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 modules
Search options not deleted
user 43107
For questions on modules over rings.
5
votes
Divisible torsion $\mathbb{Z}$-modules
Since $\mathbf{Q}/\mathbf{Z} = \varinjlim (1/n) \mathbf{Z}/\mathbf{Z}$ and $V = \varinjlim V[n]$ (as $V$ is torsion), it suffices to show that for $n > 0$ the natural map ${\rm{Hom}}(\mathbf{Q}/\mathb …
- 4,713
1
vote
Simultaneous decomposition of modules over Dedekind domains
There will then be nothing to do if $r=1$, and if $r > 1$ we use rank-induction to describe these modules compatibly as direct sums of invertible modules in such a way that for the ambient $D$-module all … Continuing via induction on $r > 1$, we get compatible decompositions of $M_2$ and $M_1$ as direct sums of invertible $D$-modules in such a way that all but one of the invertible $D$-modules using for …
- 1,623