14
votes
Accepted
Which of these 4 definitions of Galois coverings of integral schemes are equivalent?
I can prove $(3) \Rightarrow (1)$ when $X$ and $Y$ are irreducible (Lemma 1 below; no integrality assumptions needed. Irreducible is needed for the question to make sense), and I can prove $(4) \...
13
votes
Accepted
A criterion for the equation $ax^n+bx+c=0$ not solvable by radicals via $a,b,c$ and $n$
No, the conjecture is false at least for $n = 5$. The irreducible quintic trinomial $f(x) = 85x^{5} - 4x + 1$ satisfies $\gcd(b,5ac) = \gcd(-4,5 \cdot 85 \cdot 1) = 1$. However, the Galois group of $f(...
12
votes
Accepted
Galois group of a polynomial modulo $p$
This result of Dedekind is not true for every prime $p$, but only for primes not dividing the discriminant of $f(x)$.
There is no “easy” proof for someone who knows only Galois theory (the setting ...
9
votes
Accepted
On the solvability of the equation $ax^p+bx^{p-1}+cx+d=0$ by radicals
This answer is experimentally driven. I tried to construct as many as possible solvable irreducible polynomials of the shape $f=x^5+bx^4+cx+d$ with integers $b,c,d$, then search for a message or a ...
8
votes
Perfect numbers, Galois groups and a polynomial
The second factor is $P(2t)$ where $P=X^{p-1}+\cdots+X+1$, the $p$-th cyclotomic polynomial.
Hence the Galois group of $P(2t)$ is the same as the Galois group of $P(t)$, which is simply $(\mathbb{Z}/p\...
7
votes
Accepted
Common Galois extension over $\mathbb Q $
If $k$ is odd, then yes. If $L'/\mathbb{Q}$ is a cyclic extension of degree $4$, choose an extension $M/\mathbb{Q}$ that is cyclic of degree $k$. Then the compositum $L'M/\mathbb{Q}$ will have ${\rm ...
7
votes
Interesting (combinatorial) actions of the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$
You should read the paper
I. Bauer, F. Catanese, F. Grunewald: Faithful actions of the absolute Galois group on connected components of moduli spaces, Invent. Math. 199, No. 3, 859-888 (2015). ZBL1318....
4
votes
Using Jordan's theorem to find Galois group for a polynomial
Every transitive group of prime degree is primitive, so you just need to find a transposition (which you have done). So yes, the argument is correct. In fact, Jordan's theorem is the way Galois groups ...
4
votes
What does the group of automorphisms corresponding to $\mathfrak{g}$
I think you have described the morphisms in the category wrong.
Morphisms should be finite etale maps forming a commutative triangle with the structure maps to $S$.
If this is so, then we can take $...
3
votes
Galois group of a polynomial modulo $p$
I'm not a number theorist, but I did find the Tate proof enlightening because it avoids the machinery of Dedekind domains, although at heart the proof is the same as in @KConrad's linked note. To the ...
2
votes
Bounding the number of polynomials whose Galois group is a subgroup of the alternating group
The set
$$\{ \mathbf{a} =(a_1, \ldots, a_n) \in \mathbb{Z}^n : ||\mathbf{a}|| \leq B, \Delta(a_1,\ldots,a_n) \text{ is a square} \}$$
is a so-called thin set in the sense of Serre.
Serre studies ...
2
votes
Atiyah on the "Galois group of the octonions" and Physics
The video of his lecture is at the HLF website. I was unable to find it myself but thanks to Lashi Bandara we have it:
http://www.heidelberg-laureate-forum.org/blog/video/lecture-monday-september-19-...
2
votes
Question on Inverse Galois Theory
I think the main point of the question is worked out in detail in section 2 of the thesis the link points to.
http://page.math.tu-berlin.de/~kant/publications/diss/geissler.pdf
2
votes
Accepted
Moving general fibers of a fibration
Definitely not; it may be that none of the fibres are isomorphic to each other. The typical situation is that a given fibre is isomorphic to only finitely many others.
1
vote
The double cover of $[W(E_7),W(E_7)] \cong Sp_6(\mathbb F_2)$ as a Galois group over $\mathbb Q$
More details on my comment:
I mean that there is an induced map $H^2( E_7^o , \mathbb Z/2) \mathbb Z/2 = H^2( W(E_7), \mathbb Z/2) \to H^2( \pi_1(E_7^o , \mathbb Z/2)) \to H^2( E_7^o , \mathbb Z/2)$...
1
vote
Galois group of a polynomial modulo $p$
I just came across the following exposition of Tate's proof on Dedekind's theorem: https://www.scirp.org/journal/paperinformation.aspx?paperid=85772
Perhaps it is of interest (I had no closer look so ...
1
vote
Accepted
Splitting field of an intermediate field
Even when $K/F$ is quadratic (hence galois and abelian), it can happen that $M$ is smaller than $L$.
For a straightforward counterexample, take a chain of groups $\{1\}<I<H<D_8$ such that $I$...
1
vote
Splitting field of an intermediate field
Here is a direct counterexample. $L=\mathbb{Q}(\sqrt[3]{2},\xi_3)$,$F=\mathbb{Q}$, $\alpha=\sqrt[3]{2}$, $K=\mathbb{Q}(\alpha)$ and $f=x^3-2$. We have $M=K\subsetneqq L$.
This is not true even when $...
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