# Primes above 2 and 3 in a quadratic number field

I'm reading Keith Conrad's note on the discriminant of a number field. So far I've made it about 1/3rd of a page :)

In example 1.2, he considers $$K = \mathbb Q (\alpha)$$ for $$\alpha$$ a root of $$T^3 - 9T - 6$$. He determines $$N(\alpha) = N(\alpha+3) = N(\alpha-3) =6$$ and then and then says "It follows that $$(\alpha) = p_2 p_3$$, $$(\alpha - 3) = p_2' p_3$$ and $$(\alpha + 3) = p_2' p_3$$". (Here I think $$p_2, p_2'$$ are primes above $$2$$ and $$p_3$$ above $$3$$). I can see why the norms imply $$(\alpha)$$ is the product of a prime above $$2$$ and $$3$$, but I don't immediately see why $$(\alpha -3)$$ must have a different prime above $$2$$ but the same prime above $$3$$, for example.

Using Dedekind Kummer's theorem I can show there's just one prime above $$3$$, and two above $$2$$, so that would get me part of the way. But the entire point of this example seems to be computing the factorization of $$(2)$$ and $$(3)$$, so I'm guessing there's a simpler way to see this intermediate step...

## 1 Answer

$$(\alpha) = \prod \mathfrak{p}_j^{e_j}$$

$$N(\alpha)=6\implies (\alpha) = p_2p_3$$ (where $$N(p_2)=2,N(p_3)=3$$)

$$N(\alpha-3)=6\implies (\alpha-3) = p_2'p_3'$$

$$N(p_3)=3 \implies 3\in p_3 \implies (3)= p_3 I_3$$ where $$N(I_3)= 3^2$$

$$(\alpha,\alpha-3) = (3,\alpha)= (p_3I_3,p_3p_2) = p_3 (I_3,p_2)=p_3$$ thus $$\alpha-3\in p_3,\alpha-3\not \in p_2$$ and $$(\alpha-3) = p_3 p_2'$$ where $$p_2'\ne p_2$$

• Ah perfect, thank you very much! – vacant Oct 19 '19 at 4:12