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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...

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

$N(\alpha)=6\implies (\alpha) = p_2p_3$ (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 $p_2'\ne p_2$

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

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