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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 …
user26857's user avatar
  • 1,313
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 $ …
Georges Elencwajg's user avatar
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 …
Georges Elencwajg's user avatar
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 …
Georges Elencwajg's user avatar
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 …
Georges Elencwajg's user avatar
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\ …
Bogdan's user avatar
  • 335
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 …
Georges Elencwajg's user avatar
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 …

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