# 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.)

• See Keith Conrad's notes Factoring After Dedekind. He does many examples of this sort. Commented Jun 21, 2021 at 6:07

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.

• 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. Commented Jun 20, 2021 at 23:16