Let $M, N$ be modules over a commutative ring $R$ and let $\phi\colon M\to N$ a surjective $R$-linear map. Let $(x_1,...,x_n) = (x'_1,...,x'_n) \in N^n$ and let $y_i,y'_i$, for $i = 1,...,n$, be some elements of $M$ such that $\phi(y_i) = x_i$ an $\phi(y'_i) = x'_i$. Is it possible to write the difference $y_1\wedge ... \wedge y_n - y'_1\wedge ... \wedge y'_n$ as the sum of $n$-fold exterior products such that every summand will contain $y_i - y'_i$ for some $i$?
2026-03-25 17:20:58.1774459258
Difference of exterior products in an exterior algebra
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The credit is due to Ted Shifrin. We prove that $y_1\wedge y_2 \wedge ... \wedge y_n - y'_1\wedge y'_2\wedge ... \wedge y'_n = (y_1 - y'_1)\wedge y_2 \wedge ... \wedge y_n + y'_1\wedge (y_2 - y'_2) \wedge ... \wedge y_n + ... + y'_1 \wedge y'_2 \wedge ... \wedge (y_n - y'_n)$ by induction on $n$. The case $n = 1$ is trivial. We work in the exterior algebra $\bigwedge M$ with its multiplication $\cdot$. Assume that it is the case for $n \geq 1$. Then $$y_1\wedge y_2 \wedge ... \wedge y_n \wedge y_{n+1} - y'_1\wedge y'_2 \wedge ... \wedge y'_n \wedge y'_{n+1}$$ $$ = y_1\cdot (y_2 \wedge ... \wedge y_n \wedge y_{n+1}) - y'_1\cdot (y'_2 \wedge ... \wedge y'_n \wedge y'_{n+1})$$ $$ = (y_1 - y'_1)\wedge y_2 \wedge ... \wedge y_n \wedge y_{n+1} + y'_1\cdot(y_2 \wedge ... \wedge y_n \wedge y_{n+1} - y'_2 \wedge ... \wedge y'_n \wedge y'_{n+1}).$$ The rest follows by the induction hypotheses and the distributivity of $\cdot$ in $\bigwedge M$.