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Let $p \in \mathbb{Z}$ be a prime, and let $f(x) = px^n + \dots$ be an irreducible degree $n$ polynomial over $\mathbb{Z}$ with leading coefficient equal to $p$. Suppose that $f(x)$ has no repeated roots modulo $p$. Then $p$ is unramified in the number field $K = \mathbb{Q}[x]/(f(x))$ and splits into distinct prime ideals $(p) = \prod_{i = 1}^r \mathfrak{p}_i$ of ramification degree $1$.

Question: Let $\theta$ be the image of $x$ in $K$. Can anything be said about the multiplicity of each prime ideal $\mathfrak{p}_i$ in the prime factorization of $\theta$?

What I know: If $p^a$ divides the constant term of $f$, then $p^{a-1} \mid \operatorname{Norm}(\theta)$. Also, since $p \theta$ is an algebraic integer, we know that every prime factor of $p$ appearing with negative multiplicity in $\theta$ must appear with multiplicity $-1$, but I don't know what else can be deduced.

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  • $\begingroup$ You are supposed to replace $f$ by $F(x)=p^{1-n} f(px)$ to make it monic, if it has no repeated root then $F(x) = \prod_j F_j(x) \bmod p$ and $(p)= \prod_j (p,F_j(x)) \in R=\Bbb{Z}[x]/(F(x))$ the $(p,F_j(x))$ being also the maximal ideals above $p$ in $O_K$. The $p$ ramified in $R$ are those dividing $Disc(F)$ $\endgroup$ – reuns Jul 4 at 20:00
  • $\begingroup$ @reuns To know that p is unramified in $\mathcal{O}_K$, I do not need to consider the monic form $F(x)$. As long as $f$ has no repeated roots modulo $p$, then $p$ does not divide the discriminant of $f$. The discriminant of $K$ divides the discriminant of $f$, so in particular, $p$ does not divide the discriminant of $K$ and is therefore unramified (in $\mathcal{O}_K$). $\endgroup$ – Ashvin Swaminathan Jul 4 at 22:05
  • $\begingroup$ What do you get with $f(x) = p x^2-1$ $\endgroup$ – reuns Jul 4 at 22:17
  • $\begingroup$ @reuns So really one needs to consider the homogenized form $f(x,z) = px^2 - z^2$, which evidently has a double root modulo $p$ at infinity in $\mathbb{P}^1(\mathbb{F}_p)$. $\endgroup$ – Ashvin Swaminathan Jul 4 at 23:02

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