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 $mathfrak{p}$?