Reading about the fundamental theorem of alternating applications which says
Given 2 vector spaces over $K$, $(V;K)$ and $(W;K)$. If $dim\ \ V=n$ and a base of V is {$u_1...u_n$}
I saw that there is a special case when you have the destination vector space as a field, implying the expression:
$$\forall f \in A_n(V^{n}; K), f(x_1...x_n)=(\sum_{i=0}\sigma(p)a_{1p(1)}...a_{np(n)})f(u_1...u_n)=\color{red}{ f(x_1...x_n)=d(x_1...x_n) \cdot f(u_1...u_n)}$$
Being $$d(x_1...x_n)=\sum_{p \in S_n}\sigma(p)a_{1p(1)}...a_{np(n)}$$ Which as I understand comes from:
$\forall f \in A_n(V^{n};K), \exists \lambda \in K|f=\lambda \cdot d$, being $d \in A_n(V^{n};K) / d(u_1...u_n)=1 $ then the vector $d\in A_n(V^{n};K)$ engenders all space.
I understand that all of the above means that, if I have 2 vector spaces over $K$, $(V;K)$ and $(W;K)$, and a basis of the first {$u_1...u_n$}, there will be a function $f$ such that, when applied to that base, will result in some element $w$ of the second vector space $(W;K)$ but, in this particular case, the result will correspond to an element of the field itself in which both spaces $(V;K)$ and $(W;K)$ are included. Said element will be the unit element of the field and, hence the product of a scalar λ by the vector d, generated by an alternating n-linear aplication on the base {$u_1...u_n$}, is considered as “a space generated” (I suppose from the base by the action of d).
My doubts are:
a) does this represent that if vectors are passed, numbers are returned, since it is of dimension 1? If a practical example were provided it would be great to get the intuitive idea.
b) is d a vector or an aplication that acts on a vector?
I think the second question is absorbed into the first, basically it confuses me that they first present d as an alternate application and then say that it is a vector, which is why I ask for clarification.
Welcome to the site.
What's important to realise here is that $A_n(V^n;K)$ is itself a vector space (over $K$)! Alternating applications are vectors ... recall "vector" just means "an element of some (contextual) vector space". If $f,g$ are two alternating $K$-linear forms then so is $f+g$, and so is $(x_1,\cdots,x_n)\mapsto\lambda f(x_1,\cdots,x_n)$ where $\lambda\in K$ and it's pretty easy to check the axioms of a vector space are met. This could be seen as the "natural" or "canonical" vector space structure on $A_n(V^n;K)$ and it's very common too; if you have a bunch of functions, then adding and multiplying them somehow usually gives an appropriate kind of algebra structure.
What does it mean for $A_n(V^n;K)$ to be a one-dimensional vector space? It means it has a basis of one (nonzero) element, that is, there exists some $d\in A_n(V^n;K)$ such that if $f\in A_n(V^n;K)$ then $f=\lambda d$ for some unique $\lambda\in K$. So in a sense, you can view $A_n(V^n;K)$ as "just numbers", in fact any one dimensional vector space over $K$ is isomorphic with $K$ (the field of scalar "numbers"), by associating $f$ with $\lambda$. But technically speaking, "$f=\lambda d$ for some unique $\lambda$" is the correct statement.
In particular, the determinant form $d=\det$ is the canonical basis element.