Quasi-Frobenius Rings by W. K. Nicholson, M. F. Yousif

By W. K. Nicholson, M. F. Yousif

This ebook presents an basic, whole account of quasi-Frobenius jewelry at a degree permitting researchers and graduate scholars to realize access to the sector. a hoop is named quasi-Frobenius whether it is "right" or "left" selfinjective, and "right" or "left" artinian (all 4 combos are equivalent). The learn of those jewelry grew out of the idea of representations of a finite team as a gaggle of matrices over a box, and the current quantity of the idea is wide-ranging.

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We may assume that τ¯ π¯ = 0. ¯ with T a right ideal of S. ¯ Let Proof. As S¯ is regular, let π¯ S¯ ⊕ τ¯ S¯ ⊕ T = S, 2 ¯ ¯ ¯ ¯ ¯ S. By hypothesis, we may η¯ = η¯ be such that τ¯ S = η¯ S and π¯ S ⊕ T = (1 − η) assume that η2 = η in S. Note that η¯ π¯ = 0. Then γ = τ + τ η(1 − τ ) satisfies ¯ Moreover, γ¯ = η¯ γ 2 = γ , γ τ = τ , and τ γ = γ , so τ M = γ M and τ¯ S¯ = γ¯ S. because τ¯ η¯ = η¯ and η¯ τ¯ = τ¯ . Hence γ¯ π¯ = η¯ π¯ = 0, so replacing τ by γ proves Claim 1. By Claim 1 we have τ π ∈ J (S), so writing K = ker (τ π ), we have K ⊆ess M by hypothesis.

Observe first that µK = µ(α K ⊕ β K ) = α K , so K ∩ µE 1 ⊆ µK = α K ⊆ K ∩ µE 1 . Hence K ∩ µE 1 = α K ⊆ ker (β). Now suppose that x ∈ ker (α) ∩ βµE 1 , say x = βµe1 where e1 ∈ E 1 . Then αβ(µe1 ) = 0, so µe1 ∈ ker (αβ) = K . Then µe1 ∈ K ∩ µE 1 ⊆ ker (β) and so 0 = βµe1 = x. This proves Claim 2. Write ι : βµE 1 → E 1 for the inclusion map in the diagram. Since E 1 is injective there exists λ ∈ end(E 1 ) such that βµ = λαβµ on E 1 . 31) and µ2 = µ ∈ end(E 1 ), and observe that τi j (M j ) ⊆ Mi for i = j because Mi is M j -injective.

Hence η ∈ J (S), so η¯ = 0. This proves the Claim. 14 Quasi-Frobenius Rings ¯ Since ⊕i π¯ i S¯ ⊆ T it suffices (by the We can now show that T ⊆ess π¯ S. ¯ Let 0 = τ¯ ∈ T. Since ⊕i π¯ i S¯ ⊆ T it follows Claim) to show that T ⊆ π¯ S. ¯ ⊆ess τ¯ S. ¯ ⊆ τ¯ S¯ ∩ π¯ S¯ ⊆ τ¯ S, ¯ so we have ¯ Also, τ¯ S¯ ∩ (⊕i π¯ i S) that τ¯ S¯ ∩ (⊕i π¯ i S) ess ¯ ¯ ¯ ¯ ¯ τ¯ S ∩ π¯ S ⊆ τ¯ S. But τ¯ S ∩ π¯ S is generated by an idempotent ( S¯ is a regular ¯ In particular, τ¯ ∈ π¯ S, ¯ as required. ring) so it follows that τ¯ S¯ ∩ π¯ S¯ = τ¯ S.

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