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Take $V = K^{n}$. Let $\omega$ be a non-zero element of $\bigoplus_{k=1}^n \bigwedge^k V$, where we have excluded the summand $\bigwedge^0 V = K$. (1) Prove that there exists an $m > 1$ for which $\omega^m$, where the power is taken in the Grassmann algebra, is zero, and $\omega^{m-1}$ is not yet zero. (2) Find such an $\omega$ for which m of the preceding part is maximal. (Hint: pure non-zero tensors will have $m = 2$, which is certainly not maximal.)

I've figured out that Grassmann algebra is the same as exterior algebra, and that $m-1$ is probably the rank of $\omega$, and therefor in 2) we look for an $\omega$ of maximal rank. But I don't know how to prove it or how to find such a $\omega$.

Any help would be greatly appreciated.

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Is $K$ an arbitrary field of characteristic $0$, or should we worry about positive characteristic subtleties? –  Matt Nov 25 '12 at 16:49
    
I think we can assume characteristic $0$. –  Christine Nov 25 '12 at 17:23
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(1) It is enough to show that there exists $m>1$ such that $\omega^m=0$. (The next step is to choose $m$ minimal with this property.) Or this is clear: the exterior algebra is a graded algebra whose homogeneous components of degree greater than $n$ are $0$. –  user26857 Nov 25 '12 at 19:36
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