# If $\mathrm{det}(M)$ satisfies [generalization of being a unit], then $M$ satisfies [generalization of having a two-sided inverse].

(I'm interested in arbitrary commutative rings, not just integral domains or PID's.)

Let $R$ denote a commutative ring and $M$ denote an $n\times n$ matrix over $R$. Suppose we're interested in claims of the following form:

Proposition. If $\mathrm{det}(M)$ satisfies [blah], then $M$ satisfies [blurg].

Sometimes, such claims are easy to prove, just using the basic properties of determinants. For example, it is easy to see that if $\mathrm{det}(M)$ satisfies "I am not a unit," then $M$ satisfies "I have neither a left inverse, nor a right inverse." The converse is much harder. Explicitly, we're trying to show that: If $\mathrm{det}(M)$ satisfies "I am a unit", then $M$ satisfies "I have a two-sided multiplicative inverse." As I understand it, this can be proved using adjugates. But the underlying insight here is non-trivial, and so too is the proof.

Okay. Lets specialize slightly. Suppose we're interested in claims like:

Proposition. If $\mathrm{det}(M)$ satisfies [generalization of being a unit], then $M$ satisfies [generalization of having a two-sided inverse].

For example: being left-cancellative, two-sided cancellative, non-nilpotent, etc.

Question. What techniques are available to establish such claims?

• It's not exactly what you're looking for (and, philosophically, I'd say that linear algebra doesn't really work very well outside the context of a module over a field, and you might have better luck with the larger machinery of commutative algebra), but how about the Dieudonne determinant? I don't think it's used that often, but I seem to remember that Serre uses it in his book on trees. Dec 5 '15 at 3:31

The proof that $\mathrm{det}(M)$ being unit implies that $M$ has a two-sided inverse is not hard: in this case, the inverse is $\mathrm{det}(M)^{-1} M^{\#},$ where $M^{\#}$ denotes the adjugate. The fact that $M M^{\#} = M^{\#} M = \mathrm{det}(M) I$ is another way of saying that the determinant can be calculated by cofactor expansion, and this formula is valid over any commutative ring.
$M$ being surjective implies that there is a matrix $N$ such that $MN = I$, so $\mathrm{det}(M) \mathrm{det}(N) = 1$ and $\mathrm{det}(M)$ is invertible.
$M$ being injective is equivalent to $\mathrm{det}(M)$ not being a zero-divisor; this is a little harder to prove.
$M$ being non-nilpotent cannot be detected by the determinant. Examples with real coefficients will show you this.
• I'm not sure I buy the opening paragraph: how do you know that the formula $MM^\# = \mathrm{det}(M) I$ holds? This seems pretty non-trivial to me. Dec 6 '15 at 2:03