Background
I'm currently working on problem 8.7 from Alaca & Williams' Intro. Alg. Number Theory. The problem states the following:
Show that $\langle3,1+2\sqrt{-5}\rangle\mid\langle1+2\sqrt{-5}\rangle$ in $\mathbb Z[\sqrt{-5}]$. Determine the integral ideal $A$ such that $$\langle3,1+2\sqrt{-5}\rangle A=\langle1+2\sqrt{-5}\rangle.$$
My progress
Proving that one ideals divides another is immediate from the fact that to divide is to contain. However finding the ideal $A$ is my problem.
My first thought was using the following exercise which asks for the inverse of the ideal $\langle3,1+2\sqrt{-5}\rangle$. So let's assume that I've found it, right? Suppose $J$ is it's inverse, then $$J\langle3,1+2\sqrt{-5}\rangle A=A=J\langle1+2\sqrt{-5}\rangle.$$ That sounds like a plan until we try to find $J$.
The problem
I had a really nice hunch with the ideal $\langle3,1-2\sqrt{-5}\rangle$, multiplying we get
\begin{align*} \langle3,1+2\sqrt{-5}\rangle\langle3,1-2\sqrt{-5}\rangle&=\langle9,3-6\sqrt{-5},3+6\sqrt{-5},10\rangle=\langle 1\rangle, \end{align*} since $1=10-9$ is in the product ideal. Therefore since $1$ is in the ideal, it is the whole ring and we conclude that both ideals are inverses.
However, I found a worksheet which claims that the inverse ideal is $\frac13\langle3,1+\sqrt{-5}\rangle$ and by multiplying these ideals we get $$\langle3\rangle\langle3,1+\sqrt{-5},1+2\sqrt{-5},-3+\sqrt{-5}\rangle.$$ We see that the second ideal is trivial since $$1+\sqrt{-5}-3-(-3+\sqrt{-5})=1$$ and therefore the product equals a principal ideal, and by a known result (Example 2.8 in K. Conrad's notes) we see that the fractional ideal is an inverse as well.
My questions
- Is my first approach correct? Solving the equation by finding the inverse?
- The two ideals $\langle3,1-2\sqrt{-5}\rangle$ and $\frac13\langle3,1+\sqrt{-5}\rangle$ seem to be inverses to the ideal in the question. However, one is fractional and the other isn't. What's happening here? Unique factorization of ideals should guarantee that inverses are unique, shouldn't it?
$A$ has norm $21/3 = 7$ so it contains $7$ (it is one of the two prime ideals above $7$).
The inverse of $(3,1+2\sqrt{-5})$ is $\frac13(3,1-2\sqrt{-5})$, and $\langle3,1+2\sqrt{-5}\rangle\langle3,1-2\sqrt{-5}\rangle=(3)$,
as it is $\sqrt{-5}$ not $\sqrt5$.