Skip to main content
47 votes
Accepted

Results in linear algebra that depend on the choice of field

The existence of Chevalley–Jordan decompositions depends on the perfectness of the field.
37 votes

When is the tensor product of two fields a field?

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^{\...
35 votes
Accepted

Ultrafilters and automorphisms of the complex field

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), ...
Paul Larson's user avatar
  • 2,500
32 votes

Results in linear algebra that depend on the choice of field

A finite-dimensional vector space is a union of finitely many proper subspaces if and only if the underlying field is finite.
32 votes
Accepted

Are the real numbers isomorphic to a nontrivial ultraproduct of fields?

The answer is no, because such ultrapowers are always $\aleph_1$-saturated, but $\mathbb{R}$ is not. More concretely, the ultraproduct will be an ordered field with uncountable cofinality — every ...
Joel David Hamkins's user avatar
28 votes

What is a field [Körper] really?

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 ...
Qiaochu Yuan's user avatar
26 votes

Results in linear algebra that depend on the choice of field

As mentioned in the comments: when the characteristic of your field is not $2$, "skew-symmetric" and "alternating" are equivalent conditions on a bilinear form. In characteristic $...
23 votes

Results in linear algebra that depend on the choice of field

Existence of Jordan canonical form (requires algebraically closed field).
22 votes
Accepted

Biggest Field Of Characteristic $p$

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 ...
Wojowu's user avatar
  • 27.4k
22 votes
Accepted

Are there only two smooth manifolds with field structure: real numbers and complex numbers?

Here is a series of standard arguments. Let $(\mathbb{F},+,\star)$ be such a field. Then $(\mathbb{F},+)$ is a finite-dimensional (path-)connected abelian Lie group, hence $(\mathbb{F},+) \cong \...
M.G.'s user avatar
  • 6,730
21 votes

Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?

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 ...
Dmitry Vaintrob's user avatar
21 votes
Accepted

Fields for which there exist multivariable polynomials vanishing at single specified point

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 $...
Fedor Petrov's user avatar
20 votes

Results in linear algebra that depend on the choice of field

For a finite field ${\mathbb F}_q$, you may calculate the probability that the determinant of an $n\times n$ matrix is $0$. This probability has a limit $\pi_q$ as $n\rightarrow+\infty$. Amazingly, ...
19 votes

Biggest Field Of Characteristic $p$

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 ...
Alex Kruckman's user avatar
19 votes
Accepted

Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?

There are three answers. Throughout let $qcl(F)$ be the quadratic closure of a field $F$ inside $\mathbb{C}$. Part 1: Yes there is a quadratically closed field strictly between $qcl(\mathbb{Q})$ and ...
Pace Nielsen's user avatar
  • 18.2k
17 votes
Accepted

Completion and algebraic closure

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 ...
Laurent Moret-Bailly's user avatar
17 votes
Accepted

Steinberg representation for sporadic simple groups?

The approach that I have taken to generalizing the Steinberg module to finite groups other than groups of Lie type is that, in general, the object we should consider is a chain complex, rather than ...
Peter Webb's user avatar
15 votes
Accepted

Is a field that never embeds twice in another field necessarily a prime field?

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 ...
Sándor Kovács's user avatar
15 votes

Definability of the ring of integer in algebraic extensions of $\mathbb Q$

If $K$ is a finite extension of $\mathbb{Q}$, then, yes, $\mathbb{Z}$ is definable in $\langle K, +, \times, 0, 1 \rangle$. See R. Rumely, Undecidability and definability for the theory of global ...
Thomas Scanlon's user avatar
15 votes
Accepted

Is this theory the complete theory of the real ordered field?

It is not. Using set forcing, we can add 'undefinable' reals in a controlled manner, while keeping complexity of parameter-free definable sets low. Specifically, let $M$ be a countable $ω$-model of ...
Dmytro Taranovsky's user avatar
15 votes

Algebraic topology over fields other than ${\bf R}$

There is an homotopy theory associated to any geometry. This can be done in various ways. but a rather systematic point of view is the one of Morel and Voevodsky. The idea is that a geometry should ...
D.-C. Cisinski's user avatar
15 votes
Accepted

Factorization of an irreducible polynomial in the field extension it defines

Let us show that for the partition $2+1+1=4$, there is no such $f$. If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group $...
Bjorn Poonen's user avatar
  • 23.7k
13 votes

What "should" be the absolute galois group of a field with one element

The Galois group of the maximal abelian extension of $\mathbb Q$ (or any number field) is given (class field theory) as the quotient of the idele class group by the connected component of the identity ...
Felipe Voloch's user avatar
13 votes
Accepted

What is the topology on the set of field orders

The topology you are looking for is called the Harrison topology. If we denote the set of ordering of a field $F$ with $\mathrm{Sper}\,F$ (more on that in a moment), this is the subspace topology ...
Denis Nardin's user avatar
  • 16.3k
13 votes
Accepted

Can nonstandard fields contain $\mathbb R$ in different ways?

Yes, this is possible. Let $F$ be a nonarchimedean strongly $\omega$-homogeneous real-closed field such that $\mathbb R\subseteq F$ (which exists by model-theoretic general reasons). Fix an ...
Emil Jeřábek's user avatar
13 votes
Accepted

The map $k \mapsto \mathbf{PGL}_2(k)$

What's written above is true for all $n, n_1 \geq 2$ and all skew fields except two finite cases: $PSL(2, \Bbb F_7) \cong PSL(3, \Bbb F_2)$, $PSL(2, \Bbb F_4) \cong PSL(2, \Bbb F_5)$. (V. M. Petechuk, ...
Denis T's user avatar
  • 4,446
12 votes

Collecting proofs that finite multiplicative subgroups of fields are cyclic

I like the following explanation of the fact that if $G=\{g_0,\dots,g_{n-1}\}$ is your group, and $0<m<n$, then there exists $i$ for which $g_i^m\ne 1$: otherwise the Vandermonde matrix $(g_i^...
12 votes
Accepted

Division ring on a field

I assume $\text{char}\,\mathbf F=0$. Put $d:=b-a$. Because of $a^2-2ab+b^2=d^2-ad+da$ your equation is equivalent to $$ (*)\qquad d^{-1}a-ad^{-1}=1. $$ This precludes $\dim_{\mathbf F}D<\infty$ (...
Friedrich Knop's user avatar
12 votes
Accepted

Finding all automorphisms of $\mathbb{C}(x,y)$

The proof of Noether that the automorphisms of $\mathbb{C}(x,y)$ are generated by $\mathrm{PGL}(3,\mathbb{C})$ and the standard quadratic transformation had troubles because of infinitely near points. ...
Jérémy Blanc's user avatar
12 votes
Accepted

Are all real-closed subfields of $\overline{\mathbb{Q}}$ conjugate?

Any real-closed subfield $R\subseteq\overline{\mathbb Q}$ is a real closure of $\mathbb Q$ (being real closed and algebraic over $\mathbb Q$). Thus, by uniqueness of real closures, any two such fields ...
Emil Jeřábek's user avatar

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