All Questions
Tagged with axiom-of-choice ac.commutative-algebra
13
questions
21
votes
1
answer
1k
views
A Krull-like Theorem and its possible equivalence to AC
A well known equivalent of the Axiom of Choice is Krull's Maximal Ideal Theorem (1929): if $I$ is a proper ideal of a ring $R$ (with unity), then $R$ has a maximal ideal containing $I$. The proof is ...
- 528
17
votes
1
answer
991
views
Artin Rings, Noetherian Rings, and the Axiom of Choice
It is a well-known fact that in ZF, the axiom of choice is equivalent to the statement that every commutative ring has a maximal ideal. On the other hand, for Noetherian rings, this is not necessary (...
- 576
0
votes
1
answer
278
views
Noetherianess of a finite module over a noethrian ring without Axiom of Choice
All rings are assumed to be commutative with 1.
We say a module over a ring is strictly noetherian if every non-empty set of submodules has a maximal member. We say a ring is strictly noetherian if it ...
- 1,159
2
votes
1
answer
170
views
Matching power series to infinity
As pointed out by Makoto, on this question about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another ...
- 18.3k
11
votes
1
answer
798
views
Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?
Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial (scalar multiplication is not identically zero) $\mathbb{C}$-module.
Now my questions are?
0.Is it consistent with $ZF$ that $\mathbb{R}$ ...
- 2,371
10
votes
1
answer
2k
views
Can we prove that the ring of formal power series over a noetherian ring is noetherian without axiom of choice?
Let $A$ be a commutative ring with an identity.
Suppose that every non-empty set of ideals of $A$ has a maximal element.
Let $A[[x]]$ be the formal power series ring over $A$.
Can we prove that every ...
- 1,159
5
votes
1
answer
364
views
Proper-class sized "ring" with no maximal ideals
Suppose I have a collection of "elements" together with operations that satisfy the axioms for a commutative ring with identity --- except that these elements form not a set, but a proper class.
Must ...
- 22.7k
4
votes
1
answer
415
views
Is this height-transcendence-degree inequality true without AC ?
Let $R$ be a $k$-algebra ($k$ a field) and a domain of finite Krull dimension. In
$\quad$ Krull dimension less or equal than transcendence degree?
it is shown that
$$\text{Krull-dim}(R) \le \text{...
- 16.2k
4
votes
0
answers
2k
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 zorn's ...
- 1,778
9
votes
1
answer
2k
views
Minimal prime ideals and Axiom of Choice(revised version)
From the page:
Existence of prime ideals and Axiom of Choice.,
I have found that The existence of prime ideals in commutative rings is equivalent to the Boolean Prime Ideal theorem. But $BPI$ is ...
- 1,778
12
votes
3
answers
4k
views
Existence of prime ideals and Axiom of Choice.
One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple ...
- 1,778
4
votes
3
answers
1k
views
Chevalley's valuation extension theorem and the axiom of choice
Hello,
Do we know if the axiom of choice is needed for Chevalley's valuation/place extension theorem (i.e. the theorem that states that for every valued field and a field extension, one can extend ...
- 43
27
votes
4
answers
3k
views
Nilradicals without Zorn's lemma
It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.
Every proof I found (e.g. in the classical "Commutative Algebra" by ...
- 1,042