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I am working on an exercise that asks us to consider the ring $R = \mathbb{Z}[\sqrt{-3}]$ and the ideal $I = (2, 1 + \sqrt{-3})$ in $R$. Part (a) asks to show that $I^2 = (2)I$ but $I \neq (2)$, and part (b) asks to show that $I$ is the unique prime ideal in $R$ containing $(2)$, and to then conclude that $(2)$ does not factor into prime ideals in $R$.

A previous exercise asked to show that $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5}) = (2)$ in $\mathbb{Z}[\sqrt{-5}]$. Now $(2, 1 + \sqrt{-5}) = (2, 1 - \sqrt{-5})$, so then $(2) = (2, 1 + \sqrt{-5})^2$. But this should work in $R$ too, so then $I^2 = (2)$. But $I$ is prime, so then $(2)$ does factor into prime ideals in $R$. Where did I go wrong? Thanks.

Ethan Alwaise
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    In your first case you have $I^2=(2)I$, while in the second $I^2=(2)$. These two relations are different, hence lead to different results. – Crostul Aug 31 '15 at 22:01

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The equality $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5}) = (2)$ that holds in $\Bbb Z[\sqrt{-5}]$ does not carry over into $R = \Bbb Z[\sqrt{-3}]$.

Here $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3}) \ne (2)$. The inclusion $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3}) \subset (2)$ still holds, but if you compute a generating set of $(2, 1 + \sqrt{-3})(2, 1 - \sqrt{-3})$, you only get

$$4 = 2\cdot 2 = (1+\sqrt{-3})\cdot (1-\sqrt{-3})$$ and $$2 \pm 2\sqrt{-3} = 2\cdot (1 \pm \sqrt{-3}).$$ Since $2 - 2\sqrt{-3} = 4 - (2 + 2\sqrt{-3})$, this generating set can be reduced to $(4, 2 + 2\sqrt{-3}) = 2I$. However, there is no way to express $2$ as a $\Bbb Z[\sqrt{-3}]$-linear combination of $4$ and $2 + 2\sqrt{-3}$.

The difference with the $\Bbb Z[\sqrt{-5}]$ case is that in that case you see that the corresponding products of generators include both $4$ and $6 = (1+\sqrt{-5})\cdot (1-\sqrt{-5})$, and thus $2 = 6-4$ lies in $(2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5})$

Rolf Hoyer
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  • Ah, great. Thanks! I still don't see how this implies that proper ideals do not factorize into prime ideals though, since $(2)$ is not prime and cannot be factorized into prime ideals. So isn't the factorization $I^2$ then unique? – Ethan Alwaise Aug 31 '15 at 23:49
  • @EthanAlwaise $I^2$ is the unique prime factorization of $I^2 = (2)I$, but since $(2) \ne I^2$, you still have that $(2)$ doesn't have a prime ideal factorization. So the failure of unique factorization is that factorizations might not exist, rather than the fact that factorizations might not be unique. – Rolf Hoyer Sep 01 '15 at 00:20