Derivation of the determinant as a natural transformation on commutative rings Let $K$ be a commutative ring and $K^*$ the group of invertible elements of K. 
The discussion here suggests that for $K = \mathbb{R}$, any natural transformation from $\mathrm{GL}_n(\mathbb{R})$ to $\mathbb{R}^\times$ would be $\det$ up to composition with an automorphism of $\mathbb{R^\times}$.
By trying to construct a natural transformation $\mathrm{GL}_n(K) \rightarrow K^*$, is it possible to canonically arrive at an explicit formula for the determinant?
By canonical, I mean not making a priori choices of other mathematical structures in the construction, like for instance an exterior algebra.
 A: Recall that the group scheme $GL_n(-)$ can be described abstractly as the group-valued functor $R \mapsto GL_n(R)$ sending a commutative ring $R$ to the group of invertible $n \times n$ matrices over $R$. This functor happens to be affine, meaning that $GL_n(-)$ is actually represented by maps out of a commutative ring (which inherits the structure of a Hopf algebra). Explicitly, this commutative ring is given by
$$\mathcal{O}_{GL_n} \cong \mathbb{Z}[x_{ij} : 1 \le i, j \le n][\det(X)^{-1}]$$
where $X$ is the square matrix with coefficients $x_{ij}$; the easiest proof of this uses the fact that a matrix is invertible iff its determinant is invertible, but the point is that by the Yoneda lemma, once you have any way of showing that $GL_n(-)$ is affine, it is represented by a commutative ring which is unique up to unique isomorphism, so the determinant is there whether you know it's there or not. 
The easiest case of this to see is the case $n = 1$: here $GL_1(-) \cong \mathbb{G}_m(-)$ is called the multiplicative group scheme, it sends a ring $R$ to its group $R^{\times}$ of units, and it's clearly represented by the commutative ring 
$$\mathcal{O}_{\mathbb{G}_m} \cong \mathbb{Z}[x, x^{-1}]$$ 
with comultiplication $\Delta x = x \otimes x$. 
Now, a natural transformation $GL_n(-) \to GL_1(-)$ of group schemes is the same thing, by the Yoneda lemma, as a morphism of Hopf algebras
$$f : \mathcal{O}_{GL_1} \cong \mathbb{Z}[x, x^{-1}] \to \mathcal{O}_{GL_n}.$$
In particular such a morphism has the property that $f(x)$ is invertible; in addition $f(x)$ must satisfy $\Delta f(x) = f(x) \otimes f(x)$ (it must be what is called a grouplike element); this is necessary and sufficient. Now, it's not hard to check that the only invertible elements of $\mathcal{O}_{GL_n}$ are of the form $\pm \det(X)^n, n \in \mathbb{Z}$, for example via a degree argument. Moreover, the comultiplication takes the form
$$\Delta (\pm \det(X)^n) = \pm \det(X)^n \otimes \det(X)^n$$
(this follows from the multiplicativity of the determinant; I recognize that I am assuming facts you might not want to assume but I'll arrive at an alternate definition of the determinant in the end), and so in fact the grouplike elements are precisely those of the form $\det(X)^n, n \in \mathbb{Z}$. The conclusion is:

Proposition: Every morphism $GL_n \to GL_1$ of group schemes is an integer power of the determinant.

Because $\mathbb{G}_m$ is commutative, the set of such morphisms is an abelian group, and so we can now define the determinant (up to a mild ambiguity) as one of the two possible generators of this group; the determinant is distinguished by the fact that it is also well-defined on not-necessarily-invertible matrices, which corresponds to the subalgebra $\mathcal{O}_{M_n} \cong \mathbb{Z}[x_{ij}]$ of $\mathcal{O}_{GL_n}$. 
There's an analogous but harder version of this discussion where we use $\mathcal{O}_{M_n}$ and $\mathcal{O}_{M_1}$ instead; the conclusion will be that every morphism $M_n(-) \to M_1(-)$ of monoid schemes is a nonnegative integer power of the determinant, or equivalently that the grouplike elements of $\mathbb{Z}[x_{ij}]$ are nonnegative integer powers of $\det(X)$, so now we can define the determinant, with no ambiguity, as the unique generator of this monoid. 
