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Let $M$ be a module over a commutative ring $R$, and let $v_0,\dots,v_{k-1}$ be elements of $M$. If $R$ is a field then $v_0\wedge\dots\wedge v_{k-1}$ is equal to $0$ if and only if $v_0,\dots,v_{k-1}$ are linearly dependent. But if $R$ isn't a field then this needn't be true.

For example if we view $\frac{\mathbb{Z}}{2\mathbb{Z}}$ as a $\mathbb Z$-module then $1\in\frac{\mathbb{Z}}{2\mathbb{Z}}$ is linearly dependent on its own, because $2.1=0$, but it doesn't get sent to $0$ in $\Lambda^1\left(\frac{\mathbb{Z}}{2\mathbb{Z}}\right)$.

Is there a nice characterisation of which lists of vectors do get sent to $0$?

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  • $\begingroup$ Uh $\Bbb{Z}/2\Bbb{Z}$ is itself a field, and $1$ isn't linearly dependent by itself. $\endgroup$ – jgon Jun 8 at 14:39
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    $\begingroup$ Ah, I was thinking of it as a $\mathbb Z$-module. I'll edit the question to make it more clear. $\endgroup$ – Oscar Cunningham Jun 8 at 14:41
  • $\begingroup$ This is not easy. Consider $M=\Bbb{Z}/(2)\oplus \Bbb{Z}/(3)$ as a $\Bbb{Z}$ module. Then $M\otimes M\simeq M$, and by definition of the exterior algebra $\Lambda^2M=0$. For a $\Bbb{Z}$-module this suggests you should split the module into its primary components. Also a side comment, it's not clear what linear independence should mean for modules. You appear to have abstracted the statement that $\sum a_i v_i =0$ implies all $a_i=0$. But you could equally well use the statement $v_i\not \in \langle v_1,\ldots,\hat{v}_i,\ldots,v_n\rangle$ for all $i$. $\endgroup$ – jgon Jun 8 at 15:46
  • $\begingroup$ This has the benefit of making wedges of dependent vectors $0$, but vectors can still be independent by this definition and yet have wedge product $0$, as $M$ demonstrates. $\endgroup$ – jgon Jun 8 at 15:47
  • $\begingroup$ @jgon I was thinking the condition might be "$\sum a_iv_i=0$ implies $\langle a_1,\dots,a_n\rangle\neq R$". $\endgroup$ – Oscar Cunningham Jun 8 at 20:34

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