Search Results
| 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 ac.commutative-algebra
Search options not deleted
user 125073
Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
1
vote
Restricted extension of prime ideals of the ring of polynomials over $\mathbb{Q}$
I think the answer is yes when $n=2$.
Let $R=\mathbb{Q}[x,y]$; since $R$ is a two-dimensional the only non-trivial case is for height-1 primes, and since $R$ is a UFD we can suppose that the prime id …
- 101
-1
votes
UFD containing element with finite quotient
Edit: disregard this answer, I completely misread the question.
Let $D:=\mathbb{Z}[X]$, and let $S:=D\setminus(2D\cup(3,X)D)$. Let
Take a DVR $V$, and let $R=V[X]$ be the polynomial ring on $V$. Th …
- 101
5
votes
Accepted
Are unique prime ideal factorization domains locally noetherian?
No, this property does not imply that $A$ is locally Noetherian.
For example, let $F\subset L$ be an extension of fields, and let $A=F+XL[[X]]$ (that is, $A$ is the set of power series over $L$ whose …
- 101
1
vote
Accepted
Chain of closed irreducible sets on Zariski Riemann spaces
The answer to the problem is no. One reason is that the dimension of a valuation ring $V\in\mathrm{ZR}(K,A)$ may be greater than the dimension of $A$. (The supremum of the dimension of the elements of …
- 101