Yes. In any commutative ring, the set of zero-divisors is the union of the prime ideals in $\operatorname{Ass}A$. An artinian ring has Krull dimension $0$, hence a local artinian ring has only $1$ prime ideal, and $\operatorname{Ass}A=\operatorname{Max}A=\operatorname{Spec}A$ is the set of zero-divisors in $A$.
Here's a finer statement:
Lemma. Let $A$ be a ring and consider the following properties:
- $A$ is Artin local,
- $A$ has a nilpotent maximal ideal,
- $A$ has a unique proper radical ideal,
- $A$ has a unique prime ideal, and
- the nilradical of $A$ is maximal.
Then 1$\Rightarrow$2$\Rightarrow$3$\Leftrightarrow$4$\Leftrightarrow$5, and if $A$ is Noetherian, then all five stated properties are equivalent.
Without Noetherian assumptions, the implication 1$\Rightarrow$2 is strict [ref] and also 2$\Rightarrow$4 [ref].
Proof. 1$\Rightarrow$2. In an Artin ring the nilradical equals the Jacobson radical [AM, 8.2]. Since $A$ is Noetherian [AM, 8.5], the nilradical is nilpotent.
2$\Rightarrow$3. Let $\mathfrak{m}\subset A$ be a nilpotent maximal ideal and let $I\subset A$ be a radical proper ideal. Then $\mathfrak{m}^r\subset I$ for some integer $r\geq 1$, whence $\mathfrak{m}=\sqrt{\mathfrak{m}^r}\subset \sqrt{I}=I$. Thus $\mathfrak{m}=I$ by maximality of $\mathfrak{m}$.
3$\Rightarrow$4. Every prime ideal is proper radical.
4$\Leftrightarrow$5. The nilradical equals the intersection of all prime ideals.
4$\Rightarrow$3. A radical ideal equals the intersection of the prime ideals containing it.
Now suppose $A$ is Noetherian.
4$\Rightarrow$1. $A$ is local. On the other hand, it is of dimension $0$ and Noetherian, and thus, Artin [AM, 8.5]. $\square$
References
AM. Atiyah, Macdonald, Introduction to Commutative Algebra.
Best Answer
In a reduced ring the set of zerodivisors equals the union of minimal primes; see here. Since the depth is zero the maximal ideal is contained in this union, so it is minimal. This shows that $\dim A=0$. (In fact, we get $\mathfrak m=(0)$, so $A$ is a field.)