Tagged Questions

10
votes
1answer
274 views

Question about whether axiom of choice is needed in this proof

Do I need axiom of choice in this proof here? I think not: at each step we choose one element from a set $N - \langle g_1, \dots, g_k \rangle $. So while there is indeed a countable number of sets ...
2
votes
0answers
134 views

Noetherian condition on the ring of formal power series without Axiom of Choice

I use the definition of a Noetherian ring given by Qiaochu in this: A commutative ring is Noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is ...
1
vote
1answer
227 views

Existence of valuation rings in a finite extension of the field of fractions of a weakly Artinian domain without Axiom of Choice

Can we prove the following theorem without Axiom of Choice? This is a generalization of this problem. Theorem Let $A$ be a weakly Artinian domain. Let $K$ be the field of fractions of $A$. Let $L$ ...
1
vote
1answer
204 views

Dedekind's theorem on an integrally closed algebra over a commutative ring without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951. Can we prove the following theorem without Axiom of Choice? If the answer is affirmative, by using ...
3
votes
1answer
410 views

Krull-Akizuki theorem without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951. We use the definitions in my answers to this question. Can we prove the following theorem without ...
1
vote
1answer
356 views

Dedekind's theorem on a weakly Artinian integrally closed domain without Axiom of Choice

Let $A$ be a commutative ring. Let $f$ be any non-zero element of $A$. Suppose that $A/fA$ has a composition series as an $A$-module. Then we say $A$ is a weakly Artinian ring (this may not be a ...
2
votes
2answers
616 views

Nilpotency of the Jacobson radical of an Artinian ring without Axiom of Choice

Let $A$ be a commutative ring. Suppose $A$ has a composition series as an $A$-module. EDIT Since $A$ has a composition series, $A$ has a maximal ideal. Let $J$ be the intersection of all the maximal ...
-4
votes
2answers
389 views

Dedekind's theorem on an integrally closed algebra over a field without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951. Can we prove the following theorem without Axiom of Choice? Theorem Let $A$ be an integrally ...
5
votes
1answer
284 views

Hilbert's Nullstellensatz without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951. Can we prove the following theorem without Axiom of Choice? Theorem Let $A$ be a commutative ...
-1
votes
3answers
258 views

Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice

Let $A$ be an integral domain which is finitely generated over a field $k$. Let $f \neq 0$ be a non-invertible element of $A$. Can one prove that there exists a prime ideal of $A$ containing $f$ ...
7
votes
2answers
632 views

Lying-over theorem without Axiom of Choice

This question is motivated by this and this. Can the following proposition be proved without Axiom of Choice? Proposition: Let $k$ be a field. Let $A$ and $B$ be commutative algebras without ...
7
votes
2answers
180 views

Commutative Algebra without the axiom of choice

It is well known that in a commutative ring with unit, every proper ideal is contained in a maximal ideal. The proof uses the axiom of choice. This fact, and others that are proved using essentially ...
6
votes
3answers
209 views

Axiom of Choice (for example in the Snake Lemma)

If we have to make a choice, but in the end it doesn't matter what choice we made, did we really make a choice to begin with? More explicitly, somewhere in the standard diagram-chasing proof of the ...
4
votes
2answers
145 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 ...
10
votes
3answers
321 views

When to use Zorn's Lemma

I was looking at an exercise this morning which I was able to reduce to showing that the nilradical is the the intersection of the prime ideals in a ring -- a fact I remembered was true, but which I ...
5
votes
2answers
282 views

Where is the Axiom of choice used?

In Reid's commutative algebra, there is a proof of equivalent conditions of Noetherian rings, especially (1) The set of ideals of $A$ has the a.c.c. $\Rightarrow$ (2) Every ideal in $A$ is finitely ...