1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – user277182 Mar 28 at 14:29
  • $\begingroup$ $l_{\mathfrak{p}}$ 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). $\endgroup$ – Kolja Mar 28 at 15:02
  • $\begingroup$ 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. $\endgroup$ – Kolja Mar 28 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.