Proving Prime Numbers in Quadratic Imaginary Fields

[Kubik]

Hi,

I need your help with the next two problems:

1) If \(\displaystyle p\) is a prime number such that \(\displaystyle p\equiv{3}\;mod\;4\), prove that \(\displaystyle \sqrt{-p}\) is prime in \(\displaystyle \mathbb{Z}[\sqrt[ ]{-p}]\) and in \(\displaystyle \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}]\) too.

2) 2) We have \(\displaystyle d > 1\) a square-free integer. Consider the quadratic imaginary field \(\displaystyle K=\mathbb{Q}(\sqrt{-d})\). Suppose that \(\displaystyle d\equiv{1} \mod 4\) or that it is divisible by two prime numbers. Prove in both cases that in the ring of integers of \(\displaystyle K\) there exists a prime ideal which is not principal, but its square is principal.

Note

We have seen in class that if the norm is a prime then the element is irreducible.

Thanks!

 

[jvicens]

Hello,

I would be happy to help you with these problems. Here are my solutions:

1) To prove that \sqrt{-p} is prime in \mathbb{Z}[\sqrt[ ]{-p}], we need to show that it is irreducible in this ring. Let's assume that \sqrt{-p} = \alpha \beta, where \alpha and \beta are elements in \mathbb{Z}[\sqrt[ ]{-p}]. Then we have \sqrt{-p}^2 = \alpha \beta \alpha \beta = -p. This means that either \alpha or \beta must be a unit in \mathbb{Z}[\sqrt[ ]{-p}], since their product equals a prime element. However, the only units in this ring are \pm 1, and neither of these can be equal to \sqrt{-p} since p is a prime number. Therefore, \sqrt{-p} is irreducible in \mathbb{Z}[\sqrt[ ]{-p}], and hence prime.

To show that \sqrt{-p} is prime in \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}], we can use a similar argument. Assume that \sqrt{-p} = \alpha \beta, where \alpha and \beta are elements in \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}]. Then we have \sqrt{-p}^2 = \alpha \beta \alpha \beta = -p. This means that either \alpha or \beta must be a unit in \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}]. However, the units in this ring are of the form \pm \frac{a+b\sqrt{-p}}{2}, where a and b are integers. Since p is a prime number, this means that a and b must both be even, and therefore \pm \frac{a+b\sqrt{-p}}{2} cannot equal \sqrt{-p}. Therefore, \sqrt{-p} is irreducible in \mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}], and hence prime.

2) In the case where d\equiv{1}\mod 4, we can use the fact that the