Valuation of a number field element over a prime ideal in an order

Having read https://mathoverflow.net/questions/144671/number-field-sieve-for-factorization-with-non-monic-non-linear-polynomial-cant I stumbled on a problem I can't prove. Most of the questions posed I managed to prove, except for one little detail that is needed to sum everything up. The problem is the following:

Let $$f(x) = a_dx^d + a_{d-1}x^{d-1} + \ldots + a_1x + a_0$$ be a non-monic irreducible polynomial in $$\mathbb{Z}[x]$$, let $$\alpha$$ be one of it's roots. Call $$\mathbb{K}=\mathbb{Q}[\alpha]$$ the field extension induced by $$f(x)$$ and let $$\mathcal{O}$$ be the corresponding ring of integers.

The ring $$A=\mathbb{Z}[\alpha]\cap\mathbb{Z}[\alpha^{-1}]$$ is a subring of $$\mathcal{O}$$ (so an order of $$\mathbb{K}$$), and it is generated by the following algebraic integers

\begin{align} \beta _ { 0 } \quad &: = a _ { d } \alpha ^ { d - 1 } + a _ { d - 1 } \alpha ^ { d - 2 } + \cdots + a _ { 3 } \alpha ^ { 2 } + a _ { 2 } \alpha + a _ { 1 }\\ \beta _ { 1 } \quad &: = a _ { d } \alpha ^ { d - 2 } + a _ { d - 1 } \alpha ^ { d - 3 } + \cdots + a _ { 3 } \alpha + a _ { 2 }\\ \beta _ { 2 } \quad &: = a _ { d } \alpha ^ { d - 3 } + a _ { d - 1 } \alpha ^ { d - 4 } + \cdots + a _ { 3 }\\ \vdots \\ \beta _ { d - 3 } \: &: = a _ { d } \alpha ^ { 2 } \:\:\:+ a _ { d - 1 } \alpha \:\:\:\:\:+ a _ { d - 2 }\\ \beta _ { d - 2 }\: &: = a _ { d } \alpha \:\:\:\:\:+ a _ { d - 1 } \end{align}

So we also have $$A=\mathbb{Z} + \sum_{i=0}^{d-2} \beta_i\mathbb{Z}$$. Another representation of the $$\beta_i$$'s is $$\beta_{-1}=0$$ and $$\beta_i = (\beta_{i-1} - a_i)\alpha^{-1}.$$

Let $$p$$ be a prime that divides $$a_0$$, and consider the following homomorphism

\begin{align*} \varphi_0 \colon\mathbb{Z}[\alpha] &\to \mathbb{F}_p\\ \alpha &\mapsto 0. \end{align*}

Then $${\mathfrak{p}_0}=A\cap \operatorname{Ker}(\varphi_0)$$ is a first degree prime ideal of the order $$A$$. This gives rise to a valuation on $$\mathbb{K}$$ induced by $${\mathfrak{p}_0}$$, i.e. a homomorphism $$l_{\mathfrak{p}_0}\colon \mathbb{K}^*\to \mathbb{Z}$$. This homomorphism satisfies

$$l_{\mathfrak{p}_0}(x) = \sum_{\substack{ \mathfrak{q}\mid {\mathfrak{p}_0}\\ \mathfrak{q}\subseteq \mathcal{O}}} {f}(\mathfrak{q}/{\mathfrak{p}_0})\: l_\mathfrak{q}(x),$$ where $${f}(\mathfrak{q}/{\mathfrak{p}_0})= [\mathcal{O}/\mathfrak{q} : A/{\mathfrak{p}_0}]$$, and $$l_\mathfrak{q}$$ are the valuations induced by the unique factorization of ideals in $$\mathcal{O}$$.

Another way to compute $$l_{\mathfrak{p}_0}$$ is using the Jordan-Hölder theorem, if $$x\in A$$ then consider any non-refinable chain of ideals in $$A$$ $$A = \mathfrak { a } _ { 0 } \supseteq \mathfrak { a } _ { 1 } \supseteq \mathfrak { a } _ { 2 } \supseteq \cdots \supseteq \mathfrak { a } _ { t - 1 } \supseteq \mathfrak { a } _ { t } = x A.$$ Then $$l_{\mathfrak{p}_0}(x)$$ is the number of $$i$$'s in $$\{1,2,\ldots,t\}$$ such that $$A_i/A_{i-1}$$ is isomorphic to $$A/{\mathfrak{p}_0}$$ as an $$A$$-module. If $$x\not\in A$$ we can write $$x=a/b$$ with $$a,b\in A$$ and then $$l_{\mathfrak{p}_0}(x)=l_{\mathfrak{p}_0}(a)-l_{\mathfrak{p}_0}(b)$$.

