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

I am having trouble doing the following question (I'm studying for quals, it isn't homework)

If $A$ is a noetherian integral domain such that for every maximal $m\subset A$, the quotient $m/m^2$ is a one-dimensional vector space over the field $A/m$

(a) Prove every nonzero prime ideal is maximal.

(b) Prove $A$ is integrally closed.

There is a hint which says that one should localize at maximal ideals. My problem is that I'm not really sure how to use the $m/m^2$ condition. A solution or hint in the right direction using a minimal amount of commutative algebra would be much appreciated (but clearly a decent amount should be used).

-
As in the hint, you can assume $A$ is local. The dimension condition on $m/m^2$ implies $m$ is principal (use Nakayama's lemma); thus $m=(x)$, say. The Krull intersection theorem implies any prime in $m$ is then either $m$ or $0$. That's part (a). Part (b) follows because now $A$ is dimension one, and the condition on $m/m^2$ implies it's a DVR. – user641 Aug 18 '11 at 19:37

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.
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.