## Division of ideals

Hi,

I have a couple of questions concerning algebraic number theory. Why does $I$ contains $J$ imply $I$ divides $J$? Firstly assuming $I$ contains $J$ how do we know there exists a fractional ideal $I'$such that $I'I=J$ and then how do we get that $I'$ is actually an integral ideal?

Another problem I am having is in proving: if $I$ is ideal and $N(I)=p$ some prime $p$ then $I$ is prime ideal. So we assume $A,B$ are ideals such that $AB$ is contained in $I$. We want to show either $A$ is contained in $I$ or $B$ is. Assuming the above then we know $I$ divides $AB$. The proof then states that $I=ABC$ some ideal $C$. I thought $I$ divides $AB$ means $IC=AB$ for some ideal $C$ not other way round. I get that we could multiply by $C$ inverse to get $I$ on its own but then $C$ inverse is a fractional ideal.

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I’m presuming that you’ve read the course notes at least twice or so at this point. So for your first question, it’s already assumed that we’ve shown that the non-zero fractional ideals form a group. This gives the existence of the $I'$ you mention. But then
$I'I\subset J\subset I$, so each element of $I'$ stabilizes the ideal $I$. This implies that every element of $I'$ is an algebraic integer, i.e., that $I'$ is an ideal.

For the statement about prime ideals, note that we are assuming the factorization theorem. So any $I$ can be written

$I=\prod P_i$

for prime ideals $P_i$. But then

$N(I)=\prod N(P_i).$

Therefore, if $N(I)$ is prime, only one $P_i$ can occur.