1
vote
1answer
57 views

Ultrafilter Lemma and Dimension Theorem

Reading on Wikipedia I find out that (the uniqueness in) the Dimension Theorem for arbitrary Vector Spaces can be proved using just the Ultrafilter Lemma (a strictly weaker version of Axiom of ...
9
votes
2answers
414 views

Is the Axiom of Choice implicitly used when defining a binary operation on a quotient object?

Let's say you have a group $(G,\cdot)$ and you have a normal subgroup $N$ (note we are considering this only as a set). And now we want to define a binary operation $\star$ on $G/N$ such that $(G/N, ...
1
vote
2answers
158 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 ...
6
votes
1answer
212 views

Do proper dense subgroups of the real numbers have uncountable index

Just what it says on the tin. Let $G$ be a dense subgroup of $\mathbb{R}$; assume that $G \neq \mathbb{R}$. I know that the index of $G$ in $\mathbb{R}$ has to be infinite (since any subgroup of ...
5
votes
1answer
264 views

Bourbaki Proof of Zorn's Lemma in Lang's Algebra

Serge Lang in his book Algebra has a nice appendix on set theory at the end of the book. In particular, in paragraph 2, pp. 881-884 he provides a proof of Zorn's Lemma from "other properties of sets ...
3
votes
2answers
132 views

Axiom of Choice and Ascending Chain Condition

Matsumura, in his "Commutative Ring Theory" p. 14 proves that "A partially ordered set $\Gamma$ satisfies the ascending chain condition $\Leftrightarrow$ every nonempty subset of $\Gamma$ has a ...
5
votes
2answers
232 views

Need for axiom of choice?

When you invoke the axiom of choice, it is because whatever you need to choose consists of small parts you technically have to choose one by one. So if I need to use the axiom of choice, but ...
2
votes
1answer
131 views

Infinite dimensional vector spaces and inductive sets

Consider an infinite dimensional vector space $E$ and define $$S:=\left\{F \subset E\biggr| F\ne 0\text{ is a subspace of }E\right\}.$$ Endow $S$ with the reverse inclusion. Is it possible to find a ...
18
votes
4answers
939 views

Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?

I know the usual proof of the existence of an algebraic closure for any field using Zorn's Lemma. The answer to this previous question makes it clear that in general, some nonconstructive axiom (not ...
10
votes
1answer
986 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 ...
6
votes
1answer
266 views

Simple functions and axiom of choice

The question I have is more of a curiosity, and that is why I decided to post here instead of Mathoverflow. Before posing the question, let me set up some background. Background: Let $\Omega$ be a ...
4
votes
2answers
149 views

Is $\operatorname{Hom}_A(M,N)$ a set without axiom of choice?

Let $M$ and $N$ be $A$-modules, $\operatorname{Hom}_A(M,N)$ the set of all $A$-module homomorphisms $M\rightarrow N$. $\operatorname{Hom}_A(M,N)$ can be viewed as a subset of the cartesian product ...
9
votes
3answers
712 views

Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?

More specifically, I just want to prove that in a Noetherian (even Dedekind) ring, every ideal in contained in a maximal ideal. Is the axiom oh choice needed here? The usual proofs (for general rings ...
6
votes
2answers
436 views

Zorn's lemma in abstract algebra?

It is well konwn that Zorn's lemma implies: Prop.1 Every commutative unital ring has a maximal ideal. Prop.2 Every proper ideal is contained in a maximal ideal in a unital ring. Question: Can we ...
4
votes
1answer
355 views

Algebra without Zorn's lemma

One can't get too far in abstract algebra before encountering Zorn's Lemma. For example, it is used in the proof that every nonzero ring has a maximal ideal. However, it seems that if we restrict ...