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For a ring $R$ with maximal ideal $I$, is there an extension for powers of $I$ like for $n \in \mathbb{N}$ such as $$ rs \in I^n, s \not \in I \implies r \in I^n $$

I was at first looking for general prime ideals, but Not primary ideal having a prime radical which yields a prime counter-example, but the comments on the question mention maximal ideals being okay, hence why I edited the question.

Under the same assumption we clearly have $r \in I$, but I'm stumped on this extension. It clearly holds true in $\mathbb{Z}$ by factoring the elements into primes, but in a general ring we might not have a factorisation into irreducibles, so I'm wondering whether it's possible to show/disprove this.

My current thoughts are that quotienting by $I^n$, $$ (r+I^n)(s+I^n) = 0 + I^n $$ but we can't then say one of the LHS terms must be zero since $R/I^n$ isn't necessarily an integral domain... Which makes me think there's a possibly very simple counterexample that I'm missing, or a proof which has a few more details.

George
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    This is a special case of Euclid's Lemma (ideal form), viz. $\tag{}$ $(s,I)=(1) \Rightarrow \color{#c00}{(s,I^n)=(1)},,$ so $,I^n \supseteq rs, rI^n \Rightarrow I^n \supseteq (rs,rI^n) = r\color{#c00}{(s,I^n)} = (r).$ $\tag{}\ \ $ – Bill Dubuque Apr 23 '23 at 00:19
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    The use of Euclid's lemma is in the second part right? Does the first implication uses $I$ maximal and follow the reasoning in Mike Daas' answer? – George Apr 23 '23 at 19:41
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    On second thoughts the first part is the other marked duplicate - I find Mike Daas' method is quicker in this specific case, but realise it may answer fewer other questions. – George Apr 23 '23 at 19:56
  • Yes, the first inference uses the 2nd dupe (powers of coprimes are coprime), and the 2nd uses Euclid's Lemma. The argument in Mike's answer is the same as the elementwise (Bezout) form of Euclid's Lemma in the first linked dupe. – Bill Dubuque Apr 24 '23 at 17:26
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    I'm just wondering about the downvote - are duplicates a reason to downvote, or is there something I can do to improve my question? It seems odd to not delete/improve a question with a negative score, but I'm unable to delete since there have been answers, and not sure what I could do to improve it. – George Oct 15 '23 at 17:13
  • I see two upvotes and two downvotes. I wouldn't worry about it. – Gerry Myerson Oct 15 '23 at 20:47
  • For the record, the original dupe links mentioned above are here and here. It is sad that some users are destroying site organization efforts by reopening abstract dupes (likely due to OP's meta question) $\ \ $ – Bill Dubuque Oct 16 '23 at 13:41

1 Answers1

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This is true; from the facts that $s \notin I$ and that $I$ is maximal, you can deduce the existence of some $t \in R$ with $st + 1 \in I$. By raising this expression to the power $n$, you can verify that this leads to some element $t' \in R$ such that $st' + 1 \in I^n$. Hence $r(st' + 1) = rst' + r \in I^n$. Since $rs \in I^n$, it follows from this that $r \in I^n$.

Mike Daas
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 23 '23 at 00:20
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    I don't see how my answer is a duplicate. Sure, you propose an alternative proof using a lemma discussed in a decade old post, but nothing here is obviously directly related to any of that. Some results can be viewed and proved in more than one way. I appreciate your efforts but I would humbly invite you to briefly reconsider your notion of what constitutes a "duplicate". Cheers. – Mike Daas Apr 23 '23 at 00:34
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    @Mike Your argument is essentially the same as the linked (standard) proof of Euclid's Lemma, but yours is done elementwise (vs. ideal-wise). Such instantiations of Euclid's Lemma are ubiquitous (and a FAQ here). See my comment on the question, and see also abstract duplicates. – Bill Dubuque Apr 23 '23 at 00:38
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    To the untrained eye, an "essentially the same" proof can seem very different and can provide vastly different degrees of clarity. Similar for an "essentially the same" question. If you don't see this, then we can just agree to disagree. Cheers. – Mike Daas Apr 23 '23 at 00:46
  • Alas, those "untrained eyes" will remain untrained if users destroy dupe links to answers explaining how this is simply an ideal form of Euclid's Lemma - as has been done here by reopening it. – Bill Dubuque Oct 16 '23 at 15:44