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We know the theorem that an integral domain $R$ is UFD if and only if

(1) $R$ satisfies the ascending chain condition for principal ideals;

(2) every irreducible element is prime.

Now I need to give a counterexample:

$\mathbb{Z}[\sqrt{-5}]$ satisfies (1) but does not satisfy (2). Since $\mathbb{Z}[\sqrt{-5}]$ is not unique factorization domain, we only need to show $\mathbb{Z}[\sqrt{-5}]$ satisfies (1). How can I prove this?

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Dear @Ren : I've noticed you tend to use the principal-ideal-domain tag for posts which talk about principal ideals. If the post is not really about PIDs, it's not really appropriate to use the tag. If you need a substitute, I'd suggest the ideals tag. Thanks! –  rschwieb Dec 31 '13 at 13:56
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1 Answer

up vote 3 down vote accepted

$\mathbb{Z}[\sqrt{-5}]$ can be written as $\mathbb{Z}[x]/(x^2+5)$.

Hilbert's basis theorem says that, if a ring $R$ is noetherian, then so is $R[x]$. Also, if $I$ is an ideal of a noetherian ring $R$, then $R/I$ is also noetherian (this can be seen by looking at the generators of the inverse image of an ideal of $R/I$ under the standard projection map).

Since $\mathbb{Z}$ is noetherian, these two facts then imply that $\mathbb{Z}[x]/(x^2+5) \cong \mathbb{Z}[\sqrt{-5}]$ is noetherian, and so, in particular, satisfies the ACC on principal ideals.

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