bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 3 years, 1 month |
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stats | profile views | 1,646 |
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 |
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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 |
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Is the Amitsur-Levitzki identity essentially unique?
... and why "combinatorics"? |
Nov 22 |
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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
added 1 character in body |
Nov 22 |
answered | Powers of finite simple groups |
Nov 20 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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Does the linear automorphism group determine the vector space?
Thanks, @Todd. I corrected the number. |