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(I'm going to use $o$ for the lowercase mathcal(o) in the picture). In the proposition below, $o$ is a Dedekind domain, $L/K$ is a separable extension, $\mathcal{O}$ is the integral closure of $o$ in $L$, and $L = K(\theta)$ where $\theta \in \mathcal{O}$. The conductor is defined to be $$\mathfrak{F} = \{\alpha \in \mathcal{O} \mid \alpha \mathcal{O} \subset o[\theta]\}$$

enter image description here

My question is, what happens when $\overline{p}(X) = 1$?

For example, let $K = \mathbb{Q}$ and $L = K(\alpha)$ where $\alpha$ is a root of $p(X) = X^3+2X+1$. I've found the ring of integer of $L$ to be $\mathbb{Z}[\alpha]$. So the conductor is just $\mathbb{Z}(\alpha)$, and so is relatively prime to the ideal $(3)$. But $p(X)$ mod 3 is simply 1. I also know that discriminant of the field $L$ is $-59$, so 3 is unramified in $L$. So can I conclude that 3 remains prime in $L$?

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As far as I can see, $\overline p(X)=X^3+\overline 2X+\overline1=X^3-X+\overline 1$ in $\Bbb F_3$. This polynomial is irreducible over $\Bbb F_3$ so $3$ is inert in $L$ ($3\mathcal{O}_L$ is prime).

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  • $\begingroup$ Ah, yeah you're right $\endgroup$ – Aaron Johnson Jan 14 '18 at 8:54

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