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22h
comment Rotate a unit sphere such as to align it two orthogonal unit vectors
@StephenT Good job :) It's nice when we can give general ideas and still let you have the benefit of the experience working the problem.
1d
comment Looking for a terminology in ring theory (“ideal” which is not necessarily closed under addition )
"left semigroup ideal" sounds like a viable ad-hoc term to use, if nothing better appears.
1d
comment Looking for a terminology in ring theory (“ideal” which is not necessarily closed under addition )
@Pierre-GuyPlamondon I remembered "adele" but I must have forgotten "idele" :) Yes, "idel" would be altogether too close to those. The temptation to mimic "rng" and "rig" was too great for me :)
1d
comment Looking for a terminology in ring theory (“ideal” which is not necessarily closed under addition )
Candidate if there is no term: idel? Ideal without addition?
1d
comment Looking for a terminology in ring theory (“ideal” which is not necessarily closed under addition )
Interesting question. I personally haven't seen such a word used (probably because I've never seen this idea discussed.)
1d
comment Is a = 0 a valid counterexample to this statement?
@GeoffreyCritzer I don't see that the distinction is important here. Maybe you could just rephrase it as "if e is an idempotent and e is not 1 and not 0, then e is a zero divisor." That much is certainly true.
1d
comment Is a = 0 a valid counterexample to this statement?
@GeoffreyCritzer Sounds like an error then :) Still, you can fruitfully solve the modified problem.
1d
comment Is a = 0 a valid counterexample to this statement?
@marwalix Well, some authors define zero divisors to be nonzero. It just depends. Clearly some flexibility is required for terms like this one :)
1d
answered Is a = 0 a valid counterexample to this statement?
1d
answered Under what conditions does $M \oplus A \cong M \oplus B$ imply $A \cong B$?
1d
comment All simple modules are projective $\Rightarrow$ semisimple
Dear @egreg : No sarcasm was intended at all: I was just illustrating my disbelief with an example, because I did not understand the objection. I frequently use this overly formal format to avoid misunderstandings. In the future please don't assume my intentions are negative. Thanks and Regards
2d
comment Example of $Q((x))$ that doesnt match field of fractions of ring $F[[x]]$
IMO this answer (basically a flat assertion) would be greatly enhanced by a hint. At worst I can see this answer fooling beginners into thinking any "infinite" series is not in $frac(\Bbb Z[x])$.
2d
answered Rotate a unit sphere such as to align it two orthogonal unit vectors
Jun
27
comment $R$ is semisimple $\iff $ it is Artinian and $J(R)=0$
possible duplicate of Prove that $M$ is finitely generated and it is a semi-simple module.
Jun
27
comment why is a simple ring not semisimple?
@learnmore the example you gave is a product of fields, which is perfectly fine considering you can use a 1-by-1 square matrix ring.
Jun
27
revised why is a simple ring not semisimple?
added 151 characters in body; added 1 character in body
Jun
27
comment why is a simple ring not semisimple?
@egreg I've long since clarified what I meant by suitable substructure in the comments and solution, so I don't believe that objection applies. Subgroups of course do not admit good quotients.
Jun
26
comment why is a simple ring not semisimple?
@TobiasKildetoft I was taking for granted the viewpoint that substructures are suitable for quotients, so there is no conflict here (except in opinion of exposition.) But what you're saying is clearer, so I'll incorporate some, if you dont' mind.
Jun
26
revised why is a simple ring not semisimple?
added 280 characters in body
Jun
26
answered why is a simple ring not semisimple?