Skip to main content
10 events
when toggle format what by license comment
Jan 15, 2021 at 15:09 vote accept Sergey Guminov
Jan 15, 2021 at 14:09 comment added YCor OK, I wrote it as an answer then.
Jan 15, 2021 at 14:08 answer added YCor timeline score: 1
Jan 15, 2021 at 14:04 comment added Sergey Guminov @YCor Thank you! That was exactly what I was looking for. By this criterion, this is a trivial deformation after all.
Jan 13, 2021 at 19:04 comment added YCor There's a practical computer-algebra way to determine whether (a) they're all isomorphic up to finitely many exceptions or (b) there "(bounded finite)-to-one" non-isomorphic, i.e., there exists $n$ such that for each $t$ the set of $s$ such that $A_s\simeq A_t$ has cardinal $\le n$. See mathoverflow.net/questions/378149/…
Jan 13, 2021 at 17:51 comment added Benjamin Steinberg In the semigroup case you would look at what the isomorphism does to multiples of matrix units. But in your setting this may not work because there are no units in the matrix $X_t$ which we usually want in the semigroup setting
Jan 13, 2021 at 17:48 comment added Benjamin Steinberg The obvious guess is the algebras are isomorphic iff there are invertible 2x2 matrices P,Q over R and an automorphism $\phi$ of $R$ with $P\phi(X_t)Q=X_m$. At least your is what happens in the semigroup setting
Jan 13, 2021 at 17:45 comment added Benjamin Steinberg You are looking at an example of a Rees matrix ring sciencedirect.com/science/article/pii/002186938390193X over the contracted monoid algebra of the four element monoid 1,x,y,0 with all products not involving 1 equal to zero. For Rees matrix semigroups there are characterizations of isomorphism but I am not so sure in this context
Jan 13, 2021 at 17:03 answer added tim timeline score: 2
Jan 13, 2021 at 14:16 history asked Sergey Guminov CC BY-SA 4.0