Dtyjlui

Let $V$ be a rank $n$ free module over a commutative ring $R$. Let $dagger$ denote the adjoint with respect to the natural perfect pairing given by the wedge product
$$textstyle
bigwedge^kotimes bigwedge^{n-k}overset{wedge}{longrightarrow} bigwedge^n.$$

Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove
$$textstyle
(bigwedge^{n-k}f)^daggercirc bigwedge^kf=det fcdot 1_{bigwedge^kV}.$$

Now let $Voverset{f}{to}V$ be an $R$-linear endomorphism. It induces an $R[f]$-module structure on $V$ which in turn induces an $R[f,t]$-module structure on the $R[t]$-module $Votimes _RR[t]$. Define the characteristic polynomial $chi_fin R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $Votimes _RR[t]$.

By the above fact we have the following equation in the category of $R[t]$-modules.
$$textstyle
(bigwedge^{n-k}(f-t))^daggercirc bigwedge^k(f-t)=chi_fcdot 1_{bigwedge^k(Votimes_RR[t])}$$

I'm trying to follow the proof of Cayley-Hamilton along these lines given in 28.10, but I am confused by the sudden passage to the category of $R[f,t]cong R[f]otimes _RR[t]$-modules.

How to formally derive the Cayley-Hamilton theorem from the latter equation?

First let me simplify the clutter of notation a bit; set $S:=Votimes_RR[t]$ and $E:=operatorname{End}_{R[t]}(S)$.
Throughout this answers I will view elements of $R[f,t]$ as $R[t]$-linear endomorphisms of $S$, i.e. as elements of $E$. As for exterior powers; for the proof only the case $k=1$ is relevant, so I won't bother with them at all.

The idea of the proof is to show that the endomorphism $chi_fin E$ vanishes on the quotient $S/(f-t)S$, and then to show that $S/(f-t)Scong V$ as $R$-modules. The main ingredient is showing that $f-tin E$ commutes with its adjugate. This relies on the fact that $chi_f$ is not a zero divisor in $R[t]$.

The proof is a lot of commutative algebra, I have assumed everything in Atiyah-Maconald. If any part is unclear, let me know.

Step 1: The characteristic polynomial is not a zero divisor in $R[t]$.

The characteristic polynomial $chi_f$ of $f-tin R[f,t]$ is the determinant of the $R[t]$-linear map $f-tin E$. Note that $chi_fin R[t]$ is not a zero divisor because $f-tin E$ is injective, because its leading coefficient as a polynomial in $t$, i.e. as an element of $(R[f])[t]$, is a unit.

Step 2: The endomorphism $f-tin E$ commutes with its adjugate w.r.t. the given pairing.

The adjugate of $f-tin E$ with respect to the given perfect pairing is the unique $Fin E$ such that
$$Fcdot(f-t)=chi_fcdot1_S.tag{1}$$

Because $chi_fin R[t]$ is not a zero divisor, localizing at $chi_f$ yields an injection $R[t] longrightarrow R[t]_{chi_f}$. Because $V$ is a finitely generated free $R$-module, this in turn yields injections
$$S longrightarrow S_{chi_f}
qquadtext{ and }qquad
E longrightarrow E_{chi_f}.$$
By construction $chi_f$ is a unit in $E_{chi_f}$ and hence $(1)$ shows that also $f-t$ is a unit in $E_{chi_f}$, so
$$F=chi_fcdot(f-t)^{-1},$$
in $E_{chi_f}$. This shows that $F$ and $f-t$ commute in $E_{chi_f}$, because both are $R[t]$-linear and $chi_fin R[t]$. Because $E_{chi_f}$ contains $E$ as a subring, they also commute in $E$.

Step 3: On the quotient module $S/(f-t)S$ we have $chi_f(f)=0$.

Because $F$ and $f-t$ commute, for all $(f-t)sin(f-t)S$ we have
$$F((f-t)s)=(f-t)F(s)in(f-t)S,$$
so $F$ maps the $S$-submodule $(f-t)Ssubset S$ into itself. This means $F$ descends to an $R[t]$-linear map
$$S/(f-t)S longrightarrow S/(f-t)S.$$
In this quotient $f-t$ is identically zero, so identity $(1)$ shows that on the quotient
$$Fcdot0=chi_fcdot1_{S/(f-t)S},$$
and so $chi_f$ is identically zero on $S/(f-t)S$, where of course $chi_f(t)=chi_f(f)$ on the quotient.

Step 4: Also $chi_f(f)=0$ on $V$.

Because $chi_f(f)=0$ on $S/(f-t)S$ and the composition
$$V longrightarrow S longrightarrow S/(f-t)S,$$
is an isomorphism of $R[f]$-modules, it follows that $chi_f(f)=0$ on $V$.