Why is $dx_1\wedge dx_2(e_1,e_2)=1$?

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This question is about definitions and how to thinks about the exterior algebra and how it relates to the tensor algebra.

Let $V$ be a real vector space, $e_1,\dots,e_n$ a basis of $V$, and $dx_1,\dots,dx_n$ the dual basis.

There are three ways to define $dx_1\wedge dx_2$ in terms of the tensor algebra:

$(A)\quad dx_1\otimes dx_2 - dx_2\otimes dx_1$

$(B)\quad\frac{1}{2}(dx_1\otimes dx_2 - dx_2\otimes dx_1)$

$(C)\quad dx_1\otimes dx_2 \operatorname{mod} I$ (where $I$ is the ideal generated by elements of the form $\alpha\otimes\alpha$)

There are four ways to understand the argument $(e_1,e_2)$:

$(a)\quad e_1\otimes e_2 - e_2\otimes e_1$

$(b)\quad \frac{1}{2}(e_1\otimes e_2 - e_2\otimes e_1)$

$(c)\quad e_1\otimes e_2 \operatorname{mod} I$

$(d)\quad e_1\otimes e_2$

When we make choices, what does $dx_1\wedge dx_2(e_1,e_2)$ evaluate to? $$ Aa=2, Ab=1, Ac=1, Ad=1$$ $$ Ba=1, Bb=\frac{1}{2}, Bc=\frac{1}{2}, Bd=\frac{1}{2} $$ $$ Ca=1, Cb=\frac{1}{2}, Cc=\operatorname{undefined}, Cd=\operatorname{undefined}$$

So we are left with the options $Ab$, $Ac$, $Ad$, $Ba$ and $Ca$.

What should I choose / How should I think about these issues?