0

I didn't study much about Galois theory and it's been a while, so I could really use some help with a sentence in this paper about an algorithm for polynomial factorization.

Preliminaries are as in another question of mine find it here:

  • $f \in Z[X]$ 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 $$Tr_i(q/r) = Tr_i(q) - Tr_i(R)$$.

If it helps, the $f_i \in \mathbb{Q}_p[X]$ can be assumed to be irreducible.

Also, for $i \in \{1\dots,n\}$, put $$ V_i := Tr_{1\dots N}(f_i) $$ and assume $V_i \in \mathbb{Q_p}^N$ for all $i$.

Then, if $g = \prod_{i=1}^n f_i^{v_i}$ for arbitrary integers $v_i$, $g \in \mathbb{Q_p}(X)$ and using the extended definitions of the trace implicated above,

$$ Tr_{1,\,\dots N}(g) = V := \sum_{i=1}^n v_i V_i $$

Now, I already know that the $V_i$ in the case I'm looking at are linearly independent over $\mathbb{Q}(\alpha_1,\,\dots,\, \alpha_n)$ so that we have an isomorphism of groups

$$ Tr_{1\dots N}: \left(\left\{\prod_{i=1}^n f_i^{v_i};\;v_1,\,\dots,v_n \in \mathbb{Z}\right\},\,\cdot\right) \rightarrow \left(\mathbb{Z} V_1 +\,\dots+ \mathbb{Z} V_n,\, +\right) \subset \mathbb{Q}_p^N $$

Assume that furthermore, $V \in \mathbb{Z}^N$ (the author wants to show that $g \in \mathbb{Q}(X)$

Now, the statement I don't understand is

The Galois group over $\mathbb{Q}$ leaves $V$ invariant, hence it permutes the preimages of $V$, so the fact that the map is 1-1 implies that $g$ is invariant and hence defined over $\mathbb{Q}$.

I don't really see where exactly Galois theory is applied here... are we talking about the Galois group of the Galois extension $\mathbb{Q}(\alpha_1,\,\dots,\,\alpha_n)/\mathbb{Q}$? How/where is it applied, and what is the connection to preimages etc.?

I don't really know where to look/what to look for to study up on, because the sentence is written for people who already understand exactly what he is talking about here, and I do not...

polynomial_donut
  • 1,108
  • 5
  • 20
  • 1
    What is the goal here? (In the sense of, what is the author actually trying to prove?) – darij grinberg Aug 12 '19 at 11:26
  • The argument means to show that $g$ has rational coefficients, i.e. $g \in \mathbb{Q}(X)$ (vs. just being in $\mathbb{Q}_p(X)$) – polynomial_donut Aug 12 '19 at 11:45
  • 1
    By taking one of the $v_i$ to be $1$ and the others to be $0$, this is claiming that all the $f_i$ have rational coefficients. And that's not true. A monic polynomial over the integers can have more irreducible factors in the $p$-adic rationals than it has in the usual rationals. This happens even for the polynomial $x^2 - a$ when $a$ has a square root in the $p$-adics. You must be missing some context. – darij grinberg Aug 12 '19 at 11:58
  • I'll look into this later, for now I've fixed the def. of $V$ (in case you missed that, the $V_i$ and hence $V$ do have a somewhat closer relation to the $f_i$ via $Tr$, see the paper I linked to (which isn't long, I think this part was on page 5 or 6, maybe 7) – polynomial_donut Aug 12 '19 at 12:40
  • @darijgrinberg If, say $v_1 = 1$ and the other $v_i$ are zero, then this implies $\mathbb{Q}^N \ni V = V_1 = Tr_{1,,\dots,,N}(f_1)$, hence by Lemma $2.2$ from the paper we have $f_1 \in \mathbb{Z}[X]$, so I don't see a contradiction. Also, the $f_i$ may be assumed to be irreducible over $\mathbb{Q}_p[X]$ if that helps. – polynomial_donut Aug 12 '19 at 14:50
  • 1
    Ah, there was an extra assumption here that $V \in \mathbb{Q}^N$. It is much more plausible now. I don't see through the reasoning in the paper, though; too much happening in a single sentence. – darij grinberg Aug 12 '19 at 14:57
  • Is this due to how I wrote my question about said reasoning (in which case I could try to improve on that) or due to how the paper is written? – polynomial_donut Aug 12 '19 at 16:24
  • It's in the paper, too. I perfectly understand where your confusion is coming from. – darij grinberg Aug 12 '19 at 16:25
  • Hmm I'm wondering if the author just added this as a supposedly cool side-result (doesn't look like it's used later on in the paper), thus not checking thouroughly if his reasoning was correct – polynomial_donut Aug 12 '19 at 17:18

2 Answers2

1

$f \in \Bbb{Q}[X]_{monic}$ factors into irreducibles in $\Bbb{Q}_p$ $$ f = \prod_{i=1}^m f_i^{e_i} \in \Bbb{Q}_p[X]_{monic}$$

Let $K$ be the extension of $\Bbb{Q}$ generated by all the roots of $f$.

