TITLE: Exterior Algebra as quotient
QUESTION [8 upvotes]: Given a vector space $W$, I understand what the tensor algebra $T(W)$ is, and I understand that the exterior algebra $\bigwedge W$ is defined as $\bigwedge W := T(W)/N$ where $N$ is the two-sided algebra ideal generated by $\{v \otimes w + w \otimes v | v,w \in W\}$. However, some texts quotient out by the ideal generated by $\{w \otimes w| w \in W\}$ instead of $\{v \otimes w + w \otimes v | v,w \in W\}$. Can someone explain why taking a quotient of either of these things yields the same algebra?

REPLY [9 votes]: This is a variant of the polarization trick:
$$
v \otimes w + w \otimes v = (v + w) \otimes (v + w) - v \otimes v - w \otimes w
$$
Let $I$ be your ideal generated by all $v \otimes w + w \otimes v$ and let $J$ be the ideal generated by all $w \otimes w$.
The polarization identity shows that $I$ is contained in $J$.  Of course, you can go the other way around to get $J \subseteq I$ by setting $v = w$ in your ideal if you are able to divide by $2$.  The ideals are not the same if the scalar field for your vector space $W$ has characteristic $2$.  People who deal with characteristic $2$ usually use the ideal $J$ because of this.

REPLY [4 votes]: Not sure if this is a rigorous explanation but an intuitive way to think of this: The idea of quotiening out by the ideal $\{ v \otimes w+ w \otimes v | v,w \in W\}$ is that we impose the relation
$$v \wedge w = - w\wedge v$$
in the quotient where $v \wedge w$ is the image of $v \otimes w$ under the canonical map. Now what happens when we quotient out by $\{w \otimes w| w\in W\}$, do we impose the same relation? Yes we do because
$$\begin{eqnarray*} 0 &=& (v +w) \wedge (w +v)\\
&=&(v \wedge w ) + (w \wedge v).\end{eqnarray*}$$
This should give you an idea why the quotients are isomorphic.