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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.)

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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.

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  • $\begingroup$ Thank you a lot! Exactly such an answer I was hoping for! And didn't know about the book, but it explains to topic nicely - thank you for the pinpoint. $\endgroup$
    – slowpoke
    Commented Jun 20, 2021 at 23:16

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