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The first answer to this question shows an example of an infinite ascending chain of ideals in $F[X_1,X_2,...]$ where $F$ is a field.

Now consider the ideal $I=(x_1-x_2^2,x_2-x_3^2,...)$ and an integral domain $R$ (commutative, with unit) defined as $R=F[x_1,x_2,x_3,...]$. If we quotient $R$ by $I$ we get that $x_1=x_2^2=x_3^{2{^2}}=x_4^{2{^3}}=...=x_k^{2^{k-1}}=...$ in $R/I$ and $(x_1)\subsetneq(x_2)\subsetneq(x_3)\subsetneq...$ because $x_{k+1}$ is a proper divisor of $x_k$ for every $k\in\Bbb N$

Why does having such a chain imply that $R/I$ that the algorithm (*) terminates?

Algorithm (*):

INPUT: element a in R
OUTPUT: decomposition of a in irreducible factors

if a is irreducible return a
else a=a1b1 where neither a1 nor b1 are units
if a1 and b1 are irreducible return {a1,b1}
else a1=a2b2...

The algorithm continues as long as a factor is reducible.

John Cataldo
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    The algorithm terminates on a noetherian domain. – egreg Apr 11 '18 at 10:16
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    In the example, the algorithm doesn't terminate since you pass from $x_1$ to $x_2^2$ to $x_3^8$ and so on without ever hitting irreducibles. – rschwieb Apr 11 '18 at 11:05

1 Answers1

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As noted in the comments, this algorithm will not terminate for certain choices of $a$. This is due to the fact that $R/I$ is not noetherian, which is equivalent to the existence of an infinite ascending chain of ideals.

It will terminate for other choices of $a \in R/I$ though, because being noetherian does not mean that if you start with any ideal there is an infinite ascending chain. For example if you take $a = 1$, the algorithm will clearly terminate.

However, given your infinite ascending chain you can find an $a \in R/I$ such that the algorithm never terminates. For example in your case if you pick any of the generators of those principal ideals $a = x_i$, the algorithm will never terminate because $x_i^{2^{i-1}} = x_{i+1}^{2^{i-1}}x_{i+1}^{2^{i-1}}$.

One way to reinterpret the algorithm is that if you consider the ideal generated by $a$, $(a) \subsetneq (b)$ because $a = a_1b$ with $a_1$ and $b$ non-units. This means $a \in (b)$ but $b \not\in (a)$ as if it were then $b = a_2 a$ for some element $a_2$ which is not a unit because $a_1$ is not. But then $a = a_1a_2 a$, which means $(a) = (a_1a_2a)$. This is only possible in your case if $a_1a_2$ is a unit, but this is not the case because $a_1$ and $a_2$ are not units. The algorithm will never terminate if we can always choose $b$ as above so that $(b) \subsetneq (c)$ for some other non-unit $c$. So having an $a$ such that this algorithm does not terminate implies the existence of an infinite ascending chain of principal ideals.

We noted above that an infinite ascending chain of principal ideals implies the non-termination of this algorithm for some $a$. So having an $a$ such that this algorithm does not terminate is equivalent to the existence of an infinite ascending chain of principal ideals (beginning with $(a)$).

rpl19
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