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1d
reviewed Leave Open Is a group uniquely determined by the sets $\{ab,ba\}$ for each pair of elements a and b?
1d
answered Are there examples of families of objects which are canonically isomorphic, but where diagrams of canonical isomorphisms don't commute?
1d
reviewed Leave Open A metric associated with a continuous surjective map $f:X\to Y$
1d
answered A metric associated with a continuous surjective map $f:X\to Y$
1d
awarded  Enlightened
1d
awarded  Nice Answer
2d
comment How did the summation operation come into use?
What do you mean by "the summation operation" exactly?
Oct
21
reviewed Close About freeness of modules over the coordinate ring of an affine variety
Oct
19
answered Is every polynomial ring over any field regular?
Oct
18
comment “Nice” functions on infinite-dimensional space of germs of continuous functions at a point
Can you even construct any linear functional on germs of continuous functions (besides multiples of evaluation at the point) without the axiom of choice?
Oct
14
comment Connected graph as connected space
The closure of any connected set is connected.
Oct
12
comment Connected graph as connected space
Any compactification of a connected space is connected.
Oct
9
comment Classification of rings satisfying $a^4=a$
@MartinBrandenburg: It is a Stone space if you topologize it based on $\mathbb{F}_4$ being discrete, rather than only $\{0\}$ being closed.
Oct
7
awarded  Nice Answer
Oct
6
awarded  Yearling
Oct
1
answered Standard homology result on double complexes
Sep
29
comment What is the most useful non-existing object of your field?
@AndréHenriques: Actually, it has $2^8$ elements. But you are correct that free complete Boolean algebras on finite sets exist (and are the same as free Boolean algebras). Free complete Boolean algebras on infinite sets do not exist.
Sep
29
reviewed Looks OK What is the most useful non-existing object of your field?
Sep
23
comment A simple example of a ring that is an $A$-module but not an $A$-algebra?
I'm not sure why you say $\mathbb{R}$ is a $\mathbb{C}$-vector space but not a $\mathbb{C}$-algebra; any embedding of $\mathbb{C}$ in $\mathbb{R}$ makes $\mathbb{R}$ a $\mathbb{C}$-algebra. In your situation, $B$ will be an $A$-algebra iff the action of $A$ on $B$ commutes with the action of $B$ on itself. Equivalently, $B$ is an $A$-algebra iff the action of $a\in A$ coincides with multiplication by $a\cdot 1_B$ in the ring structure of $B$.
Sep
9
answered Completion of the set of subsets with half volume.