Must an irreducible element in $\mathbb{Z}[\sqrt{D}]$ have a prime norm? 
Let $D\in \mathbb{Z}$ where $D$ is not a perfect square. Prove that if $\alpha\in \mathbb{Z}[\sqrt{D}]$, and $\alpha= a+b\sqrt{D}$ with:                       $|a^2-Db^2| = p$, a rational prime, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{D}]$. Is the converse true?

So, for the if statement I figure I should assume $\alpha$ isn't irreducible, say $\alpha=\beta\gamma$ where they are not units, then $N(\beta)N(\gamma)=p$, but then I feel it is incorrect to assume a norm exists, because this would make it a Euclidean domain? is that the purpose of the absolute value?
 A: If $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{D}]$, is then $|\mathsf{N}(\alpha)| = p$ certain to be a prime number in $\mathbb{Z}$? (I'm stating the converse to make sure we're on the same page; if I've misunderstood what the converse is here then we'll never hear the end of it).
No. Consider $D = -10$. It turns out that $\sqrt{-10}$ is irreducible, yet it has a norm of 10. The formula for norm in this domain works out to be $$\mathsf{N}(a + b\sqrt{-10}) = a^2 - (-10)b^2 = a^2 + 10b^2.$$ That's all a norm is: a function that enables you to compare numbers from other domains within the familiar framework of $\mathbb{Z}^+ \bigcup \{0\}$. The Euclidean algorithm is another familiar thing from $\mathbb{Z}$, but, unlike the norm, it can't always be carried over.
The norm can never be a negative number here, so there's no need to specify absolute value (quite a different story if $D$ is positive). If $\sqrt{-10}$ is reducible, then we can solve $\mathsf{N}(\beta) = 2$ and $\mathsf{N}(\gamma) = 5$. Except we can't. The possible norms in this domain are 0, 1, 4, 9, 10, 11, 14, 16, 19, 25, ... (see Sloane's http://oeis.org/A020673). Either $\beta$ or $\gamma$ is a unit.
Two prior answerers already mentioned the norms of rational primes. I mention it again for the sake of completeness.
A: Given a number field $K$, the field norm $\mathsf{N}_{K/\mathbb{Q}}$ is defined regardless of whether the ring of integers $\mathcal{O}_K$ is a Euclidean domain. You can prove that an element $\alpha\in\mathcal{O}_K$ is a unit iff $\mathsf{N}_{K/\mathbb{Q}}(\alpha)=\pm1$.
Let $K=\mathbb{Q}(\sqrt{-5})$, so $\mathcal{O}_K=\mathbb{Z}[\sqrt{-5}]$. The element $2$ is irreducible in $\mathcal{O}_K$, but
$$\mathsf{N}_{K/\mathbb{Q}}(2)=2^2-(-5)0^2=4$$
is not a prime. You can see that $2$ is irreducible because, supposing for the sake of contradiction that $2=\beta\gamma$ for two non-units, we have
$$4=\mathsf{N}_{K/\mathbb{Q}}(2)=\mathsf{N}_{K/\mathbb{Q}}(\beta)\cdot \mathsf{N}_{K/\mathbb{Q}}(\gamma)=\pm2\cdot\pm 2$$
but there are no elements of $\mathcal{O}_K$ that are of norm $2$ or $-2$ because 
$$\mathsf{N}_{K/\mathbb{Q}}(a+b\sqrt{-5})=a^2-(-5)b^2=a^2+5b^2$$
always positive, it is $\geq 5$ if $b\neq 0$, and can only be a square number if $b=0$.
A: The converse can be true only when the norm is the square of a rational prime. If $D$ is negative, then $\alpha$ is purely real.
The norm function is available regardless of whether the domain is Euclidean or not. The purpose of the absolute value is so that the function does not return a negative number as it's result. For example, $N(1 - \sqrt{3}) = -2$, but with the absolute value, $|N(1 - \sqrt{3})| = 2$, so that it meets one of the requirements to be a Euclidean function (its a mapping to the positive integers and $0$).
Whether a particular positive integer can be a norm in a particular domain is a different issue. For example, in $\textbf{Z}[\sqrt{3}]$, no number has a norm of $5$ (and $5$ itself has a norm of $25$.
