Express the ideal $(6) \subset\mathbb Z\left[\sqrt {-5}\right]$ as a product of prime ideals. 
Express the ideal $(6) \subset \mathbb Z\left[\sqrt {-5}\right]$ as a product of prime ideals.

I know I can write $(6)=(2)(3)=\left(1+\sqrt {-5}\right)\left(1-\sqrt {-5}\right)$. But I guess these factors might not be prime. 
What's more, how to solve this kind of problem more systematically?
 A: Firstly, as you noticed
$$(6)=\left(1 + \sqrt {-5}\right)\left(1 - \sqrt {-5}\right),$$
it is sufficient to factorize 
$I_+ = \left(1 + \sqrt {-5}\right)$ and 
$I_- = \left(1 - \sqrt {-5}\right)$.
Since $\Bbb Z\left[\sqrt{-5}\right]$ is the ring of integers of $K=\Bbb Q\left(\sqrt{-5}\right)$, the quotient $\Bbb Z\left[\sqrt{-5}\right]\,\big/\,I_+$ has cardinality $N_{K/\Bbb Q}\left(1 + \sqrt {-5}\right)=1^2+5\cdot 1^2=6$. It is a square-free number, so that 
$$\Bbb Z\left[\sqrt{-5}\right]\,\big/\,I_+ \cong \Bbb Z/6\Bbb Z \qquad \text{via }
f_+:\Bbb Z\left[\sqrt{-5}\right]\,\big/\,I_+ \longrightarrow \Bbb Z/6\Bbb Z$$
Similarly:
$$\Bbb Z\left[\sqrt{-5}\right]\,\big/\,I_- \cong \Bbb Z/6\Bbb Z
\qquad \text{via }
f_-:\Bbb Z\left[\sqrt{-5}\right]\,\big/\,I_- \longrightarrow \Bbb Z/6\Bbb Z$$

Heuristically, we see that 
$\Bbb Z/6\Bbb Z \cong \Bbb Z/2\Bbb Z \times \Bbb Z/3\Bbb Z$ 
 and if $I_+ = \prod_{i=1}^r P_i^{e_i}$ (with $P_i \trianglelefteq \Bbb Z\left[\sqrt{-5}\right]$ distinct prime ideals) then
$\Bbb Z\left[\sqrt{-5}\right] \,\big/\, I_+ \cong \prod_{i=1}^r \Bbb Z\left[\sqrt{-5}\right] \,\big/\, P_i^{e_i}$.
So we can expect that $r=2,e_1=e_2=1$ (for $I_+$ but also for $I_-$).
Let $A:=\Bbb Z\left[\sqrt{-5}\right]$ and $\pi : A \to A/I_+$.
We are going to find necessary conditions on ideals $P_1,P_2$ satisfying $I_+=P_1P_2$. It will be easy to check that these conditions are sufficient, and that the $P_j$'s obtained are prime.
Let's say that $P_1/I_+ ≤ A/I_+$ corresponds to $2\Bbb Z/6\Bbb Z ≤ \Bbb Z/6\Bbb Z$. Then
\begin{align*}
P_1 &= \pi^{-1}(P_1/I_+) \\
&= \pi^{-1}(f_+^{-1}(2\Bbb Z/6\Bbb Z)) \\
&= \pi^{-1}(\{ [0]_{I_+} \;;\; [2]_{I_+} \;;\; [4]_{I_+} \})  \\
&= \pi^{-1}(\langle [2]_{I_+} \rangle)\\
&= \{x \in A \;:\; [x]_{I_+} \in \langle [2]_{I_+} \rangle \} \\
&= \left(2,1+\sqrt{-5}\right)
\end{align*}
Let's say that $P_2/I_+ ≤ A/I_+$ corresponds to $3\Bbb Z/6\Bbb Z ≤ \Bbb Z/6\Bbb Z$. Then
\begin{align*}
P_2 &= \pi^{-1}(P_2/I_+) \\
&= \pi^{-1}(f_+^{-1}(3\Bbb Z/6\Bbb Z)) \\
&= \pi^{-1}(\{ [0]_{I_+} \;;\; [3]_{I_+} \})  \\
&= \pi^{-1}(\langle [3]_{I_+} \rangle)\\
&= \{x \in A \;:\; [x]_{I_+} \in \langle [3]_{I_+} \rangle \} \\
&= \left(3,1+\sqrt{-5}\right)
\end{align*}
We would like that these results are actually sufficient for us, i.e. 
$I_+=P_1P_2=\left(2,1+\sqrt{-5}\right)\left(3,1+\sqrt{-5}\right)$ and $P_1,P_2$ are prime ideals. I let you think about it.

