In the ring $R = \Bbb Z[\sqrt{-3}]$, prove that $2+\sqrt{-3}$ is a prime element. 
In the ring $R = \Bbb Z[\sqrt{-3}]$, prove that $2+\sqrt{-3}$ is a prime element.

I know the definition of prime element and have to prove that $\forall a,b\in R, 2+\sqrt{-3}|ab \Rightarrow 2+\sqrt{-3}|a$ or $2+\sqrt{-3}|b$.
If $\forall a,b\in R, 2+\sqrt{-3}|ab$, $\exists N\in R, (2+\sqrt{-3})N=ab$.
So, $(2+\sqrt{-3})\overline {(2+\sqrt{-3})} N\bar N=a\bar ab\bar b$
Because $7N\bar N=a\bar ab \bar b$, $7|a\bar a$ or $7|b\bar b$.
I'm stuck here. I don't know about $\simeq$ polynomial ring or something so can you teach me how to prove that just by definition here?
 A: Recall that it is sufficient to show that $R/\langle 2 + \sqrt{-3} \rangle$ is a domain, so we will take this approach. 
First, I claim that $R \cong \mathbb{Z}[X]/\langle X^{2} + 3 \rangle$. Define a surjective homomorphism $\varphi \colon \mathbb{Z}[X] \to \mathbb{Z}[\sqrt{-3}]$ by $p(X) \mapsto p(\sqrt{-3})$, the so-called evaluation homomorphism at $\sqrt{-3}$. This is indeed a ring homomorphism (in fact, it is a homomorphism of $\mathbb{Z}$-algebras), and $a+bX \mapsto a+b\sqrt{-3}$ for any $a, b \in \mathbb{Z}$, so it is manifestly surjective. By the first isomorphism theorem for rings, it follows that $\mathbb{Z}[X]/\ker(\varphi) \cong \mathbb{Z}[\sqrt{-3}]$, so we must compute $\ker(\varphi)$. It is clear that $\langle X^{2} + 3\rangle \subset \ker(\varphi)$, but the other containment is trickier. Suppose $p(X) \in \ker(\varphi)$; since $X^{2}+3$ is a monic polynomial, we can perform Euclidean division to write $p(X)$ as $p(X) = q(X)(X^{2}+3) + r(X)$ for some $q(X), r(X) \in \mathbb{Z}[X]$, where $\deg(r(X)) < \deg(X^{2}+3) = 2$. Then 
$$\varphi(p(X)) = p(\sqrt{-3}) = q(\sqrt{-3})(0) + r(\sqrt{-3}) = r(\sqrt{-3}) = 0$$
and since no linear polynomial over $\mathbb{Z}[X]$ vanishes at $\sqrt{-3}$ (or else, among other things, $\sqrt{-3}$ would be rational!), we must have $r(X) \equiv 0$, i.e. $p(X) \in \langle X^{2}+3 \rangle$. Hence, $\ker(\varphi) = \langle X^{2} + 3\rangle$, so we have finally shown $R \cong \mathbb{Z}[X]/\langle X^{2} + 3\rangle$.
Then this isomorphism descends to an isomorphism of quotient rings $R/\langle 2 + \sqrt{-3} \rangle \cong (\mathbb{Z}[X]/\langle X^{2} + 3 \rangle)/\langle \overline{2+X} \rangle$, where $\overline{2+X}$ denotes the equivalence class of $2+X$ in $\mathbb{Z}[X]/\langle X^{2}+3\rangle$. But we have 
$$(\mathbb{Z}[X]/\langle X^{2} + 3 \rangle)/\langle \overline{2+X} \rangle \cong \mathbb{Z}[X]/\langle X^{2} + 3, 2+X \rangle \cong (\mathbb{Z}[X]/\langle X + 2 \rangle)/\langle \overline{X^{2}+3} \rangle \cong \\\mathbb{Z}/\langle (-2)^{2} + 3 \rangle \cong \mathbb{Z}/7\mathbb{Z}$$
since there is an isomorphism $\mathbb{Z}[X]/\langle X + 2 \rangle \to \mathbb{Z}$ arising from the evaluation map $\mathbb{Z}[X] \to \mathbb{Z}$ at $-2$ which (in particular) sends $X$ to $-2$. Since $\mathbb{Z}/7\mathbb{Z}$ is a field (hence a domain), we are done. 
It may seem like a lot of work went into this, but in truth, most of it was in justifying the isomorphism $\mathbb{Z}[X]/\langle X^{2}+3 \rangle \cong \mathbb{Z}[\sqrt{-3}]$, and this sort of computation becomes routine with practice. The nice thing about the strategy above is that it is relatively malleable in some sense - you could also use it to (say) prove that the ideal $\langle 2, 1+\sqrt{-5} \rangle$ is prime in $\mathbb{Z}[\sqrt{-5}]$. 
