Why $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)$ 
My question: In $\mathbb{Z}[\sqrt{-5}]$, why $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)$. My computation figure out $(3,\sqrt{-5}-1)(3,\sqrt{-5}+1)=(3)(3,\sqrt{-5}-1,\sqrt{-5}+1,2)$. $3$ and $2$ are prime numbers, $\sqrt{-5} -1$ and $\sqrt{-5}+1$ are irreducible.

This is just a trivial problem for many people, but it takes me to an ambiguity. Thank all for your help!
 A: $ ab=3c\, \Rightarrow\, (3,a)(3,b) = 3(3,a,b,c)\,$ [$ = (3)\,$ if $\,(3,\color{#0a0}{a,b})=1,\,$ as here by $\,2 = b\!-\!a\in (\color{#0a0}{a,b})$]
$\color{#c00}{ab = dc}\,\Rightarrow\, (d,a)(d,b) = (dd,da,db,\color{#c00}{ab}) = \color{#c00}d(d,a,b,\color{#c00}c)\,$ [$= (d)\,$ if $\,(a,b,c,d)= 1$] $ $ generally.
See this post and its links  for elaboration on this gcd arithmetic and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation and Schreier refinement.
A: Note that $(3, 2, a, b) = (1)$ for any $a, b\in \Bbb Z[\sqrt{-5}]$. So $(3)(3, 2, a, b) = (3)$.
A: If an ideal $I$ in a ring $R$ contains $1$, then it contains $r \cdot 1$ for any $r \in R$, so $I$ is the entire ring: $I=R$.
Your ideal $(3,2,\ldots)$ contains $3$ and $2$ so it also contains $3-2=1$, so it is the entire ring $\mathbb{Z}[\sqrt{-5}]$.
If $J$ is an ideal in $R$ then $JR = J$ because for $j \in J$ and $r \in R$ we have


*

*$j = j\cdot 1 \in JR$ (so $J \subset JR$),

*$j \cdot r \in J$ (so $JR \subset J$).

