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If $n+1$ is prime in $\mathbb{Z}$, then $1+\sqrt{-n}$ is prime in $\mathbb{Z}[\sqrt{-n}]$.

I believe this statement is true based on the answer here: https://math.stackexchange.com/a/3610560/

However, I'd like an elementary proof that doesn't involve ring quotients.

By adapting the proof here https://math.stackexchange.com/a/3566198 , I have an elementary proof of the following similar statement: If $n$ is prime in $\mathbb{Z}$, then $\sqrt{-n}$ is prime in $\mathbb{Z}[\sqrt{-n}]$. But I can't see how to further adapt it.

RitterSport
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1 Answers1

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Here is an elementary argument that, assuming $n+1$ is prime, if $1 + \sqrt{-n}$ divides $(a + b\sqrt{-n})(c + d\sqrt{-n})$, then $1 + \sqrt{-n}$ divides $a + b\sqrt{-n}$ or it divides $c + d\sqrt{-n}$.

Writing $(a + b\sqrt{-n})(c + d\sqrt{-n}) = (1 + \sqrt{-n})(u + v\sqrt{-n})$ and expanding, we have two equations

$$ac - bdn = u - nv, \qquad bc + ad = u + v.$$ Subtracting these equations and rearranging, we may derive

$$(a - b)(c - d) = (n+1)(bd - v)$$ and since $n+1$ is prime, it must divide either $a - b$ or $c - d$, say the former without loss of generality. Write $a - b = k(n+1)$, so that $a = b + k(n+1)$ and therefore

$$a + b\sqrt{-n} = k(n+1) + b + b\sqrt{-n} = k(n+1) + b(1 + \sqrt{-n})$$

or in other words

$$a + b\sqrt{-n} = (1 + \sqrt{-n})(k(1 - \sqrt{-n}) + b)$$ which was to be shown.

user43208
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    It's elementary, i.e., doesn't require concepts from abstract algebra, but it is arguably the sort of grungy proof that abstract algebra succeeds so well in cleaning up and clearing up, so that people don't get lost in the forest. – user43208 Feb 16 '24 at 23:07