TITLE: Tensor product of graded algebras
QUESTION [16 upvotes]: 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 
$(a\otimes b)(c\otimes d)= (ac\otimes bd)$ ?

REPLY [14 votes]: The use of the $\epsilon$ here can be to inherit structure from $A$ and $B$: If $A$ and $B$ are for example graded anticommutative algebras, and you want $A \otimes_R B$ to be graded anticommutative also, $(a \otimes b)(c \otimes d) = (ac \otimes bd)$ won't do: Say for example $a \in A_0$, $c \in A_1$, $b \in B_1$, $d \in B_0$. The you have $ac = ca$, $bd = db$ and hence
\[ (a \otimes b)(c \otimes d) = (ac \otimes bd) = (ca \otimes db) = (c \otimes d)(a \otimes b) \]
But if we want $A \otimes B$ to be graded anticommutative, we need a $-$ here, as $a \otimes b$, $c \otimes d \in (A\otimes B)_1$. 
If we instead define 
\[ (a \otimes c)(b \otimes d) = (-1)^{\deg b\deg c} (ab \otimes cd) \]
for homogeneous $b$ and $c$, the above example with $\deg a = \deg d = 0$, $\deg b = \deg c = 1$ gives 
\[ (a \otimes b)(c \otimes d) = (-1)^{1 \times 1}(ac \otimes bd) = -(ca \otimes db) = -(-1)^{0\times 0}(c \otimes d)(a \otimes b) = -(c\otimes d)(a \otimes b) \]
as wished.