Prime ideals of a certain form The author of a solution of an earlier problem indicated that, writing the ideal of the integer ring $O_K$ of $\mathbb Z[\sqrt{-5}]$, $$(3,1+\sqrt{-5})$$ in the form $$\{x+y\sqrt{-5}\mid y\equiv x\ \bmod3\}$$ we immediately see that this ideal is prime. I failed to immediately see why that is true.
 A: Both versions immediately show that the ideal $I = (3, 1 + \sqrt{-5})$ is prime: given any element $a + b \sqrt{-5}$ we find
$$ a+b\sqrt{-5} = a - b + (b + b \sqrt{-5}) \equiv a-b \bmod I. $$
Thus ${\mathbb Z}[\sqrt{-5}]/I \simeq {\mathbb Z}/3{\mathbb Z}$, hence $I$ is prime since the quotient is an integral domain (even a field, so the ideal is maximal).
A: Key Idea  the ideal $\,(3,1+\sqrt{-5})\,$ is already a module in Hermite normal form, which makes everything very easy. Write $\,R = \Bbb Z[w],\,\ \bar R = R/I,\,\ I=(3,1\!+\!w),\,\ w = \sqrt{-5}.\,$ By here, since $\,3\mid N(1+w) = 6,\,$ we deduce that $I = (3,1+w) = [3,1+w] = 3\Bbb Z + (1+w)\Bbb Z.\,$  But it is trivial to test module membership given such a triangularized basis, namely
$$\begin{align} \overbrace{a+bw}^{\large  a-b\ +\ \color{#0af}{(1+w)b}}\!\!\!\!\!\!&\in I = 3\Bbb Z + \color{#0af}{(1\!+\!w)\Bbb Z}\\
\iff\ \ \ a\!-\!b&\in I\\
\iff\ \ \ a\!-\!b &\in 3\Bbb Z 
\end{align}\qquad$$
so $\ a\!+\!bw\bmod I\, =\, a\!-\!b\,\bmod 3,\, $ so $\,\bar R = R/I\cong \Bbb Z/3\,$ (so $I$ is prime, in detail:
$\qquad\quad  h: \Bbb Z \to \bar R\,\color{#0a0}{ \ {\rm is\ surjective\  (onto)}}\,$ by $\bmod I\!:\  a\!+\!bw\equiv a\!-\!b\in \Bbb Z$
$\qquad\quad \color{#c00}{\ker h = 3\,\Bbb Z}^{\phantom{|^|}}\! $ follows by $\,b=0\,$ in above displayed equivalences).
Remark $ $ The criterion generalizes to an ideal test for modules $\rm\,[a,b\!+\!c\:\!\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 Franz Lemmermeyer's notes linked here..
This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.
A: Question: "The author of a solution of an earlier problem indicated that, writing the ideal of the integer ring $\mathcal{O}_K=\mathbb Z[\sqrt{-5}]$, $$(3,1+\sqrt{-5})$$ in the form $$\{x+y\sqrt{-5}\mid y\equiv x\ \bmod3\}$$ we immediately see that this ideal is prime. I failed to immediately see why that is true.
Answer: In the case of the ring $A:=\mathcal{O}_K=\mathbb Z[\sqrt{-5}]:=\mathbb{Z}[t]/(t^2+5)$ and the ideal  $I:=(3,1+\sqrt{-5})$ you may also use an elementary calculation with ideals in the ring $R:=\mathbb{Z}[t]$. You may check that there is an equality of ideals in $R$:
$$(t^2+5,3,t+1)=(3,t+1).$$
From this you get the isomorphisms
$$A/I \cong R/(3,t+1)\cong \mathbb{Z}/(3)$$
which is a field, hence $I \subseteq A$ must be a prime ideal.
