What are the necessary and sufficient conditions for a two-form to be an exterior product of two one-forms? Consider a two-form $\gamma \in \Lambda^2(V)$ where $V$ is a real vector space. Now I would like to know the necessary and sufficient conditions for $\gamma$ to be expressible as an exterior product of two one-forms, $\gamma=\alpha \wedge \beta,\, \alpha,\beta \in \Lambda(V)$.
Obviously, a necessary condition for the decomposition to hold is that any exterior power of $\gamma$ vanishes, $\gamma\wedge...\wedge\gamma = 0$. But it is not clear to me whether this condition is sufficient.
 A: Let $e_1$, ..., $e_n$ be a basis of $V$. Let $\gamma = \sum c_{ij} e_i \otimes e_j$, so $C = (c_{ij})$ is a skew symmetric matrix. Then $\gamma$ factors as $\alpha \wedge \beta$ if and only if $C$ has rank $\leq 2$. 
Proof: Let $C$ be a skew symmetric matrix, we must show that $C$ can be written as $a b^T - b a^T$ for vectors $a$ and $b$ if and only if $\mathrm{rank}(C) \leq 2$. The condition is clearly necessary, since $\mathrm{rank}(a b^T) = \mathrm{rank}(b a^T) \leq 1$.
Choose a $2$-dimensional subspace $\mathrm{Span}(v_1, v_2)$ containing the image of $C$. Since $C^T = -C$, this space also contains the image of $C^T$. Replacing $C$ by $SCS^T$, we may assume that $v_1$ and $v_2$ are the first two basis vectors.
Then the conditions on the images of $C$ and $C^T$ say that $C$ is $0$ outside the intersection of the first two rows and the first two columns, and skew symmetry further says that the upper left $2 \times 2$ block is of the form $\left( \begin{smallmatrix} 0&c \\ -c&0 \end{smallmatrix} \right)$. For such matrices, the claim is obvious. $\square$
The OP asks whether it is enough to ask that $\gamma \wedge \gamma$ vanish (well, the OP asks about all higher wedge powers vanishing as well, but that clearly follows from $\gamma \wedge \gamma=0$.) The answer is yes. Any $2$-form can be written as $u_1 \wedge v_1 + u_2 \wedge v_2 + \cdots + u_r \wedge v_{r}$ where $2r$ is the rank of the matrix $C$ and $u_1$, ..., $u_r$, $v_1$, ..., $v_{r}$ are linearly independent; this is more or less the classification of skew symmetric bilinear forms. Then 
$$\gamma^{\wedge k} = k! \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq r} u_{i_1} \wedge v_{i_1} \wedge u_{i_2} \wedge v_{i_2} \wedge \cdots \wedge u_{i_k} \wedge v_{i_k}.$$
This is clearly nonzero if and only if $k \leq r$. In particular, $r \leq 1$ if and only  if $\gamma \wedge \gamma=0$.
The condition that $\gamma \wedge \gamma =0$, expanded in coordinates, states that the $4 \times 4$ principal Pfaffians of $C$ vanish. It is a nice high school algebra challenge to show that this implies the $3 \times 3$ minors of $C$ vanish, which is the more standard algebraic test for a matrix to have rank $\leq 2$.
A: While Plücker relations give the general theory, a direct answer to your question is the following: a 2-form $\gamma$ is decomposable (is a product of two 1-forms) iff $(\iota_v \gamma) \wedge \gamma = 0$ for any vector $v$, where $\iota_v \gamma$ is the usual contraction of a vector with a 2-form, that is, $(\iota_v \gamma)(u) = \gamma(v,u)$ for any other vector $u$.
The proof is straightforward. Pick some vector $v$ such that $\alpha = \iota_v \gamma \ne 0$. If it doesn't exist, then $\gamma = 0$, which is the trivial case. Then $\alpha \wedge \gamma = 0$ implies* that there exists $\beta$ such that $\gamma = \alpha \wedge \beta$ and you are done.
* If the implication is not obvious, extend $e_1 = \alpha$ to a basis $e_i$ in $\Lambda(V)$ and expand $\gamma$ in the induced basis $e_i\wedge e_j$. Clearly, $e_1\wedge \gamma = 0$ iff the only non-zero coefficients in the expansion are in front of the terms $e_1 \wedge e_j$.
