0
$\begingroup$

Let $R$ be a commutative ring. Could anyone advise me on how to prove $R$ has $\text{ACCP}$ (Ascending chain condition for principal ideals) iff every collection of principal ideals of $R$ has maximal element?

Hints will suffice, thank you.

$\endgroup$
2
$\begingroup$

It is easy to see how the second implies the first. Ascending chains are certainly collections, so if they always have maximal elements then ACCP holds.

The first implies the second by definition of maximal element. Suppose towards a contradiction that R satisfies ACCP yet there is some collection of principal ideals with no maximal element. Then any ideal in this collection is not maximal. Thus there is another ideal in the collection which contains it, and so on. Which would constitute an infinite strictly increasing chain of principal ideals.

$\endgroup$
  • $\begingroup$ Thanks for great advice. Could you also advise me on how to prove if $R$ is an integral domain and has $\text{ACCP},$ then $R[X]$ has $\text{ACCP} \ ?$ $\endgroup$ – Alexy Vincenzo Mar 16 '14 at 6:19
  • $\begingroup$ This means $R[X]$ is also an integral domain... $\endgroup$ – Alexy Vincenzo Mar 16 '14 at 6:22
  • 1
    $\begingroup$ The integral domain hypothesis is important here. A principal ideal in R[X] is generated by a polynomial with coefficients in R. Note that since there are no zero divisors, the formula $deg(p\cdot q)=deg(p)+deg(q)$ holds. Hence given any principal ideal generated by a polynomial $p$ of degree $n$, there is a factorization that reduces the degree or not. If there is, keep going until the degree can't be made lower by factorization. If not, then all factorizations are just $r\cdot q$ for some $r\in R$, and you know that this eventually terminates because R satisfies ACCP. $\endgroup$ – Fred Byrd Mar 16 '14 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.