ACC on principal ideals implies factorization into irreducibles. Does $R$ have to be a domain?

Question

I am following an argument in chapter zero of Eisenbud's Commutative Algebra book. It is not clear whether or not he is assuming that $R$ is a domain. If I start the proof assuming $R$ is not necessarily a domain this is what I get:

Suppose that $R$ (commutative ring with identity) satisfies ACC on principal ideals but there exists a non-unit $a_1\in R$ such that $a_1$ does not admit a factorization into irreducibles. As $a_1$ is not irreducible, $a_1=bc$, with both $b$ and $c$ non-units.

Both $b$ and $c$ cannot have factorization into irreducibles since in that case we would have a factorization into irreducibles of $a_1$. So WLoG, assume that $b$ has no factorization into irreducibles.

Set $a_2=b$. Then we have $(a_1)\subsetneq (a_2)$.

This is where I am stuck. It is clear that $(a_1)\subset (a_2)$, since $a_1=a_2 c$. But I get stuck showing that $(a_1)\neq (a_2)$. What is frustrating me is that I can show with ease that $(a_1)\neq (a_2)$ if $R$ is a domain (or even if $a_1$ is not a zero divisor) but not otherwise.

The rest of the argument is straight forward. Continue picking the $a_i$ inductively and get a non-terminating ascending chain of principal ideals; contradicting ACC.

Answer 1

It is indeed not too clear from the context, but Eisenbud is assuming $R$ to be a domain in this argument. Here's a counterexample if $R$ is not a domain:

Take $R = \prod_{i=1}^n \mathbb{F}_2$ (the field with $2$ elements), with $n > 1$. Then $R$ has only one unit (namely $1$), and every nonunit admits nontrivial factorizations into nonunits (if $a = (a_1, \ldots, a_n)$ is not a unit, then $a_i = 0$ for some $i$, and then $a = a \cdot (1, \ldots, 1, 0, 1, \ldots, 1)$ with $0$ in the $i^{\text{th}}$ spot). This shows that $R$ has no irreducible elements. But $R$ is Noetherian, hence satisfies ACC on all ideals, and thus in particular also on principal ideals.

(Note: the exact same construction works if $R$ is any finite direct product of fields, although there are more units.)

Answer 2

Hint: Try making one of the factors of $a_1$ irreducible by extending its principal ideal to a maximal principal ideal. If you can pull off an irreducible factor at each step, then the other factors will give you a strictly increasing chain of principal ideals.

For instance, if $(a_1)=(a_2)$ then you have $a_2=xa_1$ and $a_1=ya_2$ with $y$ irreducible. Then $a_1=yxa_1$ implies $yx=1$ implies $y$ is a unit, a contradiction.

Answer 3

It is indeed false with the definition of "irreducible" you are using. This is what people who research factorization in commutative rings with zero divisors call "very strongly irreducible". (For more information and examples, I would suggest starting with the paper "Factorization in Commutative Rings with Zero Divisors" by Anderson and Valdes-Leon.) We call an element x "irreducible" if x = yz implies (x) = (y) or (x) = (z). With this weaker definition, the result is true. Proof: Suppose not. Then by ACCP there is a nonunit x that is not a product of irreducibles but every proper divisor of it is. Because x is not itself irreducible, we have x = yz with y and z proper divisors, hence products of irreducibles. So x is a product of irreducibles, a contradiction.