Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 19576

Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

0 votes
2 answers
331 views

finite global dimension vs integral Domain

For the quotient of polynomial rings over complex number field, its global dimension is finite is equivalent to it is domain. is this true?
iff's user avatar
  • 77
1 vote
1 answer
446 views

Does the global dimension gldim R equal the projective dimension of R as bimodule over its e...

I know that generally the answer is no, for example the weyl algebra。 But is this true for commutative algebra? or we may restrict to affine commutative algebras。 Maybe ,it is a classical result. So …
iff's user avatar
  • 77
5 votes
6 answers
4k views

an easy example of valuation ring which is not noetherian? [duplicate]

Is there an easy example of valuation ring which is not noetherian?
iff's user avatar
  • 77
-1 votes
1 answer
752 views

Question on an exercise on homological algebra?

Suppose $R$ has finite global dimension $n$, $N$ is a finitely generated module, $F$ is a free module, and $\operatorname{Ext}^n(N, F) \neq 0$. Then $\operatorname{Ext}^n(N, R)$ is also non-trivial. N …
iff's user avatar
  • 77