How do we write $m_1 \otimes m_2 \otimes m_1$ as a linear combination of elements of the form $m \otimes m$. I am trying to see that an element in the submodule of relations defining the 
$3^{rd}$ exterior power is contained in the ideal of relations defining the exterior algebra.
 A: Let $a$, $b$ be terms in any algebra1. Then
$$ba+ab=(a+b)^2-a^2-b^2$$
$$\begin{split}
a(ba) &=a(ba+ab) -a(ab)\\
&= a(a+b)^2 -a(a^2)-ab^2-a^2b +(a^2b-a(ab))\text{.}
\end{split}$$
Consequently, in any alternative algebra2
$$
aba = a(a+b)^2 -a^3-ab^2-a^2b\text{.}
$$



*

*algebra here means abelian group $A$ with bilinear map $A\otimes A \to A$

*an algebra is alternative if any terms $x$, $y$ satisfy $x(xy)=x^2y$, $(yx)x=yx^2$, and $(xy)x=x(yx)$.

A: I just learned this. This can be found in Dummit & Foote chapter 11, section 5.

Look at $(m+n) \otimes (m+n)$. Expanding gives us:
$$(m+n) \otimes (m+n) = m\otimes m + m\otimes n + n\otimes m + n\otimes n.$$
So, $$m\otimes n =-n\otimes m + [(m+n) \otimes (m+n) - m\otimes m - n\otimes n].$$
Therefore,
\begin{align*}m_1\otimes m_2 \otimes m_1 &= \big(-m_2 \otimes m_1 + [(m_1+m_2) \otimes (m_1+m_2) - m_1\otimes m_1 - m_2\otimes m_2]\big) \otimes m_1\\
&=-m_2\otimes m_1 \otimes m_1+(m_1+m_2) \otimes (m_1+m_2) \otimes m_1 - m_1\otimes m_1 \otimes m_1 - m_2\otimes m_2 \otimes m_1.
\end{align*}
