How does Dedekind's Theorem work for a prime dividing the discriminant of a number field? Let $f \in \mathbb{Z}[x]$ be an irreducible monic polynomial, let $N$ be its splitting field, and let $G$ be the Galois group of the extension $N/\mathbb{Q}$. Let $p$ be a prime dividing the discriminant of $N$. 

I would like to say that $f$ mod $p$ must admit a quadratic factor.

We know that $p$ ramifies in $N$ (because it divides the discriminant of $N$) and hence, provided $p$ is coprime to the conductor of $\mathbb{Z}[\theta]$ (where $\theta$ is a root of $f$), we may apply Dedekind's Theorem to obtain the result. 
However, if $p$ does divide the conductor, then I am unsure about what to do. 
 A: Yes, $\overline{f}\in\mathbb{F}_p[X]$ must indeed have duplicate factors.
Let $K=\mathbb{Q}(\theta)$, where $\theta\in N$ is a root of $f$. If $\prod_{1\leq i\leq s}\overline{g}_i^{e_i}$ is the factorization of $\overline{f}$ into irreducibles in $\mathbb{F}_p[X]$, where the $g_i$ $\in$ $\mathbb{Z}[X]$ are monic, then the primes above $p$ in $\mathbb{Z}[\theta]$ are the ideals $\mathfrak{p}_i$ $=$ $p\mathbb{Z}[\theta]$ $+$ $g_i(\theta)\mathbb{Z}[\theta]$. In $\mathbb{Z}[X]$, divide $f$ by $g_i$ and write $f$ $=$ $q_ig_i$ $+$ $r_i$ with $\deg(r_i)$ $<$ $\deg(g_i)$. Then the following quite precise form of the Kummer-Dedekind Theorem applies: $\mathfrak{p}_i$ is regular, that is, $\mathbb{Z}[\theta]_{\mathfrak{p}_i}$ is a DVR, iff $\mathfrak{p}_i$ is invertible in $\mathbb{Z}[\theta]$ iff $\mathfrak{p}_i$ is prime to the conductor $\mathfrak{f}$ $=$ $(\mathbb{Z}[\theta]:\mathfrak{O}_K)$ iff $e_i$ $=$ $1$ or $r_i$ $\not\equiv$ $0$ mod $p^2$ in $\mathbb{Z}[X]$. See Theorem (8.2) in this excellent overview.
For regular $\mathfrak{p}_i$, the ideal $\mathfrak{p}_i\mathfrak{O}_K$ is prime, and it contributes with exponent exactly $e_i$ to the prime factorization of $p$ in $\mathfrak{O}_K$. So, if all $e_i$ $=$ $1$, every $\mathfrak{p}_i$ is regular, and $p\mathfrak{O}_K$ $=$ $\prod_{1\leq i\leq s}P_i$ is the prime factorization of $p$ in $\mathfrak{O}_K$, where $P_i$ $=$ $\mathfrak{p}_i\mathfrak{O}_K$. Hence $p$ is unramified in $K$. But then $p$ is unramified in every conjugate of $K$ in $N$, so in $N$ itself - contradicting the assumption.
Incidentally, one has $\prod_{1\leq i\leq s}\mathfrak{p}_i^{e_i}\subseteq p\mathbb{Z}[\theta]$, with equality iff all $\mathfrak{p}_i$ are regular. When $\mathfrak{p}_i$ is singular (i.e. not regular, i.e. $e_i$ $\geq$ $2$ and $r_i$ $\equiv$ $0$ mod $p^2$), Theorem (8.2) provides, as a bonus, an explicit integer of $K$ that is not in $\mathbb{Z}[\theta]$, namely $\gamma$ $=$ $q_i(\theta)/p$. The order $\mathbb{Z}[\theta,\gamma]$ thus takes you one step closer to finding the ring of integers $\mathfrak{O}_K$.
