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4

Hint: let $f\colon R\times R\to R$ be a bilinear map. Then, for $(x,y)\in R\times R$ you have $$f(x,y)=xyf(1,1)$$

4

For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$. For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.) ...

2

There is also a nice description in terms of Schur functor and Young diagrams. A Young diagram $\lambda$ is a picture made by a finite set of cells, left-aligned in rows, such that the length of the rows decreases going down. Any Young diagram can be transposed (exchanging rows and columns) to obtain another Young diagram $\lambda'$. To any Young diagram ...

1

In terms of the multilinear algebra, it looks like you're on the right track. We extend $\omega$ to a basis $\omega, v_2, \dots, v_n$ and then write $$\alpha = \sum a_I \omega \wedge v_{i_2} \wedge \dots \wedge v_{i_k} +$$ $$\sum b_J v_{j_1} \wedge \dots \wedge v_{j_k}$$ Then as you've noted, wedging with $\omega$ kills everything in the first term. ...

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