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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

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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.

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  • $\begingroup$ Thank you very much ! I have to get used to think by contradiction. The hint is already helpful :) . $\endgroup$ – Jérôme Oct 4 '14 at 15:30

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