# Let's fix the notation, V = <munder> &#x2A01;<!-- ⨁ --> i &#x2265;<!-- ≥

Let's fix the notation, $V=\underset{i\ge 0}{⨁}{V}^{i}$ is a graded vector space and $\mathrm{\Lambda }V$ is the free commutative graded algebra on $V$. I have been struggling to understand this example:
Consider a graded vector space $V$ with basis $\left\{a,b\right\}$ such that $a\in {V}^{2}$ and $b\in {V}^{5}$. Now define a linear map $d$ (of degree 1) by $da=0$ and $db={a}^{3}$. It follows that d extends uniquely to a derivation .
The point of the example is to show that the derivation on $\mathrm{\Lambda }V$ is completely determined by its values on $V$. So if i understand well, he considers a linear map $d:V⟶\mathrm{\Lambda }V$ of degree one defined by
(here ${\mathrm{\Lambda }}^{k}V$ is the set of elements of word length $k$) and
${d}_{5}:{V}^{5}⟶{\mathrm{\Lambda }}^{6}V;b↦{a}^{3}$
The first question that i'm stuck on is for ${d}_{2}\left(b\right)={a}^{3}$, i mean ${a}^{3}$ is of length $3$, how it can be in ${\mathrm{\Lambda }}^{6}V$.
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percolarse2rzd
${\mathrm{\Lambda }}^{k}V$ is not necessarily the set of elements of word length $k$. Rather than just word length, you need to consider the grading as well. Because a lies in ${\mathrm{\Lambda }}^{2}V$, ${a}^{3}$ lies in ${\mathrm{\Lambda }}^{6}V.$.