ag.algebraic geometry – All Galois characters showing up in cohomology of one family of varieties

Fix a prime $p$.

Can we find a smooth proper map $Xto Y$ of $mathbb{Q}_p$-varieties such that any given representation $mathrm{Gal}(overline{mathbb{Q}_p}/mathbb{Q}_p)to mathrm{GL}_1(mathbb{F}_p)$ embeds in $oplus_{igeq 0} H^i_{mathrm{acute{e}t}}(X_ytimes overline{mathbb{Q}_p}, mathbb{F}_p)$ for some $yin Y(mathbb{Q}_p)$?

Why does the definition of a fixed field specify a subgroup of the Galois group, not just a subset?

Why does the definition for the fixed field of H specify that H must be a subgroup of the Galois group of the field extension? As far as I can see, if H is simply a subset, the ‘fixed field’ of H is still a field, as we have closure under all 4 operations.

So, am I missing something, or is the requirement for H to be a subgroup not necessary for the fixed field of H to actually be a field?

ag.algebraic geometry – A question on Galois groups and deformations of Galois coverings

I’ve recently been working on several problems related to deformations of Galois covers of projective space, and have come across the following situation: If $X$ is a projective variety and $G$ is a (finite) subgroup of automorphisms of $X$, let $pi_G:Xto X/G$ be the quotient map. We can get a family of Galois covers of $X/G$ by considering the map $pi_Gcircsigma^{-1}$ for $sigmainmathrm{Aut}^0(X)$, which is just the quotient map for the group $sigma Gsigma^{-1}$. I’m particularly interested in the case when $X/Gsimeqmathbb{P}^n$. My question is the following:

Does there exist a projective variety $X$ and a non-trivial family of Galois covers $f_t:Xtomathbb{P}^n$ with Galois groups $G_t$ such that not all the $G_t$‘s are conjugate? I see no intrinsic reason why they should be conjugate, but I can’t seem to find any examples either. For example, if $X$ is an abelian variety then all the groups are conjugate in this situation (this follows from the main result of https://arxiv.org/pdf/1801.00028.pdf, for example).

Obtaining an exact sequence of Galois modules via derived functors

This question has two parts, the first part will be to obtain the desired exact sequence while the second will be to study it in the corresponding derived category and try to obtain it from there.

Let $X$ be a smooth geometrically integral variety over a number field $k$ with canonical morphism $pi:X rightarrow mathrm{Spec},k$, I want to obtain the following exact sequence of $mathrm{Gal}(bar{k}/k)$-modules (all tensor products are over $mathbb{Z}$):

$$mu_infty rightarrow bar{k}(X)^* rightarrow bar{k}(X)^* otimes mathbb{Q} rightarrow H^1(bar{X},mu_infty) rightarrow mathrm{Pic}(bar{X}) rightarrow mathrm{Pic}(bar{X})_{free} rightarrow 0,$$

where $bar{X} := X times _k bar{k}$, $mu_infty = mathrm{colim}_nmu_n$, and $mathrm{Pic}(bar{X})_{free}$ denotes the maximal free quotient of $mathrm{Pic}(bar{X})$.

A very natural approach will be to apply the Galois cohomology functor $H^i(bar{X},-)$ to the exact sequence

$$1 rightarrow mu_infty rightarrow mathbb{G}_m rightarrow mathbb{G}_m otimes mathbb{Q} rightarrow 1.$$

To see why this sequence is exact refer to the answer of my post A Kummer exact sequence involving $mu_infty$.

So everything is fine except for the surjective map $mathrm{Pic}(bar{X}) rightarrow mathrm{Pic}(bar{X})_{free}$, which can be rewritten as

$$mathrm{Pic}(bar{X}) rightarrow mathrm{Ker}(H^1(bar{X},mathbb{G}_m) otimes mathbb{Q} rightarrow H^2(bar{X},mu_infty)).$$

Question 1. How do I show that $mathrm{Pic}(bar{X})_{free}$ is precisely the kernel written above?

