Timeline for What does the ring $K[S]$ know about a group generated by $S$?
Current License: CC BY-SA 3.0
8 events
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| Jan 7, 2015 at 22:37 | comment | added | Eric Wofsey | It's not obvious from the perspective of the universal property (or equivalently, generators and relations) of a monoid ring, but it is obvious from its explicit description as the ring of formal linear combinations (the subring of the set of formal linear combinations of $\Gamma$ generated by $S$ is just the set of formal linear combination of products of elements of $S$). | |
| Jan 7, 2015 at 22:07 | comment | added | Roland Bacher | This is not obvious to me: why can't there be some complicated expression $\sum a_i (r_i-r'_i) b_i$ with $a_i,b_i\in K[\Gamma]$ and $r_i=r'_i$ relations such that the result is a linear combination involving only monomials in $S$? | |
| Jan 7, 2015 at 20:32 | comment | added | Benjamin Steinberg | As Eric says the elements of S are linearly independent so you get the monoid ring. | |
| Jan 7, 2015 at 19:14 | comment | added | Eric Wofsey | Maybe I'm missing something, but it seems to me the answer to your first question is an obvious no. Isn't $K[S]$ just the monoid ring of the submonoid generated by $S$? | |
| Jan 7, 2015 at 19:10 | comment | added | Roland Bacher | Non-free groups generated by $S$ such that $S^*$ injects into them are easy to construct: most one-relator groups should have this property if the relation is a bit complicated and involves at least one inverse of a generator in $S$. | |
| Jan 7, 2015 at 19:04 | comment | added | Yemon Choi | For your second question: certainly one can get free non-abelian subsemigroups $S$ inside certain solvable groups, and I suspect (though I don't have a proof) that in those setting $K[S]$ would be isomorphic to the free algebra generated by elements of $S$ | |
| Jan 7, 2015 at 19:02 | history | edited | Roland Bacher | CC BY-SA 3.0 |
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| Jan 7, 2015 at 18:52 | history | asked | Roland Bacher | CC BY-SA 3.0 |