I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.

  • $\begingroup$ An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+b\sqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring. $\endgroup$ – Jyrki Lahtonen Jan 16 at 19:02
  • $\begingroup$ @JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+\sqrt{-5})(2-\sqrt(-5))$, but the norm is 9. $\endgroup$ – davidh Jan 16 at 19:09
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    $\begingroup$ What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$. $\endgroup$ – Jyrki Lahtonen Jan 16 at 19:11
  • $\begingroup$ @JyrkiLahtonen $(2+\sqrt{-5},3)$ seems to be maximal, since there's no 1 in there. $\endgroup$ – davidh Jan 16 at 19:16
  • $\begingroup$ Correct, David. It is the same ideal as $(-1+\sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime). $\endgroup$ – Jyrki Lahtonen Jan 16 at 19:19

In general, given a positive number $p$ that is prime in $\mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$\left(\frac{d}{p}\right) = 1,$$ find the smallest positive integer $n$ such that $n^2 \equiv d \pmod p$, then $\langle p \rangle = \langle p, n - \sqrt d \rangle \langle p, n + \sqrt d \rangle$.

In this instance $p$ is 3 and $d = -5$. Indeed $$\left(\frac{-5}{3}\right) = 1,$$ $-5 \equiv 1 \pmod 3$, and obviously $1^2 \equiv 1 \pmod 3$. Then $1 - \sqrt{-5}$ and $1 + \sqrt{-5}$ are both numbers with norm divisible by 3.

Furthermore, any number in $\mathbb Z[\sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $\langle 3, 1 - \sqrt{-5} \rangle$ or $\langle 3, 1 + \sqrt{-5} \rangle$. For example, $4 + \sqrt{-5} = 3 + (1 + \sqrt{-5})$ and hence $(4 + \sqrt{-5}) \in \langle 3, 1 + \sqrt{-5} \rangle$. Likewise $(4 - \sqrt{-5}) \in \langle 3, 1 - \sqrt{-5} \rangle$.

Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + \sqrt{-5}) \not\in \langle 3, 1 + \sqrt{-5} \rangle$.

Since norms are multiplicative, it's a given that all nonzero numbers in $\langle 3 \rangle$ have norms divisible by 9 and all nonzero numbers in $\langle 1 - \sqrt{-5} \rangle$ have norms divisible by 6. Same goes for $\langle 1 + \sqrt{-5} \rangle$.

Then you just need to prove that all numbers of the form $3x + y(1 \pm \sqrt{-5})$, with both $x$ and $y$ being any numbers from $\mathbb Z[\sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $\langle 3, 1 - \sqrt{-5} \rangle$ or $\langle 3, 1 + \sqrt{-5} \rangle$ is the whole ring.


The ideals $I\subset R$ containing the principal ideal $(3)\subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $\Bbb{F}_3\times\Bbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?

  • $\begingroup$ I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)\simeq\Bbb{F}_3\times\Bbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already.. $\endgroup$ – Jyrki Lahtonen Jan 16 at 19:23
  • $\begingroup$ For example, $(11)$ is a maximal ideal of $R$, and $R/(11)\simeq \Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements. $\endgroup$ – Jyrki Lahtonen Jan 16 at 19:43
  • $\begingroup$ @JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$\Bbb{Z}[\sqrt{-5}]/(3)\cong(\Bbb{Z}[X]/(X^2+5))/(3)\cong(\Bbb{Z}[X]/(3))(X^2+5)\cong\Bbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,\sqrt{-5}-1)$ and $(3,\sqrt{-5}+1)$. $\endgroup$ – Servaes Jan 16 at 20:58
  • $\begingroup$ Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker? $\endgroup$ – Jyrki Lahtonen Jan 16 at 21:02
  • $\begingroup$ Sure, I figured the isomorphism $\Bbb{F}_3[X]/(X^2-1)\cong\Bbb{F}_3\times\Bbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader. $\endgroup$ – Servaes Jan 16 at 21:06

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