# Characterize the determinant function on a commutative ring with identity.

Given a commutative ring $$A$$ with identity, the determinant function $$\det_n: \mathcal{M}_n(A)\to A$$ can be characterized as the unique $$n$$-linear, alternating function such that $$\det(\textbf{I}_n)=1_A$$. This characterization, however, involves properties relating to specific rows or columns of a matrix. For example, the property of alternating means that changing two different rows of an $$n\times n$$ matrix $$M$$ reverses the sign of $$\det_n(M)$$. If we view $$\mathcal{M}_n(A)\simeq \text{End}_A(A^n)$$, I wonder if we can give a more “global” characterization for $$\det$$ that does not involve specific elements of the ring or values of a function.

Moreover, determinants can be viewed as a natural transformation between two functors from $$\textbf{CRing}$$ to $$\textbf{Grp}$$. Can we them give another characterization for $$\det$$ in this case? What are the connections among all these characterizations?

There is a very elegant description of determinants using exterior powers. Namely, if $$V$$ is a free module of rank $$n$$ and $$T:V\to V$$ is an endomorphism, then it induces an endomorphism $$\bigwedge^n T:\bigwedge^n V\to \bigwedge^n V$$. But $$\bigwedge^n V$$ is a free module of rank $$1$$, so any endomorphism of it is multiplication by some scalar. That scalar is the determinant of $$T$$.
(This definition makes many of the basic properties of the determinant trivial; for instance, its basis-invariance, and $$\det(AB)=\det(A)\det(B)$$. However, you don't actually get to escape doing the hard work of constructing the determinant with its basic properties: the fact that $$\bigwedge^n V$$ is free of rank $$1$$ is nontrivial and to prove it you basically need to construct an $$n$$-linear alternating function which sends the identity matrix to $$1$$.)