Let $V$ be a one dimensional graded $\mathbb Q$-vector space; $V=\bigoplus_{i\geq 0}V_i$ and all $V_i$ are zero except $V_{2n+1}$ for some given $n$. Let $v$ be a generator of $V_{2n+1}$. Now take the free commutative graded algebra $\Lambda V$ on $V$. I want to understand the grading on $\Lambda V$. I know that $\Lambda V$ is graded as $\Lambda V=\bigoplus_{i\geq 0}{\Lambda^iV}$ where each $\Lambda^iV$ has a basis consisting of all possible products of $i$ elements from the basis of $V$. In our case the only basis element is $v$ and $v^2=0$ so the only product is $v$ of lenght 1, hence $\Lambda V=\Lambda^0V\oplus \Lambda^1V$ where $\Lambda^0V$ is isomorphic to $\mathbb Q$ with basis $1$ and $\Lambda^1V$ is isomorphic to $V$ with basis $v$. Is this correct? is it possible to have the following grading : $\Lambda V=\Lambda^0V\oplus \Lambda^{2n+1}V$?

I think it is not possible because we can not have an element of length $2n+1$ out of $v$ since $v^2=0$. Thank you for your help!!


The decomposition of $\Lambda V$ you give is useful when considering the length, as you refer to it, or number of factors in a wedge product $x = v_1 \wedge \ldots \wedge v_k \in \Lambda V$.

To figure out the grading, consider that the free commutative graded algebra usually has a universal property, namely being universal for maps from a graded vector space into a graded commutative, graded $\mathbb{Q}$-algebra: Given such a map of graded modules $f:V \to U(E)$, where $U$ is the forgetful functor from graded algebras to graded vector spaces, there should be a unique map of graded algebras $\hat f: \Lambda V \to E$ extending $f$.

In particular, $V$ can be taken to be contained in $\Lambda V$ and the inclusion should respect the grading, hence, the grading you propose yourself with $\mathbb{Q}$ in degree $0$ and $\Lambda^1 V = V$ (!) in degree $2n+1$ is the way to go here.

In general, elements of the form $v \in V$ keep their grading, and products or elements of degree higher than one add their gradings: $$ \operatorname{deg} ( v_1 \wedge \ldots \wedge v_k) = \sum_{i=1}^k \operatorname{deg}(v_i). $$

The confusion probably stems from the fact that if $V$ is a vector space interpreted as a graded vector space sitting in degree $1$, then in $\Lambda V$ the notion of length and degree conincide.


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