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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.

Rodrigo de Azevedo
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Karthik Kannan
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    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 https://www.cip.ifi.lmu.de/~grinberg/algebra/trach.pdf for details. – darij grinberg Jul 23 '22 at 06:41
  • @darijgrinberg thank you very much for the reference. It'll take me a while to read and understand it. – Karthik Kannan Jul 23 '22 at 07:50

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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.

Qiaochu Yuan
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