-1
votes
0answers
120 views

Construction of the field of real numbers within $ZF$ [duplicate]

I am interested in a problem whether the field of real numbers can be constructed within $ZF$. I will state the problem more precisely as follows. Definition 1 An ordered field $K$ is called ...
1
vote
2answers
144 views

Algebraic closure exists: What's wrong with this proof?

Given a field $K$, let $U = K[X] \times \mathbb{N}$. Identify each $k\in K$ as $(X-k,1) \in U$, so $K \subseteq U$. Consider fields $(S,+,\cdot)$ where $K \subseteq S \subseteq U$, and the inclusion ...
2
votes
0answers
147 views

Normal closure of field extension, axiom of choice

Update My previous proof was incorrect. This updated proof is inspired by the comment by 'MartianInvader'. Problem I can prove the statement 'Every algebraic extension $L:K$ has a normal closure ...
7
votes
3answers
240 views

Do maximal proper subfields of the real numbers exist?

To clarify the problem, consider the field ${\Bbb R}$ as a field extension of ${\Bbb Q}$ using some sort of Hamel basis. Does there exist a field $F\subsetneq{\Bbb R}$ such that $F(\sqrt{2})={\Bbb R}$ ...
3
votes
1answer
96 views

Can one construct an algebraic closure of fields like $\mathbb{F}_p(T)$ without Zorn's lemma?

I have heard that an algebraic closure of $\mathbb{Q}$ can be constructed without Zorn's lemma and so can an algebraic closure of a finite field $\mathbb{F}_p$. What about $\mathbb{F}_p(T)$? Do there ...
2
votes
0answers
82 views

Theorem's relying on algebraic closures

When working with fields, it's a usual method to work on an algebraic closure of a field to obtain results about that field. In general (i. e. unless you're explicitly considering "well-behaved" ...
10
votes
1answer
915 views

Constructive proof of the existence of an algebraic closure

It is well-known that, assuming the axiom of the choice (in the form of Zorn's lemma), one can prove that any field $F$ has an algebraic closure. One proof roughly goes as follows: consider the ...