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!


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


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

  • $\begingroup$ The question is about prime ideal not prime element. Sorry for the ambiguity. The post is now edited. Does the norm help to prove primality of an ideal too? Because an ideal generated by two prime elements may not be a prime ideal. Thanks! $\endgroup$ – Vinay Deshpande Aug 5 '20 at 13:05
  • $\begingroup$ @VinayDeshpande oh! then, the answer has to change. Wait till I try to add a new answer $\endgroup$ – vidyarthi Aug 5 '20 at 13:07
  • 1
    $\begingroup$ @VinayDeshpande see here $\endgroup$ – vidyarthi Aug 5 '20 at 13:27

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