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Let $R$ be an integral domain. $r \in R$ has a factorization into irreducibles if there exists a sequence of irreducibles $q_1,\dots,q_n$ such that $r = q_1\dots q_n$. $R$ is a domain with factorization if every non-zero element has a factorization.

Ascending chain condition. Let $X$ be a set and $\leq$ a partial order relation on $X$. $X$ satisifies $a.c.c.$ if for every chain $x_1 \leq x_2 \leq \dots$, the chain stabilizes eventually.

For any integral domain $R$, it is clear that if $R$'s ideals principal ideals satisfying a.c.c. (via inclusion), then $R$ is a domain with factorization. However, intuitively the converse does not hold. If we can find $r = \prod_{i = 1}^{\infty} r_i$ where each $r_i$ is not a unit, and if $r$ also admits a finite factorization, that would be a good example. I cannot find any, though. Could anyone give some hint?

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    Seek for `non-noetherian U.F.D.`s – 2017-02-27
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    The rings you are considering are usually called **atomic**. Every Noetherian domain is atomic, and for this reason @Bernard 's comment applies. – 2017-02-27
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    Consider the infinite product of a commutative ring that contains infinitely many maximal ideals. Then you can easily get an ascending chain that never stops. – 2017-02-27
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    @Bernard I made a change in my question. What I was seeking is a.c.c. on principal ideals rather than ideals. I am not sure if the same thing applies. – 2017-02-27
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    Possible duplicate of [Example of FD not satisfying ACCP](http://math.stackexchange.com/questions/1941913/example-of-fd-not-satisfying-accp) – 2017-02-28

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A simple example is the polynomial ring in a countable number of indeterminates over a field $K$, $K[X_1, X_2,\dots, X_n,\dots]$. It is a non-noetherian U.F.D., which is the direct limit of the system of noetherian U.F.D.s $\bigl(K[X_1, X_2,\dots, X_n], i_n\bigr )_{n\ge 1}$, where $i_n$ is the inclusion $K[X_1, X_2,\dots, X_n]\hookrightarrow K[X_1, X_2,\dots, X_{n+1}]$.

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For the modified question (an integral domain satisfying the a.c.c. for principal ideals which is not a U.F.D.), just take a noetherian domain which is not a U.F.D.

For instance, the ring of integers of a convenient quadratic extension of $\mathbf Q$ which is not principal. Such a ring of integers is a Dedekind domain, and it is principal if and only if it is a U.F.D. Such an example is the ring $\mathbf Z[\sqrt{-5}]$.

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    Sorry, I made a mistake in asking my question. The example I was looking for concerns a.c.c. on principal ideals. It is true that in $k[x_1, \dots, x_n, \dots]$ I can find an increasing chain of ideals, i.e. $(x_1) \subset (x_1, x_2) \subset \dots$, but I am not sure if the same thing applies to *principal ideals*. – 2017-02-27
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    Ah I got the idea – 2017-02-27
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    @Misheng Liu: No : U.F.D.s satisfies the a.c.c. for principal ideals. Actually it is equivalent for an integral domain to be a U.F.D. and to satisfy the a.c.c. for principal ideals + irreducible elements generate prime ideals. Please see my updated my answer – 2017-02-27