prove $\left(3, 1+\sqrt{-5}\right)$ is prime ideal of $\mathbb{Z}\left[\sqrt{-5}\right]$ How to prove that $(3, 1+\sqrt{-5})$ is prime ideal of $\mathbb{Z}[\sqrt{-5}]$?
attempt 1: use definition
Consider $a, b, c, d, k_1, k_2 \in \mathbb{Z}$ s.t. $$ac-5bd=3k_1+k_2,\, \, ad+bc=k_2.$$ To prove $\exists j_1, j_2 \in \mathbb{Z}$ s.t. $3j_1+(1+\sqrt{-5})j_2=a+b\sqrt{-5}$ or $=c+d\sqrt{-5}$.
This is a bad way.
attempt 2:
To prove $\dfrac{\mathbb{Z}\left[\sqrt{-5}\right]}{\left(3, 1+\sqrt{-5}\right)}$ is integral domain. I know how to work with quotient of polynomial ring but not how to work with quotient of $\mathbb{Z}\left[\sqrt{-5}\right]$.
attempt 3:
$$\mathbb{Z}\left[\sqrt{-5}\right]\cong \mathbb{Z}/\left(x^2+5\right)$$
When we have $\mathbb{Z}/\left(x^2+5\right)$, converting into $\mathbb{Z}\left[\sqrt{-5}\right]$ simplifies the problem. May be the other way round is useless.
Please give a hint. Please do not give solution. Thanks!

Hint: The square of norm function $(a^2-5b^2)$ is a multiplicative function in the ring $\mathbb{Z}[\sqrt{-5}]$ for a number $a+b\sqrt {-5}$. Use this to prove the primality by proving one of the factors of the norm is $1$. After showing that the numbers are primes, it is correct that the ideal you describe is prime, by using this answer
