let K be an algebraic number field and let $O_K$ be its ring of integers.
Lemma; Let $a,b$ be fractional ideals of $O_K$. If $b \subseteq a$ then there is an ideal $c$ such that $b=ac$.
I need to find a counterexample for why this fails for general orders $R$ in our number field in particular a counterexample for $K=Q(\sqrt{-3})$ and $R=Z[\sqrt{-3}]$.
I am confused as to what I should be looking for and am finding it hard to attempt even though I imagine it is probably fairly easy. I am thinking that the most likely condition that goes missing which will help is the fact that a prime ideal in R is no longer necessarily maximal. If someone could help get a counterexample and explain why its true that would be great, thanks.
Barring a mistake the following should work. Based on using the fact that $(-1+\sqrt{-3})/2$ is missing.
Let $a=\langle 2,1+\sqrt{-3}\rangle$ and $b=\langle 1+\sqrt{-3}\rangle$. Then $$ a=\{x+y\sqrt{-3}\in R\mid x\equiv y\pmod 2\} $$ and $$ b=\{x+y\sqrt{-3}\in R\mid x\equiv y\pmod 4\}. $$
Claim 1. $a$ is the only ideal properly between $b$ and $R$.
Proof. We see that $b$ is of index $4$ in $R$, a set of representative of cosets consisting of $\{0,1,2,3\}$. An ideal containing either $1$ or $3$ in addition to all the elements of $b$ will be all of $R$. The only alternative is to include $2$. But then we get $a$. Q.E.D.
Claim 2. There is no ideal $c$ such that $b=ac$.
Proof. Assume that such an ideal $c$ exists. Clearly $b\subseteq c$, so by Claim 1 we must have $c=a$. But as $(1+\sqrt{-3})^2=-2+2\sqrt{-3}$ we get $$ a^2=\langle4,2+2\sqrt{-3},-2+2\sqrt{-3}\rangle. $$ This is too small.