A polynomial in multiple variables with nice properties After answering another question (The number of values of $f(x)/x$ when $f$ is a linearized polynomial), I stumbled upon an interesting polynomial in multiple variables. Let $\mathbb{F}_q$ be the field of $q$ elements, and let $K$ be a field containing it. Then define 
\begin{equation}
L(X_1, X_2, ...) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n \text{Frob}^{i-1}(X_{\sigma(i)}).
\end{equation}
This polynomial is $\mathbb{F}_q$-multilinear, and detects whether the elements are linearly dependent over $\mathbb{F}_q$. In other words, it is nonzero exactly when they are linearly independent. I think this can be extended to the case where $K$ is an algebra, but haven't checked. 
This seems interesting enough that I am guessing it's been found and used before; does this polynomial have a name, and is it used anywhere interesting?
 A: This is a CW answer to remove this question from the unanswered list (once someone upvotes it). This is the determinant of the Moore matrix $\left( x_i^{q^{j-1}} \right)_{1 \leq i,j \leq n}$. This determinant can be expressed as a product of linear factors:
$$\det \left(x_i^{q^{j-1}} \right) = \prod_{(c_1:c_2:\cdots:c_n) \in \mathbb{F}_q \mathbb{P}^{n-1}} (c_1 x_1 + \cdots + c_n x_n),$$
up to a scalar factor depending on how we choose the representatives $(c_1, \ldots, c_n)\in \mathbb{F}_q^n$ of the points of $\mathbb{F}_q \mathbb{P}^{n-1}$.
A: That is just a generalization of the determinant. Let $n$ be the degree of $K$ over $\mathbb{F}_q$ and $F := Frob$. Then the powers $(F^1,F^2,\ldots, F^n)$ are linearly independent as elements of $End_{\mathbb{F}_q}(K)$. In fact one can even show that 
$$End_{\mathbb{F}_q}(K) \cong \bigoplus_{i=1}^n F^iK$$
(see for reference the proof of theorem 29.12 in Reiner: Maximal Orders).
On the other hand, $K$ is a vector space over $\mathbb{F}_q$ of dimension $n$ and thus
$$End_{\mathbb{F}_q}(K) \cong \mathbb{F}_q^{n \times n}.$$
Understanding and working out these isomorphisms, you should come to the conclusion that your formula is nothing but the determinant on this matrix algebra. Thus your $n$ vectors are linearly indipendant if and only this determinant is non-zero.
This can of course be extended to other algebras (that are, among other things, also vector spaces), but as it is just a nice way to write the determinant, this should not be surprising.
