Abstract Algebra by David S. Dummit, Richard M. Foote

By David S. Dummit, Richard M. Foote

Broadly acclaimed algebra textual content. This ebook is designed to offer the reader perception into the ability and sweetness that accrues from a wealthy interaction among assorted components of arithmetic. The e-book conscientiously develops the speculation of alternative algebraic buildings, starting from uncomplicated definitions to a couple in-depth effects, utilizing quite a few examples and workouts to help the reader's knowing. during this method, readers achieve an appreciation for a way mathematical buildings and their interaction bring about robust effects and insights in a couple of diverse settings.

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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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