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.
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I think that 'immediately' is a bit over the top. Maybe we can see it after writing out the details of what it would mean for the ideal to be prime or not? Did you try that? – Vincent Apr 30 '22 at 10:41
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@Vincent I do know how to show that an ideal is prime. I only asked because I wondered if the latter form allowed us to apply the same principal to other ideals of form (p,a+sqrt(d)) for arbituary p prime, a and d and immediately see that they are prime. – Cissalc Apr 30 '22 at 11:33
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That is an interesting question. Obvious it would not be the case for all ideals, so it would be nice to get a closer look at for which ideals it works or not. I'll think a bit about it – Vincent Apr 30 '22 at 12:32
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@Vincent It is fairly "immediate" if one knows the pertinent results - see my answer (that argument works very generally and only need be made once). – Bill Dubuque May 01 '22 at 09:15
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2A hyperlink for context would be nice. – Jacob Manaker May 04 '22 at 04:40
3 Answers
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).
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More generally, an ideal $I$ that has prime index $p$ in a commutative ring $R$ has to be a maximal ideal in $R$ since $R/I \cong \mathbf Z/p\mathbf Z$. The converse of course is false (such as $(3)$ in $\mathbf Z[i]$ or $(11)$ in $\mathbf Z[\sqrt{-5}]$, which are maximal ideal despite having non-prime index). – KCd May 01 '22 at 18:39
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.

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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.

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