# Exterior Algebra Graded Ring Structure?

In Hatcher's Example 3.13 he defines the Exterior Algebra $\Lambda_R[\alpha_1,...,\alpha_n]$ over a commutative ring $R$ with identity to be the free $R$-module with basis the finite products $\alpha_{i_1}\cdots\alpha_{i_k}$ with $i_1 < \cdots <i_k$ with associative, distributive multiplication defined by the rules $\alpha_i \alpha_j = -\alpha_j \alpha_i$ and $\alpha_i^2=0$, and the empty product serving as the multiplicative identity.

He then simply says that the exterior algebra becomes a commutative graded ring by specifying odd dimensions for the generators $\alpha_i$.

I understand the $\alpha_i$ need to have odd dimensions in order to satisfy the $\alpha_i \alpha_j = -\alpha_j \alpha_i$, but how exactly does the exterior algebra decompose as a direct sum, which is part of the definition of a graded algebra?

• What do you mean "specifying odd dimensions"? Aren't the products $\prod_{j=1}^n \alpha_{i_j}$ defined to have grade $n$? Then the space decomposes into direct sums of the pieces generated by elements of separate grades? – rschwieb Jun 22 '17 at 19:29
• @rschwieb I'm not 100% sure what he means by that -- that's just what he said. I've never studied exterior algebras before so I would have preferred a bit more detail from him. – TuoTuo Jun 22 '17 at 20:02

Let us write $E=\Lambda_R[\alpha_1,...,\alpha_n]$. A basis for $E$ as an $R$-module is the set of all monomials $\alpha_{i_1}\dots\alpha_{i_m}$ where $i_1<\dots<i_m$. Now for each $i$, let $d_i$ be an odd integer that will be the degree of $\alpha_i$. We can then define the degree of a monomial $\alpha_{i_1}\dots\alpha_{i_m}$ to be $d_{i_1}+\dots+d_{i_m}$. Let $E_d$ be the span of all monomials of degree $d$. Then $E$ is the direct sum of the submodules $E_d$, making $E$ a graded module. Furthermore, this grading is compatible with the ring structure making $E$ a graded algebra, and it is in fact a commutative graded algebra since each $d_i$ is odd (though this takes a little work to check).