# New PDF release: A Problem Book in Algebra By V. A. Krechmar

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Extra resources for A Problem Book in Algebra

Example text

Er } a self-hereditary family of pairwise orthogonal bricks forming the mouth of the tube T . 2). 2), given i ∈ {1, . . , r} and j ≥ 1, we have • End Ei [j] ∼ = K[t]/(tm ), for some m ≥ 1, • End Ei [j] ∼ = K if and only if j ≤ r, and • Ext1A (Ei [j], Ei [j]) ∼ = DHomA (Ei [j], τ Ei [j]) = 0 if and only if j ≤ r − 1. (c) If the tube T is homogeneous then Ext1A (M, M ) = 0, for any indecomposable M in C. Proof. Because the stable tube is standard and {E1 , . . 2. 6) applies to the extension category E = EX TA (E1 , .

Let A be an algebra, E = EX TA (E1 , . . 6), TE be the tube of rank r formed by the indecomposable modules in E, and let M be an indecomposable A-module that is not in TE . (a) If there exist i ∈ {1, . . , r} and a non-zero f : Ei −→ M then, for any j ≥ 2, there exists a gj : Ei [j] −→ M such that f = gj uij . (b) If there exist i ∈ {1, . . , r} and a non-zero f : M −→ Ei then, for any j ≥ 2, there exists a hj : M −→ Ei−j+1 [j] such that f = pi−1,2 hj . homomorphism homomorphism homomorphism homomorphism Proof.

This shows that the inclusion radp+1 i k A radpA (Ei , Ek ) ⊆ radp+1 A (Ei , Ek ) holds, for any p ∈ {1, . . , m − 1}, and consequently, we get radA (Ei , Ek ) = ∞ radm A (Ei , Ek ) = radA (Ei , Ek ) = 0, because the component T is generalised standard. Then the proof of the theorem is complete. 6. Corollary. Let A be an algebra, T a standard stable tube of the Auslander–Reiten quiver Γ(mod A) of A, and B = A/AnnA T . Then T is a hereditary standard stable tube of Γ(mod B), under the fully faithful exact embedding mod B → mod A induced by the canonical algebra surjection A −→ B. 