Here is the question I want to solve:
Showing that $(1 + \sqrt{3})$ is a prime ideal of $\mathbb{Z}[\sqrt{3}]$ but that $(2)$ is not a prime ideal.
A HINT GIVEN TO ME:
For the first part I was given this hint:
Let $I$ be a maximal ideal containing $(1 + \sqrt{3})$ then use that every ideal of $\mathbb{Z}[\sqrt{3}]$ is principal but then I know that every maximal ideal is prime by a lemma. does that prove that $(1 + \sqrt{3})$ is a prime ideal of $\mathbb{Z}[\sqrt{3}]$? But I do not understand why we should have that $(1 + \sqrt{3})$ be a principle ideal first?
For the second part, $(2)$ is not a prime ideal. I was given this hint:
Use the Norm function and here is my trial:
Let $R = \mathbb{Z}[\sqrt{3}],$ suppose $2R$ is a prime ideal. Then we have $(\sqrt{3} + 1)(\sqrt{3} - 1) = 3 -1 = 2 \in 2R.$ Now suppose $(\sqrt{3} + 1) \in 2R$ then using the norm function we should have $N((\sqrt{3} + 1)) = N(2)N(c)$ for some $c,$ but $N(2) = 4$ and N((\sqrt{3} + 1)) = 2 but 4 does not divide 2. Same argument can be used for $(\sqrt{3} - 1).$
Could anyone check if my solutions are correct or no please? and if not, I want to know the correct solution please?