# 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.

$$\frac{\mathbb{Z}\left[\sqrt{-5}\right]}{\left(3, 1+\sqrt{-5}\right)} \cong \frac{\mathbb{Z}[x]}{\left(3,1+x,x^2+5\right)} \cong \frac{\mathbb{Z}_3[x]}{\left(1+x,x^2-1\right)} \cong \cdots$$
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