Showing that an ideal is not principal 
I supposed to the contrary that $I$ is principal, so then $2$ and $1+\sqrt{-3}$ must have a common factor. However, since $2$ and $1+\sqrt{-3}$ are irreducible, the only common factors are $1$ and $-1$. Thus, $I$ is not principal.
Is this enough? Should I show that $I$ is a proper ideal? Thank you!
 A: Yes, you should also show that $I$ is proper.
A brute force computation to show that $I$ does not contain 1: Suppose $a, b, c, d \in {\mathbb Z}$ are such that $1 = (a + b \sqrt{-3}) 2 + (c + d \sqrt{-3}) (1 + \sqrt{-3})$. Then $2a + c - 3d + (2b + c + d)\sqrt{-3} = 1$. Therefore $2a + c - 3d = 1$ and $2b + c + d = 0$. Solving for $c$ in the second equation and substituting in the first equation gives $2a - 2b - 4d = 1$, but the left hand side is even and the right hand side is odd.
More elegant is to first go modulo 2 and show that the ideal $(1 + \sqrt{-3})$ of ${\mathbb Z}[\sqrt{-3}]/(2)$ is not the full ring. The $\sqrt{-3}$ might be confusing, so I'll write the ring ${\mathbb Z}[\sqrt{-3}]/(2)$ as the quotient ring ${\mathbb Z}_2[x]/(x^2 + 3)$ and the question is whether or not the ideal $(\overline{1 + x})$ is a proper ideal of this quotient ring. Since $x^2 + 3 = x^2 - 1 = (x - 1)(x+1)$, it is.
Finally, a simple example to show that you must indeed show that $I$ is proper. Consider the ideal $J = (x, x+1)$ of the polynomial ring ${\mathbb Q}[x]$. The polynomials $x$ and $x+1$ are irreducible, have no factor in common, but of course $J$ is still a principal ideal, as $J = (1)$.
