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 21793
Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
2
votes
Finding purely transcendental parts of field extensions
I don't know if this is a counterexample or not, but what would you do if $K$ is the field of fractions of the monoid ring $R = \mathbb{Z}[\prod_{i=1}^\infty \mathbb{N}]$? ($R$ is an integral domain b …
- 166
10
votes
Integrally closed
I claim the integral closure of the ring $R:=F+xk[[x]]$ is $R':=F'+xk[[x]]$, where $F'$ is the algebraic closure of $F$ inside $k$. For one, the field of fractions of $R$ is $k((x))$, and $F'$ is inte …
- 166