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 $K$:
$$
D:=disc(K)=
    \begin{cases}
        4\cdot d & \text{if $d=2,3$ mod $4$}\\
        d & \text{if $d=1$ mod $4$}
    \end{cases}
$$
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 $O_\Bbb{Q}=\Bbb{Z}$. ($O_K$ being the ring of integers of a field K)
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:
$$
2\Bbb{Z} [\sqrt{3}] = \mathfrak{p}^2
$$
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 $\mathfrak{p}$?
(Dedekind theorem also determine the prime decomposition for $p=2$, but I'm not so interested in this special case.)
 A: I will not give a proof of my claims, but I believe this can all be found in Serge Lang's Algebraic Number Theory.
Take the minimal polynomial $P$ of a integral generator $\alpha$ of $K$ and look at its reduction mod $p$. It will factor mod $p$ (ie $\bar P = \prod_i \bar P_i^{e_i}$ for some polynomials $P_i$, where $\bar P$ is the reduction mod $p$ of $P$). Then, letting $\mathfrak p = p O_K$, we find that its prime decomposition is $\mathfrak p = \prod \mathfrak P_i^{e_i}$ where $\mathfrak P_i = pO_K+ P_i(\alpha)O_K$.
In the case of $\mathbb Q[\sqrt{3}]$ we have $\alpha =\sqrt{3}$, $P = X^2-3$ and $D = 12$. The primes of interest are $2$ and $3$, as you worked out.
You'll find that the reduction mod $p$ of $P$ factors: $P \equiv (X+1)^2 \pmod 2 $ and $P \equiv X^2 \pmod 3$.
First let's look at the first case (mod $2$). This implies that
$$2\mathbb Z[\sqrt{3}] = \mathfrak P_1^2\,,$$
where $\mathfrak P_1 = 2\mathbb Z[\sqrt{3}] + (\sqrt{3}+1)\mathbb Z[\sqrt{3}]$.
In the second case (mod $3$), we find that
$$3\mathbb Z[\sqrt{3}] = \mathfrak P_2^2\,,$$
where $\mathfrak P_2 = 3\mathbb Z[\sqrt{3}] + \sqrt{3}\mathbb Z[\sqrt{3}] = \sqrt{3}\mathbb Z[\sqrt{3}]$.
This also holds for primes that do not divide the discriminant: for $p=5$, $X^2-3$ is irreducible mod $p$ (otherwise it would have a root). Thus $5\mathbb Z[\sqrt{3}]$ is a prime ideal. (as you have written $\mathfrak p = 5\mathbb Z[\sqrt{3}]$).
Note that this requires some mild assumptions on $p$, which hold for all primes in the case of quadratic extensions.
