Let $A = \bigoplus_{i \in \mathbb{Z}} A_i$ be a strongly graded unital algebra over $\mathbb{C}$, with no zero divisors. Is it always true that $$ m: A_i \otimes_{A_0} A_j \to A_{i+j} $$ is an isomorphism?
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asked Jan 6, 2020 at 20:19
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2$\begingroup$ For context, for an abelian group $G$ a $G$-graded algebra $A$ is called strongly graded if $A_tA_u=A_{t+u}$ for all $t,u\in G$. (I'm not sure what's the implicit meaning of "algebra": I just guess "associative unital algebra".) $\endgroup$– YCorCommented Jan 6, 2020 at 21:13
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$\begingroup$ and the group $G$ does not necessarily have to be abelian. $\endgroup$– Konstantinos KanakoglouCommented Jan 6, 2020 at 21:48
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$\begingroup$ yes, algebra means associative unital algebra $\endgroup$– Fofi KonstantopoulouCommented Jan 6, 2020 at 22:39
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1 Answer
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Yes this is always an isomorphism of $A_0$-bimodules.
It is a general result for strongly graded rings. It holds for an arbitrary grading group $G$ (not necessarily $\mathbb{Z}$) and does not depend on the presence of zero divisors. For a proof, see Corollary 3.1.2, p.82, from Methods of Graded Rings.
answered Jan 6, 2020 at 20:56