Noetherian integral domain such that $m/m^2$ is a one-dimensional vector space over $A/m$

Solution 1:

Recall that if $\mathfrak{m}$ is a maximal ideal of $A$ then $A_\mathfrak{m} / \mathfrak{m} A_\mathfrak{m} \simeq A / \mathfrak{m} = k(\mathfrak{m})$ and $\mathfrak{m} A_\mathfrak{m} / \mathfrak{m}^2 A_\mathfrak{m} \simeq \mathfrak{m} / \mathfrak{m}^2$. Then the condition implies that $\mathfrak{m} A_\mathfrak{m} / \mathfrak{m}^2 A_\mathfrak{m}$ is $k(\mathfrak{m})$-vector space of dimension $1$. Since $A_\mathfrak{m}$ is local, for Nakayama's lemma (Matsumura, Theorem 2.3) $\mathfrak{m} A_\mathfrak{m}$ is a non-zero cyclic $A_\mathfrak{m}$-module, i.e. a non-zero principal ideal of $A_\mathfrak{m}$. Now $A_\mathfrak{m}$ is a noetherian domain with principal maximal ideal and $A_\mathfrak{m}$ is not a field, then $A_\mathfrak{m}$ is a DVR (Matsumura, Theorem 11.2).

We have proved that $A_\mathfrak{m}$ is a DVR for every maximal ideal $\mathfrak{m}$ of $A$. From this point, it is easy to show that $A$ is a Dedekind domain.

Solution 2:

This is just a hint (since you are studying for a qualifying exam). I would suggest reading about Dedekind domains. For instance, read Dummit and Foote's Algebra. In particular, look at Section 16.3, and more concretely, the equivalence $(1) \longleftrightarrow (2)$ in Theorem 15.