# When does $v_0\wedge\dots\wedge v_{k-1}=0$ when working over a ring that's not a field?

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 $$0$$?

• Uh $\Bbb{Z}/2\Bbb{Z}$ is itself a field, and $1$ isn't linearly dependent by itself. – jgon Jun 8 at 14:39
• Ah, I was thinking of it as a $\mathbb Z$-module. I'll edit the question to make it more clear. – Oscar Cunningham Jun 8 at 14:41
• 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 $i$. – jgon Jun 8 at 15:46
• 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. – jgon Jun 8 at 15:47
• @jgon I was thinking the condition might be "$\sum a_iv_i=0$ implies $\langle a_1,\dots,a_n\rangle\neq R$". – Oscar Cunningham Jun 8 at 20:34