Does the following property of matrices hold for any commutative ring with identity?

Question

I have recently gone through the proof of the following theorem given in my book $:$

Theorem $:$

<p>Let <span class="math-container">$R$</span> be a commutative ring with identity with quotient field <span class="math-container">$K.$</span> Let <span class="math-container">$\alpha \in K.$</span> Let <span class="math-container">$R[\alpha]$</span> be a finitely generated <span class="math-container">$R$</span>-module. Then <span class="math-container">$\alpha \in K$</span> is integral over <span class="math-container">$R.$</span></p>

In the proof of the above theorem it has been used that $$A\cdot \text {adj} (A) = \det (A) \cdot I_n$$ where $A \in \Bbb M_n (R).$ I know that this is a well known result for any matrix $A \in \Bbb M_n (\Bbb R)\ \text {or}\ \Bbb M_n (\Bbb C).$ But I am not sure about whether or not this result also holds over any commutative ring instead of $\Bbb R$ or $\Bbb C.$ Would anybody please help me in understanding this result for an arbitrary commutative ring? Any help will be highly appreciated.

Thank you very much.

Answer 1

It holds for any commutative ring indeed, and there are many ways to see this.

My favourite way to do it is to deduce it from the field version, here's how (it's very related to my answer here : Is studying a free group (or other free object) equivalent to considering only the consequences of the basic axioms? )

Note that $A \mathrm{adj}(A)= \det (A) I_n$ is a collection of polynomial identities with integer coefficients relating the entries of $A$. Therefore it holds for any commutative ring and any matrix if and only if it holds for the generic matrix with coefficients $a_{ij}=X_{ij}$ over the ring $A=\mathbb{Z}[X_{ij}, 1\leq i,j\leq n]$

But this ring is an integral domain ! Therefore our generic matrix also lives in $M_n(Frac(A))$, and $Frac(A)$ is a field. It follows that the identity holds there. But $\mathrm{adj}$, $\det$ and $I_n$ don't depend on whether you're in $Frac(A)$ or in $A$, so the identity also holds in $A$; and by freeness of $A$ it holds in any commutative ring.