Help with a problem about Artinian rings from Christian Peskine's book

Question

I am stuck with problem 3, chapter 4, from the book of Peskine, An Algebraic Introduction to Complex Projective Geometry.

Let $A$ be a Noetherian ring. Assume that if $a \in A$ is neither invertible nor nilpotent, then there exist $b \in A$ such that $b$ is not nilpotent and that $ab = 0$. Show that $A$ is Artinian.

I have tried to prove that every prime is also maximal since I already have the Noetherian property and I can't see how to prove directly the dcc.

Answer 1

Here's a high-level proof - perhaps someone else will find a more elementary one. $\DeclareMathOperator{\nil}{nil}$ $\DeclareMathOperator{\Ass}{Ass}$ $\DeclareMathOperator{\Spec}{Spec}$ $\DeclareMathOperator{\Min}{Min}$

Set $B := A/\nil(A)$. The assumption on $A$ implies that every nonunit in $B$ is a zerodivisor (notice: this is stronger than saying that every nonunit in $A$ is a zerodivisor). Thus every prime ideal $\mathfrak p$ of $B$ is contained in the union of the associated primes of $B$. But $B$ is reduced, so every associated prime is minimal. Then $\displaystyle \mathfrak p \subseteq \bigcup_{\mathfrak q \in \Min(B)} \mathfrak q \implies \mathfrak p \subseteq \mathfrak q$ for some $\mathfrak q \in \Min(B) \implies \mathfrak p = \mathfrak q$. Thus every prime ideal of $B$ is minimal, so $\dim B = 0$, and so $\dim A = 0$.

Answer 2

Here is my answer. Assume that $A$ is not artinian. Then there exists a maximal ideal not contained in the union of all minimal primes. Hence there exists an element $a$ not invertible and not contained in any minimal prime.

Consequently if $ab=0$ then $b$ is contained in all minimal primes and is therefore nilpotent, contradicting the hypothesis.