As for $I_-$, this is very similar. We get:
$$I_- = Q_1Q_2$$ with $Q_1 = \left(2,1 - \sqrt {-5}\right)$ and $Q_2=\left(3,1 - \sqrt {-5}\right)$. I let you prove that the $Q_j$'s are prime ideals of $A$ (the quotient $A/Q_1$ should be isomorphic to the field $\Bbb F_2$), and that $I_- = Q_1Q_2$ indeed holds (the inclusion $\subseteq$ is not difficult to establish, and you can get the equality from some cardinality argument).

To sum up:
$$(6)=\left(2,1 + \sqrt {-5}\right) \left(3,1 + \sqrt {-5}\right)
\left(2,1 - \sqrt {-5}\right) \left(3,1 - \sqrt {-5}\right)$$
[which is the same result as user26857's result because 
$$(2,1+\sqrt{-5})=(2,-1+\sqrt{-5})=(2,1-\sqrt{-5})\\
(3,2+\sqrt{-5})=(3,-1+\sqrt{-5})=(3,1-\sqrt{-5})\\
(3,2−\sqrt{-5})=(3,-1-\sqrt{-5})=(3,1+\sqrt{-5})$$
]

Two more remarks:


*

*I should explain the step $\{x \in A \;:\; [x]_{I_+} \in \langle [2]_{I_+} \rangle \} = \left(2,1+\sqrt{-5}\right)$ more carefully. We have 
\begin{align*}
\{x \in A \;:\; [x]_{I_+} \in \langle [2]_{I_+} \rangle \} &= 
\{x \in A \;:\; [x]_{I_+} = [2k]_{I_+} \text{ for some } k \in \Bbb Z\}\\
&= \{x \in A \;:\; x = y+2k \text{ for some } k \in \Bbb Z,y \in I_+\}\\
&= I_+ + 2\Bbb Z \qquad \text{(as sets, } 2\Bbb Z \ntrianglelefteq A)\\
&= I_+ + 2A = \left(2,1+\sqrt{-5}\right)
\end{align*}
where the equality $I_+ + 2\Bbb Z = I_+ + 2A$ holds because $2A \subset I_+ + 2\Bbb Z$ since $2(a+b \sqrt{-5})=2b(1+\sqrt{-5}) + 2a-2b$.

*This method can be used in other situations. Notice that if I started with $(6)=(2)(3)$, it would have been a bit more difficult, since the norm of $2$ is not a square-free integer, so that determining the quotient is less obvious. But once you know what the quotient $A/I$ is (where are $A$ is your ring of integers and $I$ your ideal), then the things can be easier.
Other related questions are: (1), (2), (3), (4).
A: In a principal ideal domain, all ideals are principal, whether generated by prime numbers or composite numbers. But you already know that $\mathbb Z[\sqrt{-5}]$ is not a principal ideal domain.
In domains such as these, it helps to remember that all prime ideals are maximal ideals (with the possible exception of $\langle 0 \rangle$, but that's a digression you may or may not care to get into).
So clearly neither $\langle 2 \rangle$ nor $\langle 3 \rangle$ are maximal ideals of $\mathbb Z[\sqrt{-5}]$, since they are properly contained in $\langle 2, 1 \pm \sqrt{-5} \rangle$ and $\langle 3, 1 \pm \sqrt{-5} \rangle$, respectively. Might either of those ideals be principal?
Oh, wait, you said you wanted a more systematic way of solving this sort of problem. In that case, you should memorize these facts:


*

*$\langle 2 \rangle$ is prime in $\mathcal O_{\mathbb Q(\sqrt d)}$ if $d \equiv 5 \pmod 8$. 

