Grassmann algebra

In studying associative algebras' theory I was introduced to the notion of Grassmann algebra, but I don't know if I well understood how to construct this algebraic structure.

Let $F$ a field and $X=\{e_1,e_2,\dots\}$ a countable set.

Def. A word on $X$ is a finite sequence of element of $X$, ie $e_{i_1}e_{i_2}\cdots e_{i_k}$ where $k\geq 1$ and $e_{i_j}\in X$ .

Def. The sequence which does not contain any element of $X$ is called empty word and denoted by $1$.

Let $X^*$ be the set of all possible words on $X$, including $1$, $$X^*=\{w\,:\, w \,\text{is a word on}\,X\}\cup\{1\}.$$

We now introduce a binary operation on $X^*$, called juxtaposition, as $$(e_{i_1}e_{i_2}\cdots e_{i_k})(e_{j_1}e_{j_2}\cdots e_{j_s}):=e_{i_1}e_{i_2}\cdots e_{i_k}e_{j_1}e_{j_2}\cdots e_{j_s}.$$ It is clear that $X^*$, equipped with this operation, is a monoid with unity $1$.

We now consider $FX^*$, the $F$-vector space whose elements are $F$-linear combination of words of $X^*$, $$FX^*=\{\sum_{w \in X^*}\alpha_{w}w\,:\,\alpha_{w}\in F\,\text{ and}\, \alpha_{w}=0 \,\text{almost everywhere}\,\}.$$

The juxtaposition of $X^*$ induces a multiplication on $FX^*$, $$\left(\sum_{w \in X^*}\alpha_{w}w\right)\left(\sum_{w' \in X^*}\beta_{w'}w'\right):=\sum_{w,w'\in X^*}(\alpha_{w}\beta_{w'})ww'.$$ It is routine to check that this defines the structure of an $F$-algebra on $FX^*$.

If moreover we require that $FX^*$ satisfies the following condition $$e_ie_j=-e_je_i \qquad \text{for all}\, i,j\geq 1$$ then the resulting algebra is called Grassmann algebra and denoted by $G$. Supposing that $\text{car}F\neq 2$, then ${e_i}^2=0$ for all $i\geq 1$ and $$G=\text{span}_{F}\{e_{i_1}\cdots e_{i_t}\,:\,1\leq i_1<i_2<\cdots<i_t, t\geq 0\},$$ where if $t=0$ we obtain the empty word $1$.

Is this construction correct?

• If your construction is for the Grassmann algebra of the vector space $F^X$ then yes that works. On the other hand, if you pick $X\subseteq V$ where $V$ is an $F$-vector space, then no it gives a much larger algebra than the Grassmann algebra of $V$. – user10354138 Sep 18 '18 at 18:39
• My doubt comes out from the fact that my teacher simply defines the Grassmann algebra as the algebra generated by the countable set $\{e_1,e_2,\dots\}$ satisfying the following condition: $e_i e_j+e_j e_i=0\,\,\text{for all}\,i,j\geq 1$. But I don't really know what "algebra generated by a set" means! – eleguitar Sep 19 '18 at 7:15
Yes: in the first part your description of $FX^\ast$ is of the free algebra generated by $X$. This would be appropriate when constructing the tensor algebra for a vector space $V$ with dimension $|X|$.
When you quotient out by the relations $e_ie_j=-e_je_i$, you arrive at the exterior algebra (or Grassman algebra) of $V$ where $\dim(V)=|X|$.
• My doubt comes out from the fact that my teacher simply defines the Grassmann algebra as the algebra generated by the countable set $\{e_1,e_2,\dots\}$ satisfying the following condition: $e_i e_j+e_j e_i=0\,\,\text{for all}\,i,j\geq 1$. But I don't really know what "algebra generated by a set" means! – eleguitar Sep 19 '18 at 7:15