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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 $(4) \Rightarrow (1)$ when $X$ and $Y$ are normal, connected, and Noetherian (in particular integral, see Tag 033M) (Lemma 2 below).
I do not know how to approach the ...

answered Jun 18 '16 at 6:01

R. van Dobben de Bruyn

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10

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 where the result is usually first met). You can find a proof in Jacobson’s Basic Algebra I, attributed to Tate, that aims to be self-contained at the level of ...

9

I don't know how helpful this will be either. But let me modify your question a bit to ask what is the correct analogue of the Galois group (of say a finite field or number field) in Hodge theory? It is certainly true that when $X$ is a smooth complex projective variety,
$Gal(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ acts on $H=H^i(X,\mathbb{C})$,
by ...

9

I'm not sure how helpful this is, but anyway: in the étale cohomology of a variety over a finite field there is an action of $\mathrm{Gal}(\overline{\mathbf F}_q/\mathbf F_q)$ on
$$ H^\bullet_{\text{ét}}(X \times_{\mathbf F_q} \overline{\mathbf F}_q,\mathbf Q_\ell) $$
and in the Betti cohomology of a complex variety there is a Galois action of $\mathrm{Gal}(\...

answered Nov 12 '13 at 9:31

Dan Petersen

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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 $(\mathbb{Z}/p\mathbb{Z})^\times$, which is cyclic of order $p-1$.
The other factor has the form $t^p -a$. The splitting field is $\mathbb{Q}(\zeta_p, a^{1/p})$, and it is ...

7

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 ${\rm Gal}(L'M/\mathbb{Q}) \cong {\rm Gal}(L'/\mathbb{Q}) \times {\rm Gal}(M/\mathbb{Q})$ which will be cyclic of order $4k$.
If $k$ is even, the answer is no. In ...

6

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). ZBL1318.14034.
Quoting from MathSciNet review:
The authors study the action of the absolute Galois group $\mathrm{Gal}(\bar{\mathbb{Q}}/ \mathbb{Q})$ on certain ...

answered Mar 8 '18 at 14:40

Francesco Polizzi

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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 are usually calculated (by machine).

answered Jun 11 '17 at 1:43

Igor Rivin

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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 $\operatorname{Spec} B$ to be an affine open in $S$, and then $\operatorname{Spec} A$ to be the fiber product of $X$ with $\operatorname{Spec} B$ over $\...

3

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.

3

The Galois group is a subgroup of $A^4$ if and only if the discriminant is a perfect square. If you change $x^3$ to $x^2$ you get the discriminant to be:
$$\frac{1728^2 \left(b^2+3 b+9\right)^2 \left(a^3 (2 b+3)-3 a^2 \left(b^2+3 b+9\right)+\left(b^2+3 b+9\right)^2\right)^2}{\left(a^3-3 a \left(b^2+3
b+9\right)+2 b^3+9 b^2+27 b+27\right)^4},$$
so that ...

answered Oct 13 '15 at 11:54

Igor Rivin

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(A version of) Ledet's paper can be found here. The formula given in that version has $x^2$ where the question here has $x^3$. Perhaps this accounts for the difficulties.

answered Oct 12 '15 at 22:21

Gerry Myerson

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The first question seem OK. Here is a sketch. I might have overlooked something.
If $H\subset G$ is an open subgroup, then the image
of $H$ in $Gal_\mathbb{Q}$ is open so necessarily of the form $Gal_K$. So we get an extension
$$1\to H\cap \hat{\mathbb{Z}}\to H\to Gal_K\to 1$$
The kernel $ H\cap \hat{\mathbb{Z}}\subset \hat{\mathbb{Z}}$ is closed and of ...

2

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 these in detail in the book "Lectures on the Mordell-Weil theorem". He uses the large sieve in Section 13 to show that for any thin set $\Omega \subset \mathbb{Z}^...

answered Feb 1 '17 at 10:19

Daniel Loughran

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2

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-19-2016-sir-michael-atiyah/

2

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 best of my understanding, which might be faulty, the only thing it uses that is not typically seen before Galois theory is that the algebraic integers (complex ...

answered Mar 12 '21 at 13:42

Benjamin Steinberg

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1

More details on my comment:
I mean that there is an induced map $H^2( E_7^o , \mathbb Z/2) \mathbb Z/2 = H^2( W(E_7), \mathbb Z/2) \to H^2( \pi_1(E_7^o , \mathbb Z/2)) \to H^2( E_7^o , \mathbb Z/2)$ where $E_7^o$ is the open subset of regular semisimple elements and the map $\pi_1 ( E_7^o) \to W(E_7)$ is from the cover whose fiber over an element $g$ is ...

1

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 $I$ is not normal (cf. subgroups of D8). Let $L/F$ be a $D_8$ extension, and let $K$ (resp. $M$) be the fixed field of $H$ (resp. $I$). Then $K/F$ and $M/K$ are ...

1

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 $K/F$ is galoisian. A slight modification of the counterexample above gives a new counterexample.
Lets take $F=\mathbb{Q},K=\mathbb{Q}(\sqrt{2}),M=\mathbb{Q}(\...

1

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

1

Since the answer is already in the comments I'm just putting it here as community wiki.
As Noam Elkies noted:
Yes, a proper subgroup of a finite group G cannot intersect every conjugacy class. This has appeared here at least once before: mathoverflow.net/questions/2697

Only top voted, non community-wiki answers of a minimum length are eligible

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