Why is there such an ideal?

Question

I don't understand the $3$rd line in the following proof, why is there an ideal $N$ s.t. $JL=sN$. I thought ''to contain means divides'' is valid only in some special rings, for example Dedekind rings (whereas the converse is true for any commutative unital rings)

Answer 1

Let me extract the fact you'd like to prove.

Lemma: Let $R$ be a domain, $(a)$ be a principal ideal and $I$ be an ideal with $I \subseteq (a)$. Then there exists an ideal $J$ such that $I = aJ$.

Proof: Take $$ J = (I: (a)) = \{r \in R \mid r (a) \subseteq I\} = \{r \in R \mid ra \in I\} $$ to be the colon ideal. We claim that $I = aJ$.

$\supseteq$: Given $aj \in aJ$, then by definition of $(I: (a))$ we have $aj \in I$.

$\subseteq$: Given $b \in I \subseteq (a)$, then $b = ar$ for some $r \in R$. Since $ar = b \in I$ then $r \in (I : (a)) = J$. Thus $b = ar \in aJ$.

Returning to your question, the above lemma shows that we can take $N = (JL : (s))$.

Answer 2

We have $(s) \supseteq IL$ for principal ideals, so $(s)$ divides $IL$. Hence there exists an ideal $M$ with $JL=(s)M=sM$. Now $JL$ is a subset of $sM$ and an ideal, hence of the form $sN$ (see quasicoherent's answer).

Reference for "to contain is to divide" with principal ideals: P. Clark, Proposition $2$, page $2$.

Answer 3

I think we can show this directly; let the set $N \subset R$ be defined as

$N= \{ n \in R \mid sn \in JL \}; \tag 0$

then it is evident that

$sN \subset JL; \tag 1$

since

$JL \subset (s), \tag 2$

every $t \in JL$ is of the form $t = sa$, $a \in R$; since such $a \in N$, this shows that

$JL \subset sN, \tag 3$

so by (1) and (3) we know that

$JL = sN. \tag 4$

We show $N \subset R$ is an ideal: suppose

$n_1, n_2 \in N; \tag 4$

then

$sn_1, sn_2 \in JL, \tag 5$

and since $JL$ is an ideal of $R$,

$s(n_1 - n_2) = sn_1 - sn_2 \in JL, \tag 6$

and so

$n_1 - n_2 \in N, \; \forall n_1, n_2 \in N; \tag 7$

likewise for

$sn = t \in JL, \; n \in N \tag 8$

we have, again since $JL$ is an ideal,

$s(nr) = (sn)r = tr \in JL, \; r \in R, \tag 9$

which shows that

$nr \in N, \; \forall n \in N, \; r \in R; \tag{10}$

combining (7) and (10) shows that $N \subset R$ is an ideal in $R$.