I am somewhat stuck with the proof of Lemma 2.3 on page 6 in this paper about an algorithm for polynomial factorization.
The preliminaries:
We have
- $f \in Z[X]$ monic, squarefree (which here means no multiple roots) with degree $N$
- A prime number $p$ such that $f \mod p$ remains squarefree (not sure if this is important here)
- $f = \prod_{i=1}^n f_i$ in $\mathbb{Z}_p[X]$
We also define (for a field $F$ and some $i \in \mathbb{N}$) the $i$-trace of a polynomial $g \in F[X]$ as
$$
Tr_i(g) := \sum_{l=1}^{\deg(g)} \zeta_l^i
$$
where the $\zeta_l$ are the $\deg(g)$ (not necessarily different) roots of $g$, as well as, for an integer $l$
$$
Tr_{1\dots l}(g):=\left(Tr_1(g),\,\dots,Tr_l(g)\right)
$$
Since $Tr_i(gh) = Tr_i(g) + Tr_i(h)$, all these terms can also be defined more generally for $g \in F(X)$ if we put
$$Tr_i(q/r) = Tr_i(q) – Tr_i(R)$$.
Now, to the problem I'm facing:
Assume we have, for indices
$v_1,\,\dots,v_n \in \mathbb{Z}$ that $$ g = \prod_i f_i^{v_i} \in
\mathbb{Q_p}(X) $$ Let $\alpha_1,\,\dots, \alpha_N$ be the roots of
$f$ in an algebraic closure of $\mathbb{Q}_p$. We can put $$
V_i:=Tr_{1,\,\dots,N}(f_i) \in \mathbb{Q_p}^N $$ and $$
V:=\sum_{i=1}^n v_i V_i = Tr_{1\dots N} (g) =
\left(Tr_1(g),\,\dots,Tr_N(g)\right) $$ How do we show that $$ V \in \mathbb{Q}^N \implies V \in \mathbb{Z}^N $$ holds?
The author says that this follows due to the roots of $f$ being 'algebraic integers'.
I've only found that apparently, $a$ being an algebraic integer as well as a rational number implies it's an integer.
But I don't see how that would apply to the $Tr_i(g)$.
I've also never worked with algebraic integers before. I found their definition as the algebraic closure of $\mathbb{Z}$ in $\mathbb{C}$, but we're in an algebraic closure of $\mathbb{Q_p}$, (how?) can we just ignore that? Or are we really only working in the splitting field of $f \in \mathbb{Z}[X] \subset \mathbb{Q}[X]$?
What I have found, but which didn't seem to help me/probably is useless:
- The roots of $f$ are $p$-adic integers, hence so are the $Tr_i(g)$ (use the uniqueness of $p$-adic expansions and that $f$ is monic), but that only means their denominators are not divisible by $p$…
- Trivially, if $V \in \mathbb{Q}^N$, then if $b_1,\,\dots,b_\nu$ are the roots of $g$ counted with multiplicity, $\left(b_1,\,\dots,b_\nu\right)$ fulfills multiple algebraic relations with coefficients in $\mathbb{Q}$, given by the components of $V$.
Best Answer
The $\mathbb Q_p$ vs $\mathbb C$ problem doesn't really matter for various reasons, one of them being that (as abstract fields) $\mathbb Q_p$ sits inside $\mathbb C$, therefore so does one algebraic closure.
You could also argue as you do : what matters is the splitting field of $f$ and this is unique up to isomorphism, and those isomorphisms cannot move $\mathbb{Z,Q}$ so any result concerning those doesn't depend on which splitting field you choose.
Algebraic integers in a field $B$ containing $\mathbb Z$ are precisely the roots of monic integer-coefficient polynomials.
The proof that an algebraic integer that is also rational is an integer is independent of the field $B$, it's actually quite easy.
Now this applies to $Tr_i(f)$ because $f$ is a monic polynomial with integer coefficients, therefore its roots are algebraic integers, and it can be shown (this is more subtle) that algebraic integers in $B$ form a subring of $B$, that means they are stable under addition and multiplication : if $x,y$ are algebraic integers, so are $xy$ and $x+y$.
You can easily deduce from that that if $\zeta_1,...,\zeta_r$ are algebraic integers, so is $\sum_j \zeta_j^i$: this is your $Tr_i(f)$.