How does the definition of integer divisibility carry over to ideals?
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Does "ring of integers" mean $,\Bbb Z?\ \ $ – Bill Dubuque Oct 28 '18 at 23:59
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yes. filler filler – Damian Siniakowicz Oct 29 '18 at 01:33
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In the case of principal ideal domains, an ideal $I$ divides an ideal $J$ of a PID $R$ if and only if $J \subseteq I$. The intuition here is that, if $J = (a)$ and $I = (b)$,
$$ J \subseteq I \iff a \in (b) \iff a = bc \text{ for some $c \in R$} \iff b | a. $$
which matches with the idea of divisibility we already have: $(b)$ divides $(a)$ if and only if $b$ divides $a$. More generally, this concept can be carried over to Dedekind domains (in which non trivial ideals can be decomposed as product of powers of prime ideals, similar to the integers): if $I = \mathfrak{p_1}^{\alpha_1} \cdots \mathfrak{p_n}^{\alpha_n}$ and $J = \mathfrak{q_1}^{\beta_1} \cdots \mathfrak{q_n}^{\beta_n}$ with $\mathfrak{p}_i, \mathfrak{q}_j$ prime then $I$ divides $J$ if and only if each $\mathfrak{p}_i$ coincides with some $\mathfrak{q}_j$ and $\alpha_i \leq \beta_j$.
qualcuno
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Do you mean that my definition is wrong, or not general enough? I have only seen this in the context of PIDs or Dedekind domains, so maybe there is a general definition which I am unaware of. – qualcuno Oct 29 '18 at 00:05
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1$I\mid J,$ means $IK = J$ for some ideal $K$. Only in very special domains $D$ is that equivalent to $I\supseteq J$ (namely $D$ is Prufer $\iff$ the equivalence holds for finitely generated $I).\ \ $ – Bill Dubuque Oct 29 '18 at 00:15
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My bad, I was aware of the equivalence in the case of Dedekind domains but did not know about the general definition. I'm leaving the answer (but editing accordingly) in case this serves of some intuition (although depending on the OP's background, this may be utterly useless). The answer you linked is very interesting, by the way. Thanks! – qualcuno Oct 29 '18 at 00:22