# Functorial proof of Cayley-Hamilton using exterior powers

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?

• The linked proof does not work (directly) if $R$ has zero divisors. Can $R$ be assumed to be an integral domain, or are you looking for a proof of the more general (and technical) case? – Servaes Mar 12 '19 at 13:54
• Dear @Servaes, why does the linked proof not work in the presence of zero divisors? As far as the rest - I'd be interested in both! – Arrow Mar 12 '19 at 14:32
• The linked proof does in fact work with only some minor adjustments. I'll write up a proof later today. I'll rephrase it a bit because I think the proof is needlessly complicated. – Servaes Mar 12 '19 at 16:06
• @Servaes looking forward to reading it, thanks! – Arrow Mar 12 '19 at 16:19

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 $$V$$.

• Dear Servaes, thank you for your instructive answer! How can one prove that $\chi_f\in R[t]$ is monic? – Arrow Mar 25 '19 at 19:56
• Also, in the final sentence, why is an $R$-linear isomorphism sufficient? I thought we're saying that $\chi_f\in R[t]$ acts as $\chi_f(f)\in R[f]$ on the quotient, and the $R[f]$-linear isomorphism ensures it acts in the same way on $V$, namely as zero. – Arrow Mar 26 '19 at 11:09
• Also, below (1), how does finite freeness of $V$ and injectivity of $R[t]\to R[t]_{\chi_f}$ imply injectivity of $S \longrightarrow S_{\chi_f},E \longrightarrow E_{\chi_f}$? I'm guessing freeness is needed for flatness, but why is finiteness needed as well? For some distributivity of $\otimes$ over $\oplus$ perhaps? I thought finite freeness is needed in order to have $\Lambda ^n V\cong R$, which makes the determinant a well defined element of $R$... (Sorry for the many questions, I just want to make sure I understand everything.) – Arrow Mar 26 '19 at 12:22
• @Arrow I don't claim that $\chi_f\in R[t]$ is monic; only that it is not a zero divisor. This is the case because $f-t=f\otimes1-1\otimes t\in E$ is injective. As for injectivity of the induced maps; indeed finiteness is needed for the tensor product to distribute over the direct sum; it does not distribute over arbitrary direct sums. I guess it is also needed for the determinant to make sense, I haven't given that any thought. – Servaes Mar 27 '19 at 11:50
• As for the final sentence; I agree that it suffices to note that this is an isomorphism of $R[f]$-modules. The remark that it is also an isomorphism of $R$-modules only served to emphasize that the $R$-linear map $f:\ V\ \longrightarrow\ V$ is a zero of $\chi_f$. – Servaes Mar 27 '19 at 12:02