# Tagged Questions

**15**

votes

**0**answers

296 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 ...

**1**

vote

**0**answers

353 views

### 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 ...

**2**

votes

**2**answers

668 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 ...

**19**

votes

**0**answers

918 views

### Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property

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 ...

**19**

votes

**3**answers

1k 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 ...

**19**

votes

**2**answers

606 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 ...