By Ben Smith
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Extra info for Higher Algebra 1 [Lecture notes]
Example text
14. (i) A direct sum of semi-simple modules is semi-simple (this is clear from definition) (ii) The homomorphic image of semi-simple module is semi-simple. (iii) A submodule of a semi-simple module is semi-simple. (iv) Every left R module is semi-simple if and only if R is a semi-simple left R module. N is a morphism and M = ⊕Mi is semi-simple. The image of f is N Mi which is generated by {f (Mi )}. Ker(f ) ∩ Mi is either zero or Mi for each i meaning that f (Mi ) ∼ . = 0 That is f (Mi ) if not zero is simple, so N is generated by its simple submodules and by theorem N is semi-simple.
Vr be representatives for the irreducible representations of G and r, the number of matrix rings in the Artin-Wedderburn decomposition of k[G]. If this decomposition is Mn1 × · · · × Mnr , we can assume dim(Vi ) = ni , and each representation appears in k[G] with multiplicity ni . k[G] ∼ = Mn1 (k) × · · · × Mnr (k) (as rings) ∗ 0 ··· 0 ∼ = .. . .. . 0 ∗ 0 ··· 0 ⊕ ∗ 0 ··· 0 .. .. . . . 0 0 ··· ∗ ⊕ ··· ⊕ 0 ∗ 0 ··· 0 .. . .. .. ( all ∼ = as Mn1 (k)-modules so as k[G] modules) 0 0 ··· ∗ n1 terms ··· ∗ 0 ··· 0 ..
Xn ] ∼ = (R[x1 , . . , xn−1 ])[xn ]. Let I be an ideal of R[x] and show it is finitely generated. Define Jn := {a ∈ R : a = 0 or ∃f ∈ I, deg(f ) = n, a = [xn ]f } then each Jn is an ideal of R and J0 ⊆ J1 ⊆ J2 ⊆ . . Since R is Noetherian this sequence stabilizes so that there is some N with JN = JN +k for all k ≥ 0. Each Ji is finitely generated by say {ai0 , . . , airi }. Choose for each aij = 0, a polynomial fji in I with [xi ]fji = aij . claim: fji = I It is clear that I ⊇ fji : 0 ≤ i ≤ N, 0 ≤ j ≤ r(i) .
Higher Algebra 1 [Lecture notes] by Ben Smith
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