Actually, for any prime ideal $$\mathfrak{p}$$ of $$A$$, the abovementioned works, and gives rise to a homomorphism $$l_\mathfrak{p}:\mathbb{K}^*\longrightarrow \mathbb{Z}$$. All of these homomorphisms are linear combinations of valuations of the underlying primes of $$\mathcal{O}$$ weighted by $$f(\mathcal{q}/\mathcal{p})$$, the same as above. Furthermore they satisfy that for every $$x\in \mathbb{K}^*$$ only a finite number of $$l_\mathfrak{p}(x)$$ is non-zero, and $$\operatorname{Nm}(x) = \prod_{\mathfrak{p}\subseteq A}(\# A/\mathfrak{p})^{l_\mathfrak{p}(x)}.$$

The question I have is - what is $$l_{\mathfrak{p}_0}(\alpha)$$? It should be $$l_p(a_0)$$, but I don't know how to prove that. Note that $$\alpha\not\in A$$ since $$f$$ is not monic, so one can write $$l_{\mathfrak{p}_0}(\alpha) = l_{\mathfrak{p}_0}(a_d \alpha) - l_p(a_d)$$ where both $$a_d,a_d\alpha\in A$$.

A couple of more useful informations. Let $$\mathfrak{p}_\infty$$ be another prime over $$p$$. Then $$l_{\mathfrak{p}_\infty}(\alpha)\not=0$$ only if $$\mathfrak{p}_\infty=A\cap\operatorname{Ker}(\varphi_\infty)$$ where

\begin{align*} \varphi_\infty \colon\mathbb{Z}[\alpha^{-1}] &\to \mathbb{F}_p\\ \alpha^{-1} &\mapsto 0. \end{align*}

Furthermore $$\mathfrak{p}_\infty$$ is a prime of degree 1.

The primes $$\mathfrak{p}_\infty$$ and $$\mathfrak{p}_0$$ are only two primes such that $$l_\mathfrak{p}(\alpha)\not=0$$. The valuation is different from 0 if and only if $$p\mid a_d$$ or $$p\mid a_0$$ for $$\mathfrak{p}_\infty$$ and $$\mathfrak{p}_0$$ correspondingly. We may suppose $$p$$ divides both $$a_d$$ and $$a_0$$.

From the multiplicative property of the norm, and the connection with valuations, we can see that (calling $$l_p$$ the valuation in $$\mathbb{Q}$$)

$$l_p(\operatorname{Nm}(\alpha)) = l_p\big(p^{l_{\mathfrak{p}_0}(\alpha)}p^{l_{\mathfrak{p}_\infty}(\alpha)} \big)$$

Since $$\operatorname{Nm}(\alpha)=\frac{a_0}{a_d}$$, we have $$l_{\mathfrak{p}_0}(\alpha) + l_{\mathfrak{p}_\infty}(\alpha) = l_p(a_0) - l_p(a_d),$$ but this doesn't seem to be enough to prove that $$l_{\mathfrak{p}_0}(\alpha)=l_p(a_0)$$. Another thing I believe have proven is that $$l_{\mathfrak{p}_0}(\alpha) \geq 0$$ and $$l_{\mathfrak{p}_\infty}(\alpha) \leq 0$$, but this still doesn't exclude the possibility that $$l_{\mathfrak{p}_0}(\alpha) = l_p(a_0) + c$$, and $$l_{\mathfrak{p}_\infty} = -l_p(a_d) - c$$ for some $$c > 0$$.

I believe a direct calculation of $$l_{\mathfrak{p}_0}(\alpha)$$ is needed (through Jordan-Hölder), but I'm not sure how to do it.

Any help would be useful. Thanks

• How do we get a valuation from the ideal $\mathfrak{p}_0$ in $A$? If this prime is regular(nonsingular, local ring is a PID), there is a unique prime in $O_K$ lying over it, but how do we define $l_{\mathfrak{p}_0}$ if this prime isn't regular? – user277182 Mar 28 at 14:29
• $l_{\mathfrak{p}}$ is defined as the (unique) homomorphism from $\mathbb{K}^*$ to $\mathbb{Z}$ that is $\geq 0$ for $x\in A$, $>0$ for $x\in \mathfrak{p}$ and $\operatorname{Nm}(x) = \prod_{{\mathfrak{p}}\subseteq A} (\#A/\mathfrak{p})^{l_\mathfrak{p}(x)}$. One can define these homomorphisms through Jordan-Hölder and further show that $l_\mathfrak{p}$ can be written as a linear combination of $l_\mathfrak{q}$ for $\mathfrak{q}\mid \mathfrak{p}$. I'm not sure if they are valuations ( I imagine $l_\mathfrak{p}(x+y) \geq {min}\{ l_\mathfrak{p}(x), l_\mathfrak{p}(y)\}$ can fail). – Kolja Mar 28 at 15:02
• I believe I made a mistake by calling $l_\mathfrak{p}$ a valuation. It is simply a homomorphism with some good valuation-like properties, I guess this is as close to a valuation as one can get when you're in a non-maximal order. – Kolja Mar 28 at 15:05