One thought I have is that perhaps we can try to show that $H^2(bar{X},mu_infty) = 0$, and then we would need to prove that $mathrm{Pic}(bar{X}) otimes mathbb{Q} cong mathrm{Pic}(bar{X})_{free}$, which would not be surprising since we are sort of ‘killing off’ the torsion parts of the Picard group.

Now on to the next part of the question, we have that the inclusion $mu_infty rightarrow mathbb{G}_m$ induces the map

$$varphi: tau_{leq 1}Rpi_*mu_infty rightarrow tau_{leq 1}Rpi_*mathbb{G}_m$$

in the category of bounded sheaves of complexes of discrete Galois modules. Let $D := mathrm{Cone}(varphi)$, thus we have a distinguished triangle

$$tau_{leq 1}Rpi_*mu_infty rightarrow tau_{leq 1}Rpi_*mathbb{G}_m rightarrow D rightarrow (tau_{leq 1}Rpi_*mu_infty)(1).$$

This would give rise to the exact sequence

$$0 rightarrow tau_{leq 1}Rpi_*mathbb{G}_m rightarrow D rightarrow (tau_{leq 1}Rpi_*mu_infty)(1) rightarrow 0.$$

Let $h^i(A)$ denote the $i$-th cohomology group of the complex $A$, then we have a long exact sequence of cohomology groups

$$h^{-1}(D) rightarrow h^{-1}((tau_{leq 1}Rpi_*mu_infty)(1)) rightarrow h^0(tau_{leq 1}Rpi_*mathbb{G}_m) rightarrow h^0(D) rightarrow h^0((tau_{leq 1}Rpi_*mu_infty)(1))$$
$$rightarrow h^1(tau_{leq 1}Rpi_*mathbb{G}_m) rightarrow h^1(D) rightarrow h^1((tau_{leq 1}Rpi_*mu_infty)(1)).$$

Here we only consider $h^i(D)$ for $i = -1,0,1$ because by definition, the cohomology is zero outside these degrees. It is well-known that since $X$ is smooth over $k$, $tau_{leq 1}Rpi_*mathbb{G}_m$ is quasi-isomorphic to the complex $(bar{k}(X)^* rightarrow mathrm{Div}(bar{X}))$ in degrees 0 and 1. Thus one easily computes that $h^0(tau_{leq 1}Rpi_*mathbb{G}_m) cong bar{k}(X)^*$ and $h^1(tau_{leq 1}Rpi_*mathbb{G}_m) cong mathrm{Pic}(bar{X})$. Also, we have

$$h^i((tau_{leq 1}Rpi_*mu_infty)(1)) = h^{i+1}(tau_{leq 1}Rpi_*mu_infty)$$

and so the last term of the above long exact sequence is 0.

Question 2. We want to show that this long exact sequence is precisely the one mentioned in the first part of the question. All we need to do is find an injective resolution for $mu_infty$, this will enable us to get an explicit presentation of $Rpi_*mu_infty$, but this is where I have no idea how to proceed.

galois theory – Does any cubic polynomial become reducible through composition with some quadratic?

What I mean to ask is this:

given an irreducible cubic polynomial $P(X)in mathbb{Z}(X)$ is there always a quadratic $Q(X)in mathbb{Z}(X)$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)?

I did quite some testing and always found a $Q$ that does the job. For example:

$P=aX^3+b,quad Q=-abX^2,quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$

$P=aX^3-x+1,quad Q=-aX^2+X,quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$

and a particular hard one to find:

$P=2X^3+X^2-X+4,quad Q=-8X^2+5X+1,quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$

Could there be a formula for $Q$ that works for all cases?

It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can’t figure it out.

galois theory – Polynomial equation with perfect squares powers

I am currently trying to solve these kinds of equations, such as:

$$x^{16}-4x^9+9x^4-12x+1=0$$
$$3x^9+7x^4-2x+6=0$$

How can I make an approach to these kinds of questions?
What have I tried:

Solving $x^9+3x^4-5x+7=0$
Assume that the equation is
$$(x^5+a)(x^3+b)(x+c)=0$$
Or something like that, then the $x^9$ will cancel but the problem is to solve the remaining $a, b, c$

If there’s is a quicker way feel free to answer me. I have heard of the Galois theory and if that helps, please write a detaided solution.