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
$$\textstyle
\bigwedge^k\otimes \bigwedge^{n-k}\overset{\wedge}{\longrightarrow} \bigwedge^n.$$
Using naturality and uniqueness of adjoints w.r.t perfect pairings one can prove
$$\textstyle
(\bigwedge^{n-k}f)^\dagger\circ \bigwedge^kf=\det f\cdot 1_{\bigwedge^kV}.$$
Now let $V\overset{f}{\to}V$ 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 $V\otimes _RR[t]$. Define the characteristic polynomial $\chi_f\in R[t]$ of $f$ to be the determinant of the $R[t]$-linear endomorphism $f-t$ of the $R[t]$-module $V\otimes _RR[t]$.
By the above fact we have the following equation in the category of $R[t]$-modules.
$$\textstyle
(\bigwedge^{n-k}(f-t))^\dagger\circ \bigwedge^k(f-t)=\chi_f\cdot 1_{\bigwedge^k(V\otimes_RR[t])}$$
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?
Best Answer
First let me simplify the clutter of notation a bit; set $S:=V\otimes_RR[t]$ 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 $\chi_f\in E$ vanishes on the quotient $S/(f-t)S$, and then to show that $S/(f-t)S\cong V$ as $R$-modules. The main ingredient is showing that $f-t\in E$ 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 $f-t\in R[f,t]$ is the determinant of the $R[t]$-linear map $f-t\in E$. Note that $\chi_f\in R[t]$ is not a zero divisor because $f-t\in E$ 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 $f-t\in E$ commutes with its adjugate w.r.t. the given pairing.
The adjugate of $f-t\in E$ with respect to the given perfect pairing is the unique $F\in E$ such that $$F\cdot(f-t)=\chi_f\cdot1_S.\tag{1}$$
Because $\chi_f\in R[t]$ 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 $$S\ \longrightarrow\ S_{\chi_f} \qquad\text{ and }\qquad E\ \longrightarrow\ E_{\chi_f}.$$ 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 $$F=\chi_f\cdot(f-t)^{-1},$$ in $E_{\chi_f}$. This shows that $F$ and $f-t$ commute in $E_{\chi_f}$, because both are $R[t]$-linear and $\chi_f\in R[t]$. 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 $(f-t)s\in(f-t)S$ we have $$F((f-t)s)=(f-t)F(s)\in(f-t)S,$$ so $F$ maps the $S$-submodule $(f-t)S\subset S$ 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 $$F\cdot0=\chi_f\cdot1_{S/(f-t)S},$$ 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 $V$.