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Let $R$ be a commutative ring (not necessary Noetherian), $Q$ its total ring of fractions, and $I$ a finitely generated ideal of $R$ such that $\forall$ $a \in I$ we have $(a:_R I) = a$. My question is how to prove that this implies $(R:_Q I) = I$ (equivalently $Gr_RI\ge 2$, with $Gr_R$ is the polynomial grade as defined in the Chapter 5, page 149, of the book Finite Free Resolutions by D. G. Northcott.)

user26857
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Med S.
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  • I may have been negligent in the statement of the question. the element $a$ should be neither zero nor a divisor of zero. The original question (which is not really a question but a lemma) states that the following two statements are equivalent: 1- $(a:_R I) = a$ for some a $\in I -Z(R)$ 2- $(R:_K I) = R$ – Med S. Mar 16 '15 at 21:45
  • Z(R) : the set of zero divisor of R; K = the total quotient ring of R. – Med S. Mar 16 '15 at 21:47
  • You want to prove the following: if $I\subseteq Z_R(R/(a))$ then there is $\hat r\in R/(a)$, $\hat r\ne0$ such that $I\hat r=0$, and I can't see how can you do this unless $R$ is noetherian. – user26857 Mar 16 '15 at 22:42
  • According to lemma 3.1 of the article "LCM-stableness in ring extensions" by Hirohumi UDA it is true for all finitely generated ideal, he said that the result is easy so the proof is omitted.I try to understand the lemma, and it does not seem to me that easy. – Med S. Mar 16 '15 at 22:48
  • That lemma is not exactly what you asked here. Better say where you were stuck on that proof. – user26857 Mar 16 '15 at 22:51
  • how to proof that (4) is equivalent to (5) in the lemma ? – Med S. Mar 16 '15 at 22:56
  • Actually I can prove (3) equivalent to (5) :) – user26857 Mar 17 '15 at 10:28
  • I would be very grateful if you can give me some clues. – Med S. Mar 17 '15 at 11:04

1 Answers1

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Let $R$ be a commutative ring, $Q$ its total ring of fractions, and $I\subset R$ an ideal (containing a non-zero divisor). Then $(a:I)=(a)$ for each $a\in I-Z(R)$ iff $(R:_QI)=R$.

Let $x=u/v\in Q$ such that $xI\subset R$. Then $uI\subset (v)$. Since $v\in R-Z(R)$, and $I$ contains a non-zero divisor, say $b$, we have $bv\in I-Z(R)$. From $uI\subset (v)$ we get $buI\subset (bv)$, hence $bu\in(bv:I)=(bv)$. It follows $u\in (v)$, that is, $x\in R$.

Conversely, let $r\in (a:I)$. We get $rI\subset (a)$, hence $(r/a)I\subset R$. Then $r/a\in R$, so $r\in (a)$.

user26857
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