# algebraic number theory – Example of prime decomposition using dedekind theorem in \$Bbb{Q}(sqrt{3})\$

So, I’m trying to understand the prime decomposition of a prime number, $$p$$, in $$K=Bbb{Q} (sqrt{d})$$ with $$d=3$$. First we have to calculate the discriminant of $$K$$:

$$D:=disc(K)= begin{cases} 4cdot d & text{if d=2,3 mod 4}\ d & text{if d=1 mod 4} end{cases}$$

So in our case we have $$D=12$$. Now Dedekind theorem says:
$$p|D quad Leftrightarrow quad pO_K=mathfrak{p}^2$$
for a primeideal $$mathfrak{p}$$ in $$O_Bbb{Q}=Bbb{Z}$$.

We clearly see that $$p|D$$ if $$p=2,3$$. So now consider $$p=2$$. How would one write the exact prime decomposition? I was thinking something like:
$$2Bbb{Z} (sqrt{3}) = mathfrak{p}^2$$
but how do I know what $$mathfrak{p}$$ is?

If we on the other hand have that $$p$$ odd and not dividing $$D$$, Dedekind theorem gives us that:
$$pO_K= begin{cases} mathfrak{p}_1 mathfrak{p}_2 & text{if D a square mod p}\ mathfrak{p} & text{if D not a square mod p} end{cases}$$
Clearly this holds when $$p=5,7,11$$. Lets now consider $$p=5$$. I see that $$12=2 text{ (mod 5)}$$ and therefore not a square. So I suppose:
$$5 Bbb{Z}(sqrt{3})=mathfrak{p}$$
But again, how do I determine $$mathfrak{p}$$?