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Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
an arXiv tag added
Nov
22
suggested approved edit on Is the Amitsur-Levitzki identity essentially unique?
Nov
22
awarded  Nice Answer
Nov
22
comment bounding from below the cardinality of a set of generators of the $n$-fold cartesian product group of a finite group
See (my answer to) a similar question: mathoverflow.net/q/187736/24165 .
Nov
22
revised Powers of finite simple groups
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Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
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Nov
22
comment Is the Amitsur-Levitzki identity essentially unique?
... and why "combinatorics"?
Nov
22
comment Is the Amitsur-Levitzki identity essentially unique?
Why this is tagged "commutative algebra"?
Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
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Nov
22
answered Is the Amitsur-Levitzki identity essentially unique?
Nov
22
revised Powers of finite simple groups
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Nov
22
answered Powers of finite simple groups
Nov
20
comment Normal Covering of a Finite Group
This Corollary 5.5 follows immediately from the theorem of Brodie, Chamberlain and Kapp, PAMS 1988, see Nick Gill's answer: mathoverflow.net/a/185604/24165 .
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
Thomas, every finitely generated associative commutative ring is residually finite. (For free rings (= polynomial rings), this is almost obvious.)
Nov
9
comment Possible cardinality and weight of an ordered field
Taras, you may write an answer and accept your own answer to make the question "closed".
Nov
9
comment Explicitly showing that a free group is LERF
Surely, there is an algorithm for finding this finite-index subgroup and the free complement in this subgroup. Just look at the proof of Hall's free-factor theorem. It is quite easy using Stallings graphs..
Nov
9
comment Explicitly showing that a free group is LERF
Pablo, every f.g. subgroup $M$ of a f.g. free group $F$ is a free factor of a finite-index subgroup of $F$. So, the problem is reduced to the case where $M$ is a free factor of $F$, where it is quite easy.
Nov
9
comment Fantastic properties of Z/2Z
I added these features, thanks, @Sam.
Nov
9
revised Fantastic properties of Z/2Z
added 189 characters in body
Nov
9
comment Does the linear automorphism group determine the vector space?
Thanks, @Todd. I corrected the number.