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YCor
YCor
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Can all finite dimensional non commutative-dimensional noncommutative algebras with trace be trace-preserving embedded into matrix rings?

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Asvin
Asvin
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Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I always find square matrices $A_1,\dots,A_n$ (of some dimension $r$) so that I have an embedding $f: R \to M_r(k)$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2\dots a_n) = t(a_2\dots a_na_1)$. Is this the only restriction?

Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I find square matrices $A_1,\dots,A_n$ (of some dimension $r$) so that I have an embedding $f: R \to M_r(k)$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2\dots a_n) = t(a_2\dots a_na_1)$. Is this the only restriction?

Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I always find an embedding $f: R \to M_r(k)$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2\dots a_n) = t(a_2\dots a_na_1)$. Is this the only restriction?

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Asvin
Asvin
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  • 52

Can all finite dimensional non commutative algebras be embedded into matrix rings?

Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I find square matrices $A_1,\dots,A_n$ (of some dimension $r$) so that I have an embedding $f: R \to M_r(k)$ compatible with the trace functions on both sides?

One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2\dots a_n) = t(a_2\dots a_na_1)$. Is this the only restriction?