Superalgebra

Remarks

A superalgebra is supercommutative if $ab = (-1)^{\abs{a}\abs{b}} ba$ for all homogeneous $a, b$. Example: the exterior algebra $\Lambda(V)$ of a vector space $V$, with $A_0$ the sum of the even powers $\Lambda^{2m} V$ and $A_1$ the sum of the odd powers. The product maps $\Lambda^p V \times \Lambda^r V$ into $\Lambda^{p+r} V$, and $p + r$ has the parity of the sum, so the condition holds; moreover $\Lambda(V)$ is supercommutative. Example: the matrix ring $M_2(K)$, with $A_0$ the diagonal matrices and $A_1$ the off-diagonal ones, is a superalgebra that is not supercommutative. Non-example: $K[x]$ with $A_0 = K$ and $A_1 = xK[x]$ fails the condition, since $x \in A_1$ but $x \cdot x = x^2 \notin A_0$. Equivalently, a superalgebra is a monoid object in the category of $\Z/2\Z$-graded modules, with the tensor product of graded modules.