156

The answer is 'no'. Making the substitution
$$
x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2},
$$
one finds
$$
{\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x}
= \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}.
$$
Denote the right hand side of the above equation by $\beta$. Now, setting
$$
Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3},
...

answered Jun 15 '14 at 4:14

Robert Bryant

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103

I'm adding a separate answer for the general question that the OP asked, which settles the question in the negative for all $n>2$ (and gives an alternate proof for $n=3$ to the one I gave above).
Recall that the OP defined a sequence of algebraic functions $f_n$ by the rule $f_0(x) = 1$, $f_1(x) = \sqrt{x+1}$, and $f_{n+1}(x) = \sqrt{x + f_n(x)}$ for all ...

answered Jun 17 '14 at 18:24

Robert Bryant

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69

The theory of differential Galois theory is used, but in algebraic, not differential geometry, under the name of D-modules. A D-module is an object that is somewhat more complicated than a representation of the differential Galois group, in the same way that a sheaf is a more complicated than just a Galois representation, but I think it is cut from the same ...

45

The answer to the question is yes (so the answer to the title is no) and I will give an example later.
Let me first recall a couple of results. The first one is the following, that can be found in [Cox, Galois Theory, Theorem 8.6.5].
Theorem 1. Let $F$ be a subfield of $\mathbb{R}$ and let $f \in F[x]$ be an irreducible polynomial with splitting field $F \...

answered Sep 30 '14 at 18:41

Francesco Polizzi

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42

Suppose $k$ is a field, not necessarily algebraically closed. $\text{Spec } k$ fails to behave like a point in many respects. Most basically, its "finite covers" (Specs of finite etale $k$-algebras) can be interesting, and are controlled by its absolute Galois group / etale fundamental group. For example, $\text{Spec } \mathbb{F}_q$, the Spec of a finite ...

answered Mar 17 '16 at 21:11

Qiaochu Yuan

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35

As indicated by KConrad in his comments, differential Galois theory is used in the part of transcendental number theory that tries to establish algebraic/linear independence of values of special functions at algebraic numbers. Examples are given by the theorems of Siegel-Shidlovski, Nesterenko, etc.
Roughly speaking, its rôle is to guarantee that an ...

35

My co-authors and I introduced a notion of dimension for forcing
extensions in the following paper:
Hamkins, Joel David; Leibman, George; Löwe, Benedikt, Structural connections between a forcing class and its modal logic, Isr. J. Math. 207, Part 2, 617-651 (2015). ZBL1367.03095, arxiv:1207.5841, blog post.
Specifically, for any forcing extension $V\subset ...

answered May 31 '18 at 11:44

Joel David Hamkins

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34

The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where
$K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group
is
$$
[K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}
= \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}.
$$
It remains to compute $m:=[K\cap L:\mathbb{Q}]$.
First show that $K\cap ...

34

$\def\QQ{\mathbb{Q}}$Building on YCor's construction, your field is $\overline{\mathbb{Q}}$. Let $L$ be any finite Galois extension of $\QQ$ with Galois group $G$; I will show that $L$ is contained in your field.
We can find an element $x$ in $L$ so that, for any nonempty subset $S$ of $G$, the product $\prod_{\sigma \in S} x^{\sigma}$ is not square. For ...

answered Jun 4 '14 at 0:59

David E Speyer

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31

If the ODE is linear --and the notion of «explicit» refers to Liouvillian solutions (towers of iterated quadrature and exponential of meromorphic functions)-- then its differential Galois group (Picard-Vessiot theory) must be a solvable algebraic subgroup of $GL_n (\mathbb C)$. Such subgroups are rare: they define a proper algebraic subvariety. The defining ...

28

This is true; it is due to Selmer. Ljunggren (On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 1960) has obtained the complete list of reducible trinomials with $\pm 1$ coefficients. For more details see Gerry Myerson's answer to this question, which contains a review of Ljunggren's paper: About irreducible trinomials . To this I ...

answered Aug 4 '14 at 18:19

Vesselin Dimitrov

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27

This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvability of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.
...

26

Two clarifications:
For anabelian geometry, you should ask how much information about a variety is contained in the Galois action on its etale fundamental group.
While it's true that the motivic Galois group is a higher-dimensional analogue of the Galois group, it also should be true that motives are "just" a special kind of Galois representation, i.e. ...

26

The action of the monodromy group of $w(z)$ on the fiber $p^{-1}(a)$ for a non-critical value $a$ of $p$ (that is $|p^{-1}(a)|=\deg p$) is the same as the action of the Galois group of $p(x)+z$ over $\mathbb C(z)$ on the roots of $p(x)+z$ in some splitting field. One can see this by comparing each of these groups with the deck transformation group of the ...

