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), ...
- 2,570
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 ...
- 225k
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 ...
- 115k
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
Tensor product of fields over integers
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 $\...
- 115k
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 ...
- 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 \...
- 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 ...
- 7,902
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 $...
- 103k
20
votes
Accepted
Tensor product of fields over integers
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\...
- 16.2k
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 ...
- 4,431
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 ...
- 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 ...
- 19.3k
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 ...
- 186
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 ...
- 42.5k
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 ...
- 2,971
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 ...
- 6,112
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 ...
- 13.3k
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 $...
- 23.6k
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 ...
- 30.2k
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 ...
- 16.2k
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 ...
- 44.8k
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, ...
- 4,436
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 $\dim_{\mathbf F}D<\infty$ (...
- 15.3k
Only top scored, non community-wiki answers of a minimum length are eligible