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 22989

For questions requesting examples of a certain structure or phenomenon

8 votes
Accepted

Rank versus free-rank of a module

There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$. Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the catego …
Jeremy Rickard's user avatar
8 votes

Abelian category which is not well-powered

Here's a simpler, but less consequential, example. Take the category of "eventually constant" functors from ordinals (considered as a category with a single morphism $\alpha\to\beta$ when $\alpha\leq …
Jeremy Rickard's user avatar
17 votes
Accepted

Torsion-free abelian group $A$ such that $A \not \simeq A \oplus \Bbb Z \simeq A \oplus \Bbb...

$\mathbb{Z}$ is cancellable for abelian groups. This was proved in the 1950s by Walker and Cohn (independently) and is often called "Walker's cancellation theorem". The proof is only a few lines. So …
Jeremy Rickard's user avatar
18 votes
Accepted

Are all vector-space valued functors on sets free?

This is probably an absurdly over-complicated answer, but ... Let $$J(X)=\left\{\sum_{x\in X}a_xx\in GX: \sum_{x\in X}a_x=0\right\}.$$ I claim that $J$ is not of the form $H\circ G\circ F$. Suppos …
Jeremy Rickard's user avatar
22 votes
Accepted

Example of an abelian category with enough projectives and injectives which are not dual

The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respe …
Jeremy Rickard's user avatar
80 votes
Accepted

$A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$

that $A\cong B\oplus B$ and $B\cong B\oplus B\oplus B$, so $B$ is an example of an abelian group $B$ with $B\cong B\oplus B\oplus B\not\cong B\oplus B$, I think rather simpler to describe than other examples … with $A\oplus A\cong B\oplus B$, which is one of Kaplansky's "test problems" for abelian groups in his famous 1954 book on Infinite Abelian Groups, which also seems to be simpler to describe than other examples
Jeremy Rickard's user avatar