Answer on Optics Question for sid
A · U = eRe · eV = eReV ⊆ eV = U, so U is indeed an A-module.Let a = ere ∈ A, where r∈ R. Then ae = ere2 = a.If aU = 0, then 0 = aeV = aV implies that a = 0, soAU is faithful. To check that AU is simple, let us show that for0 <> u ∈ U and u'∈ U, we have u' ∈ Au. Note that u, u' ∈ eV implies u = eu, u' = eu'.We have u' = ru for some r ∈ R, so u' = eu' = eru = (ere)u∈ Au, as desired.
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