Answer to Question #17356 in Abstract Algebra for sanches
Show that, for any direct product of rings Ri, rad ((direct product)Ri) = (direct product) rad Ri.
Let y = (yi) ∈(product)Ri. Since y ∈rad ((product)Ri) amounts to 1 − xy being left-invertible for any x = (xi) ∈ (product)Ri. This, in turn, amounts to 1 − xiyi being left-invertible in Ri, for any xi ∈ Ri (and any i). Therefore, y ∈rad ((product) Ri) iff yi ∈rad Ri for all i.
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