Basis-free, field-independent definition of determinants? Let $T$ be a linear operator on a finite-dimensional vector space $V$ over the field $K$, with $\dim V=n$. Is there a definition of the determinant of $T$ that (1) does not make reference to a particular basis of $V$, and (2) does not require $K$ to be a particular field?

As motivation, if $K=\mathbb C$, I know of three ways to define $\det(T)$, two of which refer to a choice of basis, and the other of which relies on $\mathbb C$ being algebraically closed:

*

*Choose an ordered basis $B$ of $T$, and let $\mathcal M(T)$ denote the matrix of $T$ with respect to this basis. Then apply any of the formulas/algorithms for calculating a determinant to $\mathcal M(T).$ [This works for any field, but requires choosing a basis to express $\mathcal M(T)$.]

*Choose an ordered basis $B$ of $T$, and let $\det_n$ be an alternating multilinear map from $V^n\to K$. Then the determinant of $T$ can be defined as $\det_n(TB)/\det_n(B)$. [This works for any field, but requires choosing a basis to extract "column vectors of the matrix of $T$."]

*Define $\det(T)$ as the product of eigenvalues of $T$, repeated according to their algebraic multiplicity. [This makes no reference to a basis, but only works because $\mathbb C$ is algebraically closed.]

 A: 
This makes no reference to a basis, but only works because C is algebraically closed.

Not really. Every field has an extension that's algebraically closed, so you can do the math in the extension, and the result is guaranteed to be in the original field (this can be proven either by noting that the result is the same as the basis-dependent result, or by noting that the result must be unchanged by all of the members of the Galois group of the extension).
A: An endomorphism $T$ of a vector space $E$ yields endomorphisms on various vector spaces deduced from $E$. For instance on the dual space $E'$, the space of linear forms on $E$, the endomorphism $T$ induces the transposition $\hbox{}^tT$ of $T$ defined by $(\hbox{}^tTf)(x) = f(T(x))$.  Similar constructs are possible on cartesian powers of $E$, tensor products, symmetric and alternate powers of $E$ and there are a great many functorial constructions besides the few ones I quoted. Look up Schur's polynomial and plethysm in a representation theory book to find more of them.
When $E$ has finite dimension $n$, the $n$-th external power $\Lambda^n E$ of $E$ is a line and the induced endomorphism $\Lambda^n T$ on $\Lambda^n E$ is therefore a dilatation. The determinant of $T$ is (definition) the coefficient of this dilatation.
With a little bit of care, this definition works well when $E$ is a module (when coefficients are in a ring rather than in a field).  This is pretty much standard algebra, it's the way determinants are introduced in Algèbre générale by Denis Allouch and Bernard Charles and I am confident there are plenty of good references in other languages.
A: *

*Giving an endomorphism $T:V\to V$ of a dimensional vector space over a field $k$ is the same as making $V$ a module over the polynomial ring $k[t]$. (Edit/Clarification: A polynomial $p(t)$ acts on $V$ by $p(T):V\to V$ under this identification.)


*This is a finitely generated torsion module if (and only if) $V$ is finite dimensional.


*By the structure theorem of finitely generated modules over $k[t]$ we see that $V$ is uniquely expressible as direct sum $\oplus_i k[t]/(p_i(t))$ where $p_i(t)$ divides $p_{i+1}(t)$ and all are monic polynomials.


*The determinant of $T$ is defined as $(-1)^n\prod_i p_i(0)$ where $n=\dim(V)$.
The proof that this is the "usual" determinant can be given by applying @Ruy's argument to the matrix $t\cdot 1-A$ and using the elementary operations (over $k[t]$) to reduce this to the Smith Normal Form.
A: See my question here: Does any normalized function $D$ other than determinant of matrix satisfy $D(AB) = D(A)D(B)$?
I think it might be the case that the determinant is the unique function satisfying the following. Suppose $V$ is an $n$-dimensional vector field over $\mathbb{K}$ and $M_{n\times n}(\mathbb{K}$) is the set of $n\times n$ matrices over $\mathbb{K}$. Let $D:M_{n\times n}(\mathbb{K}) \to \mathbb{K}$. Let $k\in \mathbb{K}$.
(1) $D(I)=1$
(2) $D(AB) = D(A)D(B)$
(3) $D(kA) = k^n D$
(4) $D$ is continuous
Unfortunately, similar to the answer by Ruy, I think a basis-dependent version of the determinant is required to prove existence of such a function.
