# Determinant of classical adjoint over commutative ring with identity

This is a problem from Hungerford's Algebra (p. 354, Problem 4):

If $$A \in \text{Mat}_{n}(R)$$ show that $$\det(\text{adj}(A)) = \det(A)^{n-1}$$.

Here $$R$$ is assumed to be a commutative ring with identity. Note that the problem is easy if we assume $$R$$ to be a field or an integral domain, in which case we can use the standard trick of computing the determinant of both sides of the identity $$A\cdot\text{adj}(A) = \det(A)I_{n}$$, but I don't know how to deal with the general case when $$R$$ is an arbitrary commutative ring. Any help would be appreciated.

• There is a standard trick to cancel the $\det A$ factor by replacing $A$ by $A + tI_n$, where $t$ is an indeterminate. See Theorem 5.12 (a) in cip.ifi.lmu.de/~grinberg/algebra/trach.pdf for details. Jul 23, 2022 at 6:41
• @darijgrinberg thank you very much for the reference. It'll take me a while to read and understand it. Jul 23, 2022 at 7:50

Actually once you've proved this identity over, say, $$\mathbb{C}$$ you've proven it over any commutative ring, because it's just a collection of polynomial identities over $$\mathbb{Z}$$ in variables given by the coefficients of $$A$$. See this discussion of the universal matrix for more details. The cancellation of $$\det$$ by working over $$\mathbb{Z}[a_{ij}]$$ because it's an integral domain trick also works here.