Let A and B be finite-dimensional algebras over a field k. We denote the enveloping algebra of A by A^e, which is the tensor product (over k) of the algebra A and its opposite algebra. Suppose A^e and B^e are isomorphic, can we say that the algebras A and B are isomorphic?
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2$\begingroup$ Counterexamples are given by nontrivial central simple algebras, e.g. the quaternions over the reals. $\endgroup$– Qiaochu YuanCommented Sep 15, 2020 at 4:59
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1$\begingroup$ @YCor I would think it is true for algebraically closed fields and that case can probably be reduced to quiver algebras. $\endgroup$– MareCommented Sep 15, 2020 at 9:39
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1$\begingroup$ @Mare where could I find a definition of "quiver algebra"? I can find the definition of algebra associated to a quiver, but this seems pretty specific (there are uncountably many non-isomorphic finite-dimensional algebras, so won't be described with combinatorial means) $\endgroup$– YCorCommented Sep 15, 2020 at 9:45
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1$\begingroup$ @YCor A quiver algebra (or often more precise: "bound quiver algebra") is an algebra isomorphic to $KQ/I$ where $Q$ is a finite quiver and $I$ an admissible ideal. (some use the term quiver algebra also just for the algebras $KQ$, which I call "path algebras") See for example the book "Frobenius algebras I" for a good introduction to finite dimensional algebras with many quiver examples. See also springerprofessional.de/bound-quiver-algebras/2212132 $\endgroup$– MareCommented Sep 15, 2020 at 9:46
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1$\begingroup$ Thanks. BTW, in the commutative case, it's a "jet" version of the question asking whether $X\times X$ and $Y\times Y$ are isomorphic varieties implies that $X$ and $Y$ are isomorphic. $\endgroup$– YCorCommented Sep 15, 2020 at 9:49
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