Why the tensor product of graded algebras is defined with a commutation $\epsilon $ like this : $(a\otimes b)(c\otimes d)= \epsilon(ac\otimes bd)$ ? what is the usefulness of the commutator $\epsilon$ here ? why that tensor product is not defined simply by $(a\otimes b)(c\otimes d)= (ac\otimes bd)$ ?
Tensor product of graded algebras
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abstract-algebra
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0Where did you get this definition? I have never seen the tensor product of algebras defined with that element. I even double checked: https://en.wikipedia.org/wiki/Tensor_product_of_algebras Maybe for clarity you should add something like: "I have two $R$-algebras $A$ and $B$, etc..." – 2012-09-26
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0see Bourbaki, Algebras chapters 1 to 4 – 2012-09-26
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0Sorry, I had missed the graded bit. – 2012-09-26
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0Even taking into account superalgebras and the Koszul sign rule, I don't think $(a\otimes b)(c\otimes d)= \epsilon(ac\otimes bd)$ is anywhere near the correct formula. – 2015-10-08