*$\displaystyle \langle 2 \rangle = \left\langle 2, \frac{1}{2} - \frac{\sqrt d}{2} \right\rangle \left\langle 2, \frac{1}{2} + \frac{\sqrt d}{2} \right\rangle$ if $d \equiv 1 \pmod 8$ 

*$\langle 2 \rangle = \langle 2, 1 + \sqrt d \rangle^2$ if $d \equiv 3 \pmod 4$

*$\langle 2 \rangle = \langle 2, \sqrt{d} \rangle^2$ if $d$ is even


and if $p$ is an odd prime in $\mathbb Z$,


*

*$\langle p \rangle$ is prime if $\gcd(p, d) = 1$ and $d$ is not a square modulo $p$

*but if $x^2 \equiv d \pmod p$ can be solved ($\gcd(p, d) = 1$ still holds), then $\langle p \rangle = \langle p, x - \sqrt d \rangle \langle p, x + \sqrt d \rangle$

*or if $\gcd(p, d) > 1$, then $\langle p \rangle = \langle p, \sqrt d \rangle^2$.


These are well known facts, but the only book I've ever seen them neatly summarized like this is the algebraic number theory book by Alaca and Williams.
Then, given non-prime $n$ an algebraic integer of degree 1 (that is, purely real, and rational), then the factorization of $\langle n \rangle$ into ideals of $\mathcal O_{\mathbb Q(\sqrt d)}$ is a simple matter of factorizing the ideals generated by the prime factors of $n$ in $\mathbb Z$.
So in this particular example we have $n = 6$ and $d = -5$. Remember that $-5 \equiv 3 \pmod 8$, not 5, and $-5 \equiv 3 \pmod 4$, not 1. Then $\langle 2 \rangle = \langle 2, 1 + \sqrt{-5} \rangle^2$. So 2 is a square? I know, that seems super weird. The thing is that $\langle 2, 1 - \sqrt{-5} \rangle = \langle 2, 1 + \sqrt{-5} \rangle$. Verify that $1 - \sqrt{-5} = 2 (-\sqrt{-5}) + (1 + \sqrt{-5})1$.
As for 3, we see that $-5 \equiv 1 \pmod 3$ and $1^2 = 1$, so $\langle 3 \rangle = \langle 3, 1 - \sqrt{-5} \rangle \langle 3, 1 + \sqrt{-5} \rangle$.
A: My answer is shorter than that of Watson, but since mine is informed by more advanced knowledge, you may prefer his answer to mine.
Because $6=2\cdot3$, it’s only necessary to express the ideals $(2)$ and $(3)$ of $R=\Bbb Z\sqrt{-5}$ as products of primes, since we know that prime-ideal decomposition is unique.
I know that the ramified primes are just $2$ and $5$, and because $-5$ is a square modulo $3$, $(3)$ should split. Thus I should find $(2)=\mathfrak p_2^2$ and $(3)=\mathfrak p_3\mathfrak p_3'$. Indeed,
\begin{align}
(2,1-\sqrt{-5})^2&=(4,2-2\sqrt{-5},-4-2\sqrt{-5})=(4,6,-4-2\sqrt{-5})=(2)\\
(3,1-\sqrt{-5})((3,1+\sqrt{-5})&=(9,3+3\sqrt{-5},3-3\sqrt{-5},-6)=(3)\,
\end{align}
so that the factorization of $6$ is $\left(2,1-\sqrt{-5}\right)^2\left(3,1-\sqrt{-5}\right)\left(3,1+\sqrt{-5}\right)$.
