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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
31
votes
2
answers
2k
views
Should Krull dimension be a cardinal?
A totally ordered finite set $\quad \mathcal P_0 \varsubsetneq \mathcal P_1\varsubsetneq \dots \mathcal \varsubsetneq \mathcal P_n \quad$ of prime ideals of a ring $A$ is said to be a chain of lengt …
42
votes
4
answers
3k
views
What is the Krull dimension of the ring of holomorphic functions on a complex manifold?
Consider a connected holomorphic manifold $X$ and its ring of holomorphic functions $\mathcal O(X).$
My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known?
…
12
votes
Are quotients of polynomial rings almost UFDs?
No, quotients of polynomial rings are definitely not "almost UFDs".
Any finitely generated ring over $K$ is such a quotient and this means a lot of non UFDs.
Said differently, any algebraic variety in …
132
votes
3
answers
21k
views
When is the tensor product of two fields a field?
Consider two extension fields $K/k, L/k$ of a field $k$.
A frequent question is whether the tensor product ring $K\otimes_k L$ is a field. The answer is "no" and this answer is often justified by so …
21
votes
5
answers
1k
views
Computation of fraction field of formal series over the integers
What is the fraction field $K$ of the domain $\mathbb Z[[X]]$?
It is strictly smaller than the field of Laurent series $L=\operatorname {Frac}\mathbb Q[[X]]$, since $\sum_{i\geq 0}\frac {X^i}{i!}\in …
12
votes
2
answers
1k
views
Is every locally free module of rank $1$ over a commutative ring concretely invertible?
Since this subject is full of misunderstandings (see here, here, here, and here) let us fix a precise terminology.
Let $A$ be a commutative ring and $P$ an $A$-module.
I) We'll say that $P$ is a lo …
43
votes
Accepted
Rank of a module
Since your profile says you are interested in Algebraic Geometry, here are geometric considerations that might appeal to you.
Consider a projective module $P$ of finite type over a commutative ring $ …
8
votes
What is the right definition of the Picard group of a commutative ring?
1) About the second definition:
$\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring …
12
votes
1
answer
408
views
Is height preserved in a normalization?
Let $R$ be a domain and $\tilde R$ its integral closure in its fraction field: $R\subset \tilde R\subset Frac(R)$.
Is it true that a prime ideal $ \tilde {\mathfrak p} \subset \tilde R$ and its trac …
4
votes
1
answer
1k
views
Is the transcendence degree of a domain over a subfield the same as that of the fraction fie...
Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$.
Obviously if a family $(a_i)_{i\in I}$ of elements $a_i \in A$ is algebraically indep …
13
votes
Accepted
Examples of Noetherian overkill
1) It is sometimes stated that a finitely generated module $M$ over a Noetherian commutative ring $R$ is projective if for all maximal ideals $\mathfrak m\subset R$ the localized module $M_\mathfrak m …
26
votes
What makes a theorem *a* "nullstellensatz."
What I find intriguing is that the Nullstellensatz is underappreciated in the sense that many people appeal to a variation of it without saying (or realizing) they do.
For example, Hadamard's lemma …
18
votes
Duals and Tensor products
For those interested in general statements, here is a summary of assumptions under which the canonical morphisms of $A$-modules below are isomorphisms:
If $P$ is finitely generated projective:
$$P\ …
18
votes
Accepted
Is the tensor product of a power series ring and a field noetherian?
The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $ k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theore …
21
votes
2
answers
2k
views
What is the dimension of the product ring $\prod \mathbb Z/2^n\mathbb Z$ ?
In an anwswer to a question on our sister site here I mentioned that a reduced commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r …