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6 votes
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Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?

$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need ...
Jeremy Rickard's user avatar
2 votes
Accepted

Wedderburn theorem for finite-dimensional algebras over the complex numbers

[Elaborating as an answer b/c the OP asked for more details that didn't fit into a comment] Suppose $\mathcal{A} \subseteq M_d(F)$ with $F$ algebraically closed and $\mathcal{A}$ semisimple. To keep ...
Joshua Grochow's user avatar
3 votes
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Minimal ideals and subalgebras of semisimple algebras

For a not necessarily unital ring $R$, a left $R$-module $S$ is simple if $RS\neq 0$ and $S$ has no proper submodule. A simple right module is defined dually. For a not necessarily unital ring the ...
Benjamin Steinberg's user avatar
3 votes

Minimal ideals and subalgebras of semisimple algebras

In this answer, I was assuming the naive definition of a simple module $M$ for a nonunital ring $R$ as one with no proper nonzero submodules. It seems to be common to also insist that $RM\neq0$, in ...
Jeremy Rickard's user avatar

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