# Prime Factors of a Primitive Element

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.

• 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 $Disc(F)$ – reuns Jul 4 at 20:00
• @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 $\mathcal{O}_K$). – Ashvin Swaminathan Jul 4 at 22:05
• What do you get with $f(x) = p x^2-1$ – reuns Jul 4 at 22:17
• @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 $\mathbb{P}^1(\mathbb{F}_p)$. – Ashvin Swaminathan Jul 4 at 23:02