apply this for $A = ℤ$, $\mathfrak a = (m)$ and $M_1 = ℤω_1$, …, $M_n = ℤω_n$, noting that, as abelian groups,

• $\mathcal O_k \cong ℤω_1 × … × ℤω_n$, and
• $ℤω_i/mℤω_i \cong ℤ/m$ for $i = 1, …, n$.