64

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 ...

39

The answer is "yes", see below.
Dieudonné in his book "La géométrie des groupes classiques"
considers the abstract group $SL_n(K)$ for a field $K$, not necessarily commutative,
and writes $PSL_n(K)$ for $SL_n(K)$ modulo the center.
In Ch. IV, Section 9, he considers the question whether $PSL_n(K)$ can be isomorphic to $PSL_m(K')$ for $n\ge 2,\ m\ge 2$.
He ...

38

A nice theorem is: $\{\pm 1\}$ is the only group that can act freely on a sphere of even dimension. In contrast: There are infinitely many groups acting freely on every odd-dimensional sphere.

33

Polynomials are, essentially by definition, precisely the operations one can write down starting from addition and multiplication. More formally, polynomials with coefficients in a commutative ring $R$ are precisely the morphisms in the Lawvere theory of commutative $R$-algebras. So in some sense caring about polynomials is equivalent to caring about ...

31

It seems not.
It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), then after forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ (the infinite subsets of $\omega$, ordered by mod-finite containment) to produce a selective ...

31

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 ...

27

Fields are the simple (no nontrivial quotients) commutative rings. Grothendieck told us to work in nice categories with nasty objects rather than nasty categories with nice objects; fields are the nice objects, and the nice category they live in is all commutative rings. You want to work in all commutative rings and sometimes study fields the same way you ...

answered May 20 '16 at 21:51

Qiaochu Yuan

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25

I think you are lumping too many disparate kinds of fields together under the heading "zero-dimensional". As Jason says in his answer, there are some precise definitions of dimensions of fields (e.g. cohomological dimension but also other definitions of a field-arithmetic nature).
Another important comment is that in modern algebraic / arithmetic geometry ...

answered Jul 27 '13 at 8:12

Pete L. Clark

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24

This is more of a joke than a serious example. Let $K$ be a field, $K_+$ its additive group, and $K_*$ its multiplicative group. Thus $\mathbb{R}_*\cong \mathbb{R}_+\times (\mathbb{Z}/2\mathbb{Z})$. What fields have the "opposite" property, that is, $K_+\cong K_*\times (\mathbb{Z}/2\mathbb{Z})$? Answer: only $\mathbb{Z}/2\mathbb{Z}$.

23

The answer to the question is yes, though I don't have all the old literature at my fingertips. This kind of question for various classes of linear groups has a long history in the study of homomorphisms and isomorphisms of classical groups and then other algebraic groups (van der Waerden, Dieudonne, ...) The most comprehensive treatment was given by Borel ...

answered Sep 10 '12 at 20:05

Jim Humphreys

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22

Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map
$$K \otimes_k K \xrightarrow{m} ...

answered Feb 15 '16 at 4:26

Qiaochu Yuan

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21

Conway's nimbers form an interesting answer for $p=2$. That every Field of characteristic $2$ embeds into it follows from the fact they form an algebraically closed Field and that they contain arbitrarily large sets of algebraically independent elements (which is immediate because the Field is proper-class-sized).
This has been generalized by DiMuro to ...

20

I suspect the question is not answerable because utility can be subjective and rationalizing
subjective notions leads to more arguments and less elucidation. However, this gives
me an opportunity to mention a couple aspects of polynomials that deserve more
press.
Polynomials are a generalization of number representation, replacing base 10 or
base 2 by base ...

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 ...

20

If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post).
We may assume that $a_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider ...

answered Nov 14 '16 at 8:29

Fedor Petrov

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19

I already wrote this in the comments but I think this might be worth of an answer. I think we can classify all fields $K$ such that $K\otimes K$ is a field.
Claim If $K$ is a field such that $K\otimes_\mathbb{Z}K$ is a field then the multiplication map $K\otimes_\mathbb{Z} K\to K$ is an isomorphism
In fact the multiplication map is always a surjection ...

19

An algebraically closed field is determined up to isomorphism by its characteristic and its transcendence degree over its prime field. So every algebraically closed field of characteristic $p$ is isomorphic to the algebraic closure of $\mathbb{F}_p(X)$, where $X$ is some set of variables.
This suggests that the "biggest field of characteristic $p$" should ...

18

If you're taking the definition of rational to be: birational to $\mathbb{P}^1$ over the field $k$, then the stated property is not even true. There are conics which have no rational points, and so are not rational, but are rational over a quadratic extension. For example, the affine conic $x^2 + y^2 + 1 = 0$ over the field $\mathbb{Q}$.
Added: since you ...

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 ...

17

This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the separable algebraic and algebraic closures of $k$ in $K$ and $L$. The result is the following.
Theorem. Let $k \subseteq K$...

17

It is the only non-trivial group whose free square ($G*G$) satisfies a non-trivial identity (or is solvable, or is amenable...)
Edit (Nov 9, 2014), suggested by Sam Nead
... or is virtually cyclic, or is two-ended, or contains no nonabelian free subgroups...

17

At risk of being overly bold, allow me to suggest:
Polynomials are useful because quadratic polynomials are useful.
If we can all agree that linear algebra is an indispensable tool in mathematics then it's hard to argue with the success of equipping vector spaces with quadratic structures - this is the starting point of nearly all of geometry and large ...

15

From Dirk van Dalen's Logic and Structure: the theory of algebraically closed fields is not finitely axiomatizable (see page 109 and preceding).

answered Oct 18 '15 at 14:56

15

It seems that indeed only prime fields are unrepeatable.
Proof:
Let $k$ be unrepeatable and $F\subseteq k$ denote the prime field of $k$. Let $T\subseteq k$ be a transcendence base of $k/F$ and let $G=F(T)$. If $T\neq\emptyset$, then $G/F$ has non-trivial automorphisms (say take one element $t\in T$ to $t+1$). Since $k/G$ is algebraic this extends to a ...

14

First, let me say that the set/class issue is not a problem to deal with properly, and so one shouldn't be very worried about it. It is true as you say that the surreal numbers No are a
proper class, and they do not form a set. So in a purely technical sense, they are not a field. But nevertheless, they do satisfy all the field axioms and have all the usual ...

answered Dec 8 '12 at 0:29

Joel David Hamkins

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14

I think the explanation is that the concept arose somewhat independently with English-speaking mathematicians. See the discussion of the earliest known use at http://jeff560.tripod.com/f.html

14

Let $\zeta = e^{2\pi i/p}$ be a primitive $p$th root of unity. Then
$2 \cos (2\pi k/p) = \zeta^k + \zeta^{-k}$. The Galois group of $\mathbb Q(\zeta)$
is isomorphic to $(\mathbb Z/p \mathbb Z)^\times$ and acts transitively on
the powers $\zeta^k$ with $1 \le k \le p-1$. What you want is that the Galois
action fixes the set $\{x_1, x_2, x_3\}$. Now $x_j = \...

14

Disclaimer: The following perhaps isn't an answer to your question as stated, so my apologies if this answer is useless to you. However, you're asking for how to treat this problem "honestly", and I think that adding the right kind of historical perspective falls under the heading of honesty.
Anyway, I think it is important to observe here that the ancient ...

14

First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism.
Now, since $\hat{\bar{K}}$ is complete and $K$ is dense in $\hat{K}$, the inclusion $K\subset \hat{\bar{K}}$ extends continuously to $K\subset \hat{K}\subset \hat{\bar{K}}$ (in fact you can identify ...

answered Aug 16 '16 at 19:09

Laurent Moret-Bailly

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