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Mar 3, 2020 at 14:07 comment added Johannes Hahn He meant "a direct product of matrix algebras", i.e. the skew fields $D_i$ in the Wedderburn decomposition $A/J \cong \prod_{i=1}^n D_i^{k_i\times k_i}$ are all equal to $K$ itself. This is always the case for algebraically closed $K$ (because the $D_i$ are of finite dimension over $K$) but not in general.
Mar 3, 2020 at 13:37 comment added 57Jimmy Thanks a lot for your answer. With the sentence "$A/J$ is isomorphic to matrix algebras over $K$" do you mean that the quotient of $A$ by its Jacobson radical is isomorphic to some subalgebra of $M^{n \times n}(K)$ for some $n$?
Mar 3, 2020 at 13:03 vote accept 57Jimmy
Mar 3, 2020 at 12:31 history edited Mare CC BY-SA 4.0
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Mar 3, 2020 at 12:26 history answered Mare CC BY-SA 4.0