$a = e_1\wedge e_2 + e_3\wedge e_4$ is not decomposable I'm reading about exterior algebra and wikipedia gives this example of a non decomposable 2 vector:
$a = e_1 \wedge e_2 + e_3\wedge e_4$
But it does not tell the reason. I know that decomposable means that it cannot be written as a product, but why?
Wikipedia: https://en.m.wikipedia.org/wiki/Exterior_algebra
 A: The answer follows in the sentence after what you are looking at:

(This is a symplectic form, since α ∧ α ≠ 0.)

Meaning that if $a$ could be written as a single wedge product of $2$ vectors, then its wedge-square would be zero.
But $a\wedge a=(e_1 \wedge e_2 + e_3\wedge e_4)\wedge(e_1 \wedge e_2 + e_3\wedge e_4)=\\e_1\wedge e_2\wedge e_3\wedge e_4+e_3\wedge e_4\wedge e_1\wedge e_2=2(e_1\wedge e_2\wedge e_3\wedge e_4)\neq 0$
A: I will give here a "direct proof", which does not rely on the "trick" of showing that $a \wedge a \neq 0$, where $a = e_1 \wedge e_2 + 
e_3\wedge e_4 $.
Assume by contradiction that $a = e_1 \wedge e_2 + e_3\wedge e_4 =v \wedge w$, for some $v,w \in V$. Then, we have
$$ v \wedge w \wedge e_3= e_1 \wedge e_2 \wedge e_3$$
$$ v \wedge w \wedge e_4= e_1 \wedge e_2 \wedge e_4$$
$$ v \wedge w \wedge e_1= e_3 \wedge e_4 \wedge e_1$$
$$ v \wedge w \wedge e_2= e_3 \wedge e_4 \wedge e_2$$
which imply $v \in \text{span}(e_1,e_2,e_3) \cap \text{span}(e_1,e_2,e_4)\cap\text{span}(e_3,e_4,e_1)\cap\text{span}(e_3,e_4,e_2)=\{0\}$.
Since $a\neq 0$, we have reached a contradiction.
