# Lack of unique factorization of ideals I'm aware of the result that integral domains admit unique f

Lack of unique factorization of ideals
I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.
It's clear that $\mathbb{Z}\left[\sqrt{-3}\right]$ is not a Dedekind domain, as it is not integrally closed. I'm having difficulty demonstrating directly that $\mathbb{Z}\left[\sqrt{-3}\right]$ does not admit unique factorization of ideals.
Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.
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Maeve Holloway
In the ring $\mathbf{Z}\left[\sqrt{-3}\right]$, the ideal $P=\left(2,1+\sqrt{-3}\right)$ is prime since it has index 2 in the ring. Note ${P}^{2}=\left(4,2+2\sqrt{-3},-2+2\sqrt{-3}\right)=\left(4,2+2\sqrt{-3}\right)=\left(2\right)\left(2,1+\sqrt{-3}\right)=\left(2\right)P$ the principal ideal generated by 2 in $\mathbf{Z}\left[\sqrt{-3}\right]$
If there were unique factorization of ideals into products of prime ideals in $\mathbf{Z}\left[\sqrt{-3}\right]$ then the equation ${P}^{2}=\left(2\right)P$ would imply P=(2), which is false since $1+\sqrt{-3}$ is in P but it is not in (2).
In fact we have ${P}^{2}\subset \left(2\right)\subset P$ and this can be used to prove the ideal (2) does not even admit a factorization into prime ideals, as follows. If Q is a prime ideal factor of (2) then $\left(2\right)\subset Q$, so $PP={P}^{2}\subset Q$, which implies $P\subset Q$ (from Q being prime), so Q=P since P is a maximal ideal in $\mathbf{Z}\left[\sqrt{-3}\right]$.
If (2) is a product of prime ideals then it must be a power of P, and ${P}^{n}\subset {P}^{2}$ for $n\ge 2$, so the strict inclusions ${P}^{2}\subset \left(2\right)\subset P$ imply (2) is not ${P}^{n}$ for any $n\ge 0$.
The "intuition" that the ideal $P=\left(2,1+\sqrt{-3}\right)$ is the key ideal to look at here is that P is the conductor ideal of the order $\mathbf{Z}\left[\sqrt{-3}\right]$. The problems with unique factorization of ideals in an order are in some sense encoded in the conductor ideal of the order. So you want to learn what a conductor ideal is and look at it in several examples.
For example, the ideals I in $\mathbf{Z}\left[\sqrt{-3}\right]$ which are relatively prime to P (meaning I+P is the unit ideal (1)) do admit unique factorization into products of prime ideals relatively prime to P. That illustrates why problems with unique factorization of ideals in $\mathbf{Z}\left[\sqrt{-3}\right]$ are closely tied to the ideal P.
If you look at the ideal notation P${P}^{\prime }=\left(2,1+\sqrt{-3}\right)$ in the larger ring $\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]$, which we know has unique factorization of ideals, then we don't run into any problem like above because ${P}^{\prime }=2\left(1,\left(1+\sqrt{-3}\right)/2\right)=\left(2\right)$ in $\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]$, so P′ is actually a principal ideal and the "paradoxical" equation $PP=\left(2\right)P$ in $\mathbf{Z}\left[\sqrt{-3}\right]$ corresponds in $\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]$ to the dumb equation ${P}^{\prime }{P}^{\prime }={P}^{\prime }{P}^{\prime }$. (The ideal P′ in $\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]$ is prime since the quotient ring mod P′ is a field of size 4: $\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]$ is isom. as a ring to $\mathbf{Z}\left[x\right]/\left({x}^{2}+x+1\right)$, so Z$\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]/{P}^{\prime }=\mathbf{Z}\left[\left(1+\sqrt{-3}\right)/2\right]/\left(2\right)$ is isom. to $\left(\mathbf{Z}/2\mathbf{Z}\right)\left[x\right]/\left({x}^{2}+x+1\right)$, which is a field of size 4.)