2009
DOI: 10.1016/j.jalgebra.2008.08.023
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Some remarks on the ring of twisted tilting modules for algebraic groups

Abstract: In a recent paper [S. Doty, A. Henke, Decomposition of tensor products of modular irreducibles for SL 2 , Q. J. Math. 56 (2005) 189-207], Doty and Henke give a decomposition of the tensor product of two rational simple modules for the special linear group of degree 2 over an algebraically closed field of characteristic p > 0. In performing this calculation it proved useful to know that the simple modules are twisted tensor products of tilting modules. It seems natural therefore to consider the ring of twisted … Show more

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Cited by 2 publications
(4 citation statements)
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“…As in Section 3, we identify X + (T ) n , X + (T ) ∞ , X + s (T ) n , X + s (T ) ∞ as subsets of X + (T ) × X + ( T ) ∞ in the usual way. By [ [4], p.no 72 (5)] and [ [4] (iii), (iv), (v) and (vi) follows from Proposition of Section 1 of [7].…”
Section: Now By [[7]mentioning
confidence: 99%
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“…As in Section 3, we identify X + (T ) n , X + (T ) ∞ , X + s (T ) n , X + s (T ) ∞ as subsets of X + (T ) × X + ( T ) ∞ in the usual way. By [ [4], p.no 72 (5)] and [ [4] (iii), (iv), (v) and (vi) follows from Proposition of Section 1 of [7].…”
Section: Now By [[7]mentioning
confidence: 99%
“…We let A C (n) = C ⊗ Z A n . By [7], we know that the polynomials Q p−1 (X i+1 − h(X i ))), i ≥ 0, has no repeated roots. Hence to show that the ring A C (n) is reduced, it is enough to show that (P l−1 (X −1 )(X 0 − g(X −1 )) has no repeated roots.…”
Section: Now By [[7]mentioning
confidence: 99%
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