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Let $A = \bigoplus_{k \in \mathbb{Z}} A_k$ be a not necessarily commutative $\mathbb{Z}$-graded unital algebra over a field $\mathbb{K}$, and assume that it is strongly graded: $$ A_kA_l = A_{k+l}. $$ In general can it happen that the multiplication does not give an isomorphism $$ A_k \otimes_{A_0} A_l \simeq A_{k+l}? $$ The map will be surjective since we are assuming the gradind to be strong, but will it be injective?

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    $\begingroup$ What graded algebras do you know? Injectivity is false in almost every example one can think of. $\endgroup$
    – Vladimir Dotsenko
    Commented Dec 14, 2020 at 13:28
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    $\begingroup$ @VladimirDotsenko Most algebras I can think of are not strongly graded anyway (for instance if $A_k=0$ for some (possibly negative) $k$, then $A_0=A_kA_{-k}=0$ and hence $A=0$). Anyway I can see a few ad-hoc counterexamples. $\endgroup$
    – YCor
    Commented Dec 14, 2020 at 13:30
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    $\begingroup$ (Actually I was too quick and have no counterexample in mind at the moment) $\endgroup$
    – YCor
    Commented Dec 14, 2020 at 13:53
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    $\begingroup$ Take your favourite group with a homomorphism to $\Bbb Z$ and look at its group algebra. $\endgroup$
    – Denis T
    Commented Dec 14, 2020 at 16:02
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    $\begingroup$ Possible duplicate of: mathoverflow.net/q/349864/85967 $\endgroup$
    – Konstantinos Kanakoglou
    Commented Dec 14, 2020 at 17:46

1 Answer 1

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No it cannot happen.

And not only for strongly $\mathbb{Z}$-graded rings; this is always the case for any strongly $G$-graded ring, where $G$ is a group. $A_k \otimes_{A_0} A_l \simeq A_{k+l}$ is an isomorphism of $A_0$-bimodules.

(See: Corollary 3.1.2, p.82, from Methods of Graded Rings).

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    $\begingroup$ Great, I didn't hope there would be a such nice answer! Thanks a lot! $\endgroup$
    – Piet Bongers
    Commented Dec 14, 2020 at 17:52
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    $\begingroup$ You are welcome. And by the way, welcome to MO. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Dec 14, 2020 at 23:25
  • $\begingroup$ To be clear, you are saying that it cannot happen that multiplication does not give an isomorphism? $\endgroup$
    – Zach Teitler
    Commented Dec 16, 2020 at 1:24
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    $\begingroup$ @Zach Teitler, yes that is right. This is what i am saying. (Following the wording of the OP: "...can it happen that the multiplication does not give an isomorphism..") $\endgroup$
    – Konstantinos Kanakoglou
    Commented Dec 16, 2020 at 1:42

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