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!!