Simple root of characteristic polynomial of matrix over a commutative ring I'm struggling with §5 of Lombardi & Quitte's Commutative algebra: constructive methods. Specifically, I'm having a hard time with the proof of Lemma 5.1.
5.1 Lemma. Let $n\geq 2,a\in \mathbf A$ and $A\in \mathrm M_n(\mathbf A)$ be a matrix whose characteristic polynomial $f$ admits $a\in \mathbf A$ as a simple zero. Let $g=f/(x-a)$ and $K=\operatorname{Ker} (A-a\mathrm I)$ and $I=\operatorname{Im}(A-a\mathrm I)$.

*

*We have $\operatorname{Ker} (A-a\mathrm I)=\operatorname{Im}g(A)$ and $\operatorname{Im}(A-a\mathrm I)=\operatorname{Ker}g(A)$.

*The matrix $g(A)$ is of rank $1$ and $A-a\mathrm I$ is of rank $n-1$.

*If a polynomial $R$ annihilates $A$ then $R(a)=0$ i.e $R$ is a multiple of $x-a$.

*The principal minors of order $n-1$ of $A-a\mathrm I$ are comaximal. We localize by inverting such a minor, the matrix $g(A)$ becomes simple of rank $1$, the modules $K=\operatorname{Ker} (A-a\mathrm I)$ and $I=\operatorname{Im}(A-a\mathrm I)$ become free of rank $1$ and $n-1$.

I understand the proof of the first assertion, but I am struggling with the rest.


*The authors show $f^\prime (0)=g(0)=\pm \sum _i \mu_i$ is the (signed) sum of the principal minors $\mu_i$ of $A$. They claim this proves the rank of $A-a\mathrm I$ is $n-1$ and also that the rank of $g(A)$ is at least one. How does this follow?

*This is okay for me.

*The authors write that we have already seen the $\mu_i$'s are comaximal (not a typo). First of all I don't understand why this is, and second I don't understand how it shows the minors of the matrix $A-a\mathrm I$ are comaximal.

 A: I can help you with part 2 for now; part 4 involves too much terminology I have forgotten or never learned.
Just as Lombardi and Quitté do, I suppose that $a = 0$, since the general case can be transformed into this one by replacing $A$ by $A - aI_n$. Thus, the characteristic polynomial $f$ of $A$ admits $0$ as a simple zero, and we have $g = f/X$ and $h=X$ and $K = \operatorname{Ker} A$ and $I = \operatorname{Im} A$.


*So you know that $\sum_i \mu_i = \pm f'\left(0\right)$ is invertible (since $0$ is a simple root of $f$). This shows that $\left< \mu_1, \mu_2, \ldots, \mu_n \right> = \left< 1 \right>$. Hence, the ideal generated by the $\left(n-1\right)\times\left(n-1\right)$-minors of $A$ is $\left< 1 \right>$ (since the principal minors $\mu_1, \mu_2, \ldots, \mu_n$ are among these minors). In other words, $A$ has rank $\geq n-1$. However, $A$ has rank $\leq n-1$, since $\det A = 0$ (because $0$ is a zero of the characteristic polynomial of $A$). Thus, $A$ has rank $n-1$. In other words, $h\left(A\right)$ has rank $n-1$ (since $h = X$). It remains to prove that $g\left(A\right)$ has rank $1$. However, $g\left(A\right) = \pm \widetilde{A}$ (this follows from part 6 of Lemma 1.4, because $\mathrm{C}_A = f$ satisfies $f\left(0\right) = 0$ and thus $g = \pm \Gamma_A$). Hence, we need to prove that $\widetilde{A}$ has rank $1$. Well, we know that $\widetilde{A}$ has rank $\geq 1$, since the ideal generated by the entries of $\widetilde{A}$ is $\left< 1 \right>$ (indeed, this ideal is precisely the ideal generated by the $\left(n-1\right)\times\left(n-1\right)$-minors of $A$; but we have seen that the latter ideal is $\left< 1 \right>$). It remains to show that $\widetilde{A}$ has rank $< 2$. In other words, it remains to show that any $2 \times 2$-minor of $\widetilde{A}$ is $0$. This holds for any matrix $A$ whose determinant is $0$, and follows from part 8 of Lemma 1.4 (because the $2 \times 2$-minors of $\widetilde{A}$ are the entries of $\bigwedge^2 \widetilde{A}$). Note that this also follows from the Desnanot-Jacobi identity (see, e.g., Theorem 2 in my answer at Products of adjugate matrices ).

My impression is that the Schur complement will do the trick for part 4, but I'd have to look up the exact definition of "simple".
