All Questions
Tagged with axiom-of-choice fields
15
questions
6
votes
1
answer
486
views
Automorphisms of algebraically closed fields without the Axiom of Choice
In the paper Algebraische Konsequenzen des Determiniertheits-Axioms (U. Felgner – K. Schulz, Arch. Math. (Basel) 42 (1984), pp. 557–563), the authors show that in models of Zermelo-Fraenkel set theory ...
- 4,503
4
votes
1
answer
328
views
Automorphisms of vector spaces and the complex numbers without choice
In Zermelo-Fraenkel set theory without the Axiom of Choice (AC), it is consistent to say that there are models in which:
there are vector spaces without a basis;
the field of complex numbers $\mathbb{...
- 4,503
5
votes
0
answers
279
views
Reference for countable and uncountable algebraic closures of $\mathbb{Q}$ in ZF
The following facts seem to be part of the folklore (where $\mathsf{ZF}$ means Zermelo-Fraenkel set theory with no axiom of choice):
it is consistent with $\mathsf{ZF}$ that there exists an ...
- 30.8k
0
votes
0
answers
133
views
Axiom of Choice and bases of $k$-vector spaces, $k$ fixed
I know that from ZF + the Axiom of Choice (AC) follows that every vector space has a basis.
And, conversely, Blass proved that in ZF set theory, the assumption that every vector space has a basis ...
- 4,503
5
votes
0
answers
168
views
Copies of the reals in $\mathbb{C}$ without the Axiom of Choice
Suppose we work in a model in which the Axiom of Choice does not hold, and in which $\mathbb{C}$ only has one nontrivial automorphism (such models exist).
Question: "how many" subfields of $\...
- 4,503
7
votes
0
answers
592
views
Automorphism groups of the complex numbers, and other fields
If one accepts the Axiom of Choice (AC), then the automorphism group of $\mathbb{C}$ is a huge and wild group, very poorly understood.
But apparently if one does not accept the Axiom of Choice, then ...
- 4,503
48
votes
0
answers
2k
views
How many algebraic closures can a field have?
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is ...
- 38.5k
12
votes
0
answers
591
views
Is there a particular field that cannot be proven to have an algebraic closure in ZF?
Proofs that every field has a unique (up to isomorphism) algebraic closure use some form of the axiom of choice. For uniqueness this is provably necessary: there are models of ZF in which $\mathbb{Q}$ ...
- 8,961
12
votes
5
answers
1k
views
Does k(X) have a k-basis for every set X, without AC?
This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.
For any field $k$, the field $k(x)$ of rational functions in one variable has an ...
- 34.4k
22
votes
0
answers
574
views
Are the reals really a fraction field?
In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$.
A crucial point in the ...
4
votes
0
answers
2k
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Existence of algebraic closure and Axiom of choice [duplicate]
Possible Duplicates:
Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
algebraic closure of commuting pairs of matrices
we need zorn's ...
- 1,778
2
votes
2
answers
1k
views
Countable Fields with No Countable Extension
Let $\mathscr{S}$ be the set of all countable subfields of $\mathbb{C}$. Clearly, $\mathscr{S}$ is a partially ordered set under inclusion, and if $K_1\subseteq K_2 \subseteq \cdots$ is an ascending ...
- 5,639
23
votes
0
answers
2k
views
Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice
While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then ...
user5810
32
votes
3
answers
6k
views
Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?
The existence and uniqueness of algebraic closures is generally proven using Zorn's lemma. A quick Google search leads to a 1992 paper of Banaschewski, which I don't have access to, asserting that ...
- 116k
34
votes
3
answers
2k
views
How much choice is needed to show that formally real fields can be ordered?
Background: a field is formally real if -1 is not a sum of squares of elements in that field. An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you ...
- 65.1k