25

Lehmer's polynomial is symmetrical,
so $x + x^{-1} =: y$ satisfies a polynomial of half the degree.
It turns out that this is the quintic $y^5 + y^4 - 5y^3 - 5y^2 + 4y + 3 = 0$,
whose Galois group is the unsolvable $S_5$ (for instance, it's irreducible
$\bmod 2$ and decomposes as $(y^2-2y-1)(y^3-2y^2+2y+2)$ $\bmod 5$,
so the Galois group is a subgroup of $...

answered Dec 1 '15 at 21:59

Noam D. Elkies

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24

Thank you for calling this problem to my attention.
I computed $K$ en route to AWS (though this year's
topics
are a rather different flavor of number theory...).
After some simplification (gp's $\rm polredabs$), it turns out that
the field $K$ is generated by a root of
$$
f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7
$$
$$
\...

answered Mar 14 '15 at 4:01

Noam D. Elkies

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23

I addressed this exact question in my American Mathematical Monthly paper,
What is a closed-form number? Corollary 1 in that paper states that if Schanuel's conjecture holds, then the EL numbers (i.e., the numbers expressible according to your list of rules) that are algebraic are precisely the numbers expressible using radicals. So any algebraic number ...

answered Aug 19 '14 at 20:03

Timothy Chow

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23

Many of the polynomials in the Klueners-Malle database and also in my
database with John Jones come from families in the way you correctly
describe. So you have "reverse engineered" the source family.
This particular source family is the first of two similar families
described in Section 13 my Galois number fields with small root discriminant with Jones. ...

23

If I calculated correctly, $$\begin{pmatrix}2 & 1 & 0 & 0 & 0\newline 1 & 3 & 1 & 0 & 0\newline 0 & 1 & 1 & 1 & 0 \newline 0 & 0 & 1 & 1 & 1 \newline 0 & 0 & 0 & 1 & 1 \end{pmatrix}$$ has characteristic polynomial $-x^5+8x^4-20x^3+15x^2+4x-5$ which is irreducible mod $3$ and has $...

22

The same Puiseux-series tactic works (effectively) to list all $y$
for which $p(X,y)$ has a quadratic factor. For example, the linear
coefficient of such a factor is $a = -(r_i+r_j)$ for some distinct $i,j$
where $r_1,\ldots,r_5$ are the roots of $p(X,y)$. But for large $y$
there are three real and two imaginary roots, and thus four choices of
$\{i,j\}$ ...

answered Apr 16 '14 at 5:39

Noam D. Elkies

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22

This article from Le Monde (in French) and this blog (in English) are recent and seem to accurately sum up the state of affairs: In April 2012 Jean Malgoire (see the video interview in the blog) donated the manuscript of "La longue Marche" (700 handwritten pages written by Alexander Grothendieck in just 20 days in 1981) to the library of Montpellier ...

answered Oct 2 '15 at 11:36

Carlo Beenakker

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22

This is a very good question which is a big open problem. There are a number of theorems, some of them easy and some very difficult, and also a number of conjectures, restricting the class of groups which may turn out to be absolute Galois groups. But it seems that nobody has any idea about how a precise description of the class of absolute Galois groups ...

answered Mar 6 '16 at 21:54

Leonid Positselski

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22

Of course, Galois theory intervenes as a basic tool in the study of some diophantine equation. But deeper aspects, in the form of Galois representations, were crucial for the proof of at least 4 fantastic theorems in the last century.
The Mordell-Weil theorem concerning rational points of elliptic curves over the field of rational numbers (Mordell, 1922) or ...

22

There's a very deep connection between motives and Grothendieck-Teichmüller theory but it isn't well-understood yet. I can't even frame it precisely in higher genus, but at least I can frame a precise conjecture in genus zero. It has to do with motives that are connected to periods of moduli spaces on the one hand, and Grothendieck-Teichmüller being ...

21

The most direct answer I know for this is the following:
Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 ...

20

If you do not impose an algebraically closed condition, no two are equivalent. This basically follows from your (3). Namely, observe that
An extension is finite if and only if it has finitely many subextensions (otherwise, it contains $\mathbb{F}_p(x)$, which then contains $\mathbb{F}_p(x^n)$ for all $n$).
$\bar{\mathbb{F}}_p$ is the unique extension ...

18

If such polynomials exist, there will only be finitely many of them, up to composing on both sides with scalar polynomials $\alpha x$ with $\alpha\in\mathbf{Q}$. More generally, Guralnick and Shareshian proved that if $d=7$ or $d>8$ then there are only finitely many equivalence classes of irreducible degree-$d$ trinomials in $\mathbf{Q}[x]$ whose Galois ...

18

I guess you won't be satisfied with the answer $f=0$ and $g=1$. :)
But the answer is yes even if you assume that $f$ and $g$ are nonconstant. For example, consider $f(x)=2x^3$ and $g(x)=(x^3-2)^3$. If $f(a)=g(b)$ for some $a,b \in \mathbb{Q}(\zeta)$, then one finds that $2$ is a cube in $\mathbb{Q}(\zeta)$, which is a contradiction.

18

Any number field $K$ which has a real place is generated by a root of a Pisot polynomial. So the answer is no. For example, we can take $p$ a prime $\geq 11$, and use the real subfield of $\mathbb{Q}(\zeta_p)$, to get a Pisot polynomial whose splitting field is cyclic of degree $(p-1)/2$.
Suppose $K$ has $r$ real places, and $2s$ complex places. Let $V_i$ ...

answered Aug 24 '15 at 15:52

David E Speyer

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18

Dubickas and Smyth (On the Remak Height, the Mahler Measure and Conjugate Sets of Algebraic Numbers Lying on Two Circles, 2001) discuss what they call extended Salem numbers.
Moreover, they present results in the direction about which you ask, i.e., conditions under which certain conjugates all fall on two (not one) circles, thereby forcing their associated ...

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