Completely confused about extremely basic differential forms

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I am having trouble understanding what should be extremely simple. I am comfortable with the more algebraic definition of a differential form as in Bott and Tu as the tensor product of the algebra of smooth functions on $U \subseteq \mathbb{R}^{n}$ and the anti-symmetric tensor algebra as in, \begin{equation} \label{algebraicdefinition} \mathbb{C}^{\infty} \left( U \right) \otimes \bigwedge E \tag{1} \end{equation} where $E$ has a basis $dx_{i}$. My confusion is in relating this to the definition in terms of a map from $U$ to the algebra of alternating functions. Let's specialize to the case of a $2$-form for now. I am given the definition of a differential form $\omega$ as a smooth map $$ \omega: U \subseteq \mathbb{R}^{n} \to Alt^{2}\left( \mathbb{R}^{n} \right) $$ Now in terms of $\eqref{algebraicdefinition}$, this might look like $$ \left( x^{2} + y^{2} + 3xyz \right)dx \wedge dy $$ for example. So say I take the case $n=3$ and consider $\left( a, b, c \right) \in U$ and so evaluating $\omega$ there gives, say $$ \omega_{\left(a, b, c \right)} = \left(a^{2} + b^{2} + 3abc \right)dx \wedge dy.$$ My question is, how is this an alternating function on $\mathbb{R}^{2}$? If took a concrete example and took two elements of $\mathbb{R}^{2}$, say, $u = (1,2)$ and $v=(3,4)$, how would I actually calculate the value of $\omega_{\left(a, b, c \right)}\left( u, v \right)$?

I realize this is an extremely simple question, but I am finding the literature on differential forms very unhelpful. I'm still not entirely sure what "$dx$" actually is as a mathematical object.

Any help would be appreciated.

Thanks

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In the example you give: $dx\wedge dy (u,v)= dx(u)dy(v)-dx(v)dy(u)=1\cdot 4-2\times 3$. The prefactor (with a,b,c) is just a prefactor.

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This may not be the best answer, but you said any help...So $dx$ is the dual to $\frac{\partial}{\partial x}$. Thus

$$dx(\frac{\partial}{\partial x})=1$$ $$dx(\frac{\partial}{\partial y})=0$$

Then $dx\wedge dy$ is defined as in multilinear algebra via a matrix so you should get

$$ dx\wedge dy (\frac{\partial}{\partial x},\frac{\partial}{\partial y})=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1$$

$$ dy\wedge dx (\frac{\partial}{\partial x},\frac{\partial}{\partial y})=\begin{vmatrix}0&1\\1&0 \end{vmatrix}=-1$$