How do we write $m_1 \otimes m_2 \otimes m_1$ as a linear combination of elements of the form $m \otimes m$.

Question

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.

Answer 1

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

$$ aba = a(a+b)^2 -a^3-ab^2-a^2b\text{.} $$

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1. algebra here means abelian group $A$ with bilinear map $A\otimes A \to A$
2. an algebra is alternative if any terms $x$, $y$ satisfy $x(xy)=x^2y$, $(yx)x=yx^2$, and $(xy)x=x(yx)$.

Answer 2

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

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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*}