Let $Tr(f_i) \in \Bbb{Q}_p \cap K$ be the sum of the roots of $f_i$.

Assume the $Tr(f_i) $ are $\Bbb{Q}$-linearly independent (in general it has no reason to hold)

For $s,r \in \Bbb{Q}^m$, that $\sum_i s_i Tr(f_i) = \sum_i r_i Tr(f_i) \in K\cap \Bbb{Q}_p$ implies $r= s$.

Now we have the $\sigma\in G= Gal(K/\Bbb{Q})$ action on the roots of $f$.

Assume $K\cap \Bbb{Q}_p / \Bbb{Q}$ is Galois

So that $\sigma$ sends $K\cap \Bbb{Q}_p$ to itself, thus it must also permute the $f_i$ (by applying $\sigma$ to the coefficients of $f_i$) thus sending $f_i$ to $f_i^\sigma = f_{\sigma_i}$.

If $\sum_i s_i Tr(f_i) \in \Bbb{Q}$ then $\sum_i s_i Tr(f_{\sigma_i}) = \sigma(\sum_i s_i Tr(f_i))=\sum_i s_i Tr(f_i)$ which implies $s_i = s_{\sigma_i}$.

If $s \in \Bbb{Z}^m$ it implies $(\prod_i f_i^{ s_i})^\sigma = \prod_i f_i^{ s_i} \in K(X)$.

And $\forall \sigma\in G$, $(\prod_i f_i^{ s_i})^\sigma = \prod_i f_i^{ s_i}$ implies $\prod_i f_i^{ s_i} \in K^G(X) = \Bbb{Q}(X)$

where $K^G=\Bbb{Q}$ is the subsfield of $K$ fixed by $G$.

reuns
  • 77,999
  • Thank you, I will look at it later in detail... some questions/comments:
    • $\mathbb{Q}_p$-linear independence is proven in the paper in a correct manner, I think (first paragraph of the proof, page 6)
    • I assume this also works with $Tr_{1\dots N}(f_i)$ instead of the $Tr$ (?)
    • I assume you're identifying $\sigma$ with the permutation for the indices, as well as for vector coefficients (?) (?), because you're using $\sigma$ in a lot of different contexts it seems
    • what is $d$?
     – polynomial_donut Aug 12 '19 at 17:34
  • $\sigma$ is permuting the roots thus the $f_i$ thus the indices $i$. $Q_p$-linear independence is stronger than $Q$-linear independence. – reuns Aug 12 '19 at 17:37
  • Ok, to clarify my question: You have $\sigma \in Gal\left(K/\mathbb{Q}\right)$, so it's an automorphism defined on $K$. Hence, $\sigma(i)$ is probably not what you state it means, because $\sigma$ leaves all elements of $\mathbb{Z}\subset \mathbb{Q}$ fixed.

    Also, isn't it trivial that $\sigma(\sum_i s_i Tr(f_i)) = \sum_i s_i Tr(f_i)$, because $\sum_i s_i Tr(f_i) \in \mathbb{Q}$ and $\sigma$ leaves all elements of $\mathbb{Q}$ fixed?

     – polynomial_donut Aug 12 '19 at 18:23
  • Apply $\sigma$ to the roots of the $f_i$, it is the same as applying $\sigma$ to the coefficients of the $f_i$, you'll obtain $f_i^\sigma = f_{\sigma_i}$ for some permutation of the indices that of course we call $\sigma$. Sure it is trivial that $\sigma(\sum_i s_i Tr(f_i)) = \sum_i s_i Tr(f_i)$ when $\sum_i s_i Tr(f_i) \in \Bbb{Q}$, that's the point. – reuns Aug 12 '19 at 18:25
  • I guess $\sigma$ permutes the $f_i$, because $f_i ;\text{irreducible}; \implies f_i^\sigma ; \text{irreducible}$ (and $\mathbb{Q}_p[X]$ is a UFD)? (Is this easy to show (sorry, haven't seen Galois theory in years)?) Because otherwise, I wonder if some $f_i$ could be mapped to a reducible factor somehow, although intuivively I would say that's impossible. – polynomial_donut Aug 12 '19 at 18:30
  • 1
    I realize I need to assume $K\cap \Bbb{Q}_p / \Bbb{Q}$ is Galois to obtain that $\sigma$ permutes the $f_i$. I don't know if it can be repaired when it is not Galois. – reuns Aug 12 '19 at 18:49
  • Is it not enough that $K/\mathbb{Q}=\mathbb{Q}(\alpha_1,,\dots,,\alpha_n)/\mathbb{Q}$ is Galois, since it is the splitting field of a polynomial $f$ (hence normal) and $f$ is squarefree (hence if I remember correctly, the extension should be separable)? – polynomial_donut Aug 12 '19 at 18:56
1

