6
votes
0answers
232 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 ...
5
votes
1answer
195 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 ...
5
votes
1answer
228 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 <= transcendence degree? it is shown that $$\text{Krull-dim}(R) \le \text{trans.deg}_k ...
1
vote
0answers
332 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 ...
4
votes
1answer
693 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 ...
5
votes
2answers
812 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 ...
4
votes
3answers
614 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 ...
12
votes
4answers
1k 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 ...