Let $R = \mathbb{Z}[\sqrt{-n}]$ where $n$ is a square free integer greater or equal than $3$. Prove that $2$ and $\sqrt{-n}$ are irreducible in $R$.

Proof. Recall that if $R$ is an integral domain, and $r \in R$ and is not a unit, then $r$ is called irreducible in $R$ if whenever $r = ab$ with $a,b\in R$, at least one $a$ or $b$ must be a unit in $R$.

Then let $2 = xy$. Then taking the norm, we have $4 = N(x)N(y)$.

Then we need to show that $N(x)=1$ and $N(y) = 4$ or $N(y)=1$ and $N(x) = 4$.

Suppose $x = a + b\sqrt{-n}$. Then assume $N(x) = 2$. Then $x = a + b\sqrt{-n}$ implies $N(x)= 2 = a^2 + nb^2$ . Since $n \ge 3$, then equation only makes sense if $b = 0$ and $a = \sqrt2$ . So we have a contradiction, so $N(x) $ = $4$ and $N(y)$ must be $1$, so that $y$ is a unit. So $2$ is irreducible.

Now for $\sqrt{-n}$, suppose $\sqrt{-n} = xy$. Then taking the norm, we have, $n = N(x)N(y)$. Then let $\sqrt{-n} = xy = (a + b\sqrt{-n})(c + d\sqrt{-n})$. Then $n = (a^2 + nb^2)(c^2 + nd^2)$ .

Can someone please help me for $\sqrt{-n}$ ? I am stuck. Thank you!


If $b\neq 0$, then $a^2+nb^2 \geq n$. The equality occurs only when $a=0$ and $b=\pm 1$. In which case the factorization you have is the trivial one $\sqrt{-n}=\pm\sqrt{-n}(c+d\sqrt{-n})$. Thus $c=\pm 1,d=0$.

If $b=0$, then $n=(ac)^2+(ad)^2n$, using the same reasoning as above, we can conclude that $ac=0$ and $ad=1$. Consequently $c=0$ and $a=d=\pm 1$. Which again gives a trivial factorization.

Note: In the second case $d \neq 0$, otherwise $n$ will not be square free.

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  • $\begingroup$ So we would have $n = n(c^2 + nd^2) = nN(y)$ which implies $N(y) = 1$ so $y$ is a unit? and $N(x) $ must be $\sqrt-n$? $\endgroup$ – Mahidevran Nov 4 '15 at 6:18
  • $\begingroup$ @Mahidevran I have edited my solution to add more details. Hopefully this will be helpful. $\endgroup$ – Anurag A Nov 4 '15 at 6:36

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