Characteristic polynomial of the unique automorphism of the zero module

Question

Is there any convention which makes sense of the characteristic polynomial of the unique automorphism of the zero module?

This might seem like an odd question but it matters to me. The background (for those who understand):

I am studying the K-theory of the category of pairs $(P,f)$ where $P$ is some projective $R$-module and $f$ is an automorphism. One can show* that this category is equivalent to the category of finitely generated $R[U]$-modules of projective dimension one and which are of $g$-torsion for some polynomial $g\in S$, where $S$ is the set of all monic polynomials which have a unit as a constant term. The set $S$ consists in fact of all characteristic polynomials of all possible projective modules. Since I want to localise the ring $R[U]$ at $S$ later it seems crucial whether $S$ contains 1 or 0.

I assume that I will find a unique way to make everything work, but it would be great to have an intrinsic explanation.

*The proof works for a more general setting and doesn't use all properties of $S$. So we can't really deduce an answer from here.

Answer 1

The minimal and characteristic polynomials are both $f(x) = 1$.

Let A be the unique endomorphism: i.e. the unique $0 \times 0$ matrix. It is both a zero matrix and an identity matrix.

For the minimal polynomial, the usual definition is clear, just degenerate. Because the $0 \times 0$ identity matrix is zero, we have $f(A) = 0$. Since $f$ is monic and divides every polynomial, it is clearly the minimal polynomial of $A$.

To compute the characteristic polynomial, we need $\det(A - xI)$. The "right" definition of determinant in the $0 \times 0$ case, as I recall, is $\det A = 1$, so we see that $f$ is the characteristic polynomial as well.

Why is $\det A = 1$ the "right" definition? Unfortunately, off hand I can't think of any explanation other than the fact that every heuristic argument I'm aware of that would suggest a value suggests the value should be 1.

Answer 2

The determinant of a $n \times n$ matrix $(a_{i,j})$ is defined by the Leibniz rule $$\det((a_{i,j})) := \sum_{\pi \in \mathfrak{S}_n} \mathrm{sgn}(\pi) \prod_{i=1}^{n} a_{i,\pi(i)}.$$

For $n=0$ the symmetric group $\mathfrak{S}_n$ is the trivial group (this is not a convention, it follows from the definitions and basic set theory). The signature of the identity is always $1$, so we arrive at $\prod_{i=1}^{n} a_{i,i}$. But this is $1$ (empty products are 1 by convention; of course this is a good one because it makes $\prod_{i \in A \sqcup B} x_i = \prod_{i \in A} x_i \cdot \prod_{i \in B} x_i$ true when $A$ or $B$ are empty).

So the determinant of the unique $0 \times 0$-matrix is $1$. You can also get that from any other definition of the determinant. You can also derive it from the characteristic properties for the determinant (but, of course, then you have to be careful that in their development the case $n=0$ is also allowed): The unique $0 \times 0$-matrix $M$ (you can pictures this as $\begin{pmatrix} ~ \end{pmatrix}$ :-) ) is the identity matrix, so its determinant has to be $1$.

In particular, the characteristic polynomial of every $0 \times 0$-matrix is $1$. The minimal polynomial is also $1$ (here I don't agree with Hurkyl's answer):

By definition, the minimal polynomial of a $n \times n$ matrix $A$ is the unique normed generator of the ideal $\{p : p(A)=0\} \subseteq k[t]$. When $n=0$, the condition $p(A)=0$ is empty. So the ideal is generated by $1$, qed.

Many texts make the reader believe that these vacuities don't follow from the definitions and therefore are a matter of convention. But this is totally wrong. For many more examples of this type, see MO/45951.