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Answer by Jeremy Rickard for Does hereditary and connected imply that the...
$R$ 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...
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All rings are assumed to be associative and have a 1. Let $k$ be a commutative artininan ring and $R$ a finitely generated $k$-algebra. Is it true that if $R$ is connected and hereditary, then $k$ is a...
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