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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.

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2 Answers 2

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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.)

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I was trying other counterexamples and getting stuck showing that some element cannot be factored into irreducibles. This seems like a non-trivial task in general. This counterexample is good, because there are no irreducibles, so cannot have factorization into irreducibles. –  john w. Jun 22 at 19:47
    
@johnw. To alternatively see why nonunits are not irreducible, just notice that $a=aa$ for every $a$ in the ring. (This reason does not generalize to constructions outside of direct products of $F_2$, though.) And actually, this ring is also a principal ideal ring (being a finite product of PIRs.) –  rschwieb Jun 24 at 12:56

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.

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This is my argument for when $R$ is a domain. Since then I can say $a_1=yxa_1$ implies $a_1(1-xy)=0$ which implies $1=xy$. How do you show that $a_1=yxa_1$ implies $yx=1$ in the case $R$ is not a domain? –  john w. Jun 22 at 18:51

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