Ideals, Dedekind domain and $\mathbb{Z}[\sqrt{-3}]$

Question

I have the ideal $\mathfrak{a} = (2, 1 + \sqrt{-3})$ in $\mathbb{Z}[\sqrt{-3}]$. I have to show that $\mathfrak{a} \neq (2)$ but $\mathfrak{a}^{2} = (2)\mathfrak{a}$ and then conclude that ideals do not factor uniquely into primes ideals in this ring.

I think I did it for the inequality and the equality. I'll post my answer here (don't hurt me if it's too naive, I've had zero example before trying). If you would be kind to correct me if needed.

The inequality $\mathfrak{a} \neq (2)$. I supposed I just had to find an element which was only in one of those two sets. I found $1 + \sqrt{-3} \in \mathfrak{a}$ but $\notin (2)$. Done.

The equality, I began with $\mathfrak{a}^{2}$.

\begin{aligned} \mathfrak{a}^{2} & = \mathfrak{a}\mathfrak{a}\\ & = \{4a + 2(1 + \sqrt{-3})b + 2(1 + \sqrt{-3})c + (2\sqrt{-3} - 2)d \; | \; a, b, c, d \in \mathbb{Z}[\sqrt{-3}]\}\\ & = \{2 \left(2a + (1 + \sqrt{-3})b + (1 + \sqrt{-3})c + (\sqrt{-3} - 1)d\right) \; | \; a, b, c, d \in \mathbb{Z}[\sqrt{-3}]\}\\ & = \{2 \left(2a + (1 + \sqrt{-3})b\right) \; | \; a, b \in \mathbb{Z}[\sqrt{-3}]\}\\ & = (2) \mathfrak{a} \end{aligned}

And... I'm stucked to prove that ideals do not factor uniquely into primes ideals in this ring. I know $\mathbb{Z}[\sqrt{-3}]$ isn't a Dedekind domain, otherwise we could use the cancelation law, which would be a contradiction with the inequality above.

Side question : how to prove $\mathbb{Z}[\sqrt{-3}]$ isn't a Dedekind domain only with the definition (noetherian, integrally closed in its field of fractions, every nonzero prime ideal of $\mathbb{Z}[\sqrt{-3}]$ is maximal) ?

Main question : How to prove that ideals do not factor uniquely into primes ideals in the ring $\boldsymbol{\mathbb{Z}[\sqrt{-3}]}$ ?

Thank you a lot in advance,

Jérôme

Answer 1

Hint for the side question: $\frac{1+\sqrt{-3}}{2}$ is integral over $\mathbb{Z}$ and in the fraction field of $\mathbb{Z}[\sqrt{-3}]$.

Main question: Your work so far is correct and you're almost done. Suppose you had unique factorisation into prime ideals. Since $\mathfrak{a}$ and $(2)$ are different they have different factorisations. But you showed that $I=\mathfrak{a}^2=(2)\mathfrak{a}$, so the ideal $I$ would have two different factorisations into prime ideals.

Answer 2

This result is not specific to $\mathbf Z[\sqrt{-3}]$: everything works for $\mathbf Z[\sqrt{d}]$ when $d$ is an integer that's not a perfect square and $d \equiv 1 \bmod 4$: for the ideal $\mathfrak a = (2,1+\sqrt{d})$ in $\mathbf Z[\sqrt{d}]$, we have $\mathfrak a^2 = (2)\mathfrak a$ and $\mathfrak a \not= (2)$. As in the other answer, if there were unique factorization of ideals then the equation $\mathfrak a^2 = (2)\mathfrak a$ implies $\mathfrak a = (2)$, which is false.

Here is a more subtle way to see $\mathbf Z[\sqrt{d}]$ doesn't have unique factorization of ideals. It can be shown that an integral domain with unique factorization of ideals must be integrally closed, and $\mathbf Z[\sqrt{d}]$ is not integrally closed since $d \equiv 1 \bmod 4$ (the number $(1+\sqrt{d})/2$ is integral over $\mathbf Z[\sqrt{d}]$ and is in the fraction field of this ring but it is not in the ring).