Suppose we have a ring $ R $ and some $ R-$module $ N. $ Let $ a \in \bigwedge^{k} N, $ and $ b \in \bigwedge^{l} N. $

Why is it that $ a \wedge b = (-1)^{kl} \; b \wedge a? $


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If $a = e_1 \wedge e_2 \wedge \ldots \wedge e_k$, and $b = f_1 \wedge f_2 \wedge \ldots \wedge f_l$, then in order to rearrange $a \wedge b$ into $b \wedge a$, I have to move each of the $l$ terms $f_1, \ldots, f_l$ "across" the $k$ terms $e_1, \ldots, e_k$. Each time I move an $f$ "across" an $e$, I introduce a minus sign. So in all there are $kl$ minus signs introduced.

By moving "across", I mean using the identity $e_i \wedge f_j$ = $-f_j \wedge e_i$.

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    $\begingroup$ I was thinking the same, but I don't know how to justify the identity $ e_{i} \wedge f_{j} = - f_{j} \wedge e_{i} $ in the first place. $\endgroup$ – Confused Student Apr 6 at 18:02
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    $\begingroup$ @AddledStudent: That's one of the axioms of the wedge product. $\endgroup$ – Henning Makholm Apr 6 at 18:11
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    $\begingroup$ (Or when it's not an axiom then it is an axiom that $v\wedge v=0$, and therefore $$0=(v+w)\wedge(v+w) = (v\wedge v)+(w\wedge w)+(v\wedge w)+(w\wedge v) = (v\wedge w)+(w\wedge v) $$ so $v\wedge w$ and $w\wedge v$ are always additive inverses.) $\endgroup$ – Henning Makholm Apr 6 at 18:15
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    $\begingroup$ @AddledStudent are you also wondering why $v\wedge v = 0$ (or equivalently $v\wedge w = -w\wedge v$) is a useful axiom, i.e. why the wedge algebra is a useful thing? $\endgroup$ – Ben Blum-Smith Apr 6 at 20:20
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    $\begingroup$ It should be mentioned that $a$ and $b$ may not have this form, but can instead be sums of expressions of this form. $\endgroup$ – Eric Wofsey Apr 6 at 21:51

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