First note that$\mathbb{Q}(α_1,\,\dots,α_n)/\mathbb{Q}$ is a Galois extension:

  • It is normal, as it's the splitting field of $f$
  • It is obviously algebraic
  • It is obviously separable ($f$ was assumed to be squarefree)

$\newcommand{Gal}{\text{Gal}\!\left(\mathbb{Q}(α_1,\,\dots,α_n)/\mathbb{Q}\right)}$ Hence, we know that $\mathbb{Q}$ is the fixed field of the Galois group $\Gal$.

$\newcommand{gFactorSet}{\left\{\prod_{i=1}^n f_i^{v_i};\;v_1,\,\dots,v_n \in \mathbb{Z}\right\}}$

$\newcommand{gFactorSetGroup}{\left(\gFactorSet,\,\cdot\right)}$ $\newcommand{zSpanExponents}{\left(\mathbb{Z} V_1 +\,\dots+ \mathbb{Z} V_n\right)}$

$\newcommand{zSpanExponentsGroup}{\left(\zSpanExponents,\, +\right)}$ We also have a group action of $\Gal$ on $\gFactorSet$ defined by $$ \sigma_\cdot\left(\sum_{i=1}^{\deg(f)} f_i X^i\right)=\sum_{i=1}^{\deg(f)} \sigma(f_i) X^i $$ (i.e. the Galois morphisms act on the coefficients of the polynomials) as well as an action on $\zSpanExponents$ given by $$ \sigma_\ast\left(\sum_{i=1}^n v_i V_i\right)=\sum_{i=1}^n \sigma(v_i) \left(\sigma((V_i)_j)\right)_{j=1,\,\dots,\,N}=\sum_{i=1}^n \sigma(v_i) \left(\sigma(\text{Tr}_j(f_i))\right)_{j=1,\,\dots,\,N} =\sum_{i=1}^n v_i \left(\sigma(\text{Tr}_j(f_i))\right)_{j=1,\,\dots,\,N} $$ for $\sigma \in \Gal$.

$\sigma_\text{poly}:=\sigma_\cdot\left(\,\cdot\,\right)$ and $\sigma_\text{lin}:=\sigma_\ast\left(\,\cdot\,\right)$ are set (even group) automorphisms for each $\sigma$. It is easy to check that this yields a commutative diagram of sets - in fact, simply using that $\sigma$ is a $\mathbb{Q}(\alpha_1,\,\dots,\,\alpha_n)/\mathbb{Q}$-endomorphism yields that we have a commutative diagram of groups with the groups structures $\gFactorSetGroup$ and $\zSpanExponentsGroup$, but this is not needed here - as follows

$\require{AMScd}$ \begin{CD} \gFactorSet @>{Tr_{1\dots N}}>> \zSpanExponents\\ @VV\sigma_\text{poly}V @VV\sigma_\text{lin}V\\ \gFactorSet\mathbb{Z} @>{Tr_{1\dots N}}>> \zSpanExponents \end{CD} $\newcommand{Trace}{Tr_{1\dots N}}$ and we have $$ g=\Trace^{-1}(V)={\sigma_\text{poly}}^{-1}\Trace^{-1}\sigma_\text{lin}(V) $$ Since $\sigma \in \Gal$, the right hand side along with $V \in \mathbb{Q}^N$, equals $$ {\sigma_\text{poly}}^{-1}\Trace^{-1}(V) = {\sigma_\text{poly}}^{-1}(g) $$ so all in all, $$ g = {\sigma_\text{poly}}^{-1}(g) $$ which, looking at the coefficients and considering that $\sigma$ can be any element of $\Gal$, noting that ${\sigma_\text{poly}}^{-1}=\left({\sigma^{-1}}\right)_\text{poly}$, yields that they are $\Gal$-invariants and hence must be in $\mathbb{Q}$.

polynomial_donut
  • 1,108
  • 5
  • 20
  • I'm not sure we need that the diagram actually commutes in the category of groups. But I'll leave this answer for now to recheck later – polynomial_donut Aug 13 '19 at 08:33