Show that $\sqrt{-n}$ and $\sqrt{-n} +1$ are not prime in $\mathbb{Z}[\sqrt{-n} ]$

Question

What I have done is that tried writing $\sqrt{-n} =(a+b\sqrt{-n} )(c+d\sqrt{-n} ) $ .

And also $-n =(a^2+b^2(-n)) (c^2+d^2(-n)) $. (Using the definition of norms.) This however didn't yield me something useful. And also how do you attempt such problems in general?

So $\sqrt{-n}|(a+b\sqrt{-n})$ hence $n|a$ .So suppose $a =n$ and $(n+b\sqrt{-n})(c+d\sqrt{-n})$. — Guria Sona, Apr 05 '20 at 07:58
So if I calculate I am getting $b=c/(c^2+d^2n)$ how do I proceed after this @user26857 — Guria Sona, Apr 05 '20 at 08:08
There is another characterization which is more useful here: p is prime iff the factor ring R/(p) is an integral domain. — user26857, Apr 05 '20 at 08:38

user26857 · Accepted Answer · 2020-04-05T08:40:51.600

2

Start with $\mathbb Z[\sqrt{-n}]\simeq\mathbb Z[X]/(X^2+n)$ and show:

$\mathbb Z[\sqrt{-n}]/(\sqrt{-n})\simeq \mathbb Z/n\mathbb Z$
$\mathbb Z[\sqrt{-n}]/(\sqrt{-n}+1)\simeq \mathbb Z/(n+1)\mathbb Z$

edited Apr 05 '20 at 08:40

answered Apr 05 '20 at 08:31

user26857

52,094

I suppose that $n$ is a positive integer. – user26857 Apr 05 '20 at 08:35
None of your claims is actually true, in general. – user26857 Apr 05 '20 at 08:38
One more doubt I have. How would the ideal generated by (2,$\sqrt{-n}$)look like? Will it be a principal ideal?@user26857 – Guria Sona Apr 05 '20 at 10:16
Again too general question. If $n=1,2$ the ring is a principal ideal domain, so every ideal is principal. See also here: https://math.stackexchange.com/questions/1711611 – user26857 Apr 05 '20 at 10:41

Show that $\sqrt{-n}$ and $\sqrt{-n} +1$ are not prime in $\mathbb{Z}[\sqrt{-n} ]$

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