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) \...
R. van Dobben de Bruyn's user avatar
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(...
Jeremy Rouse's user avatar
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 ...
KConrad's user avatar
  • 49.6k
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 ...
dan_fulea's user avatar
  • 1,821
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\...
GreginGre's user avatar
  • 1,661
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 ...
Jeremy Rouse's user avatar
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....
Francesco Polizzi's user avatar
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 ...
Igor Rivin's user avatar
  • 95.6k
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 $...
Will Sawin's user avatar
  • 135k
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 ...
Benjamin Steinberg's user avatar
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 ...
Daniel Loughran's user avatar
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-...
Vít Tuček's user avatar
  • 8,159
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
Stephan's user avatar
  • 21
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.
Richard Thomas's user avatar
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 $H^2( E_7^o , \mathbb Z/2) \mathbb Z/2 = H^2( W(E_7), \mathbb Z/2) \to H^2( \pi_1(E_7^o , \mathbb Z/2)) \to H^2( E_7^o , \mathbb Z/2)$...
Will Sawin's user avatar
  • 135k
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 ...
Brauer Suzuki's user avatar
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 $I$...
GH from MO's user avatar
  • 98.2k
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 $...
tanjia's user avatar
  • 337

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