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?

  • $\begingroup$ 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? $\endgroup$ – Servaes Mar 12 at 13:54
  • $\begingroup$ 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! $\endgroup$ – Arrow Mar 12 at 14:32
  • $\begingroup$ 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. $\endgroup$ – Servaes Mar 12 at 16:06
  • $\begingroup$ @Servaes looking forward to reading it, thanks! $\endgroup$ – Arrow Mar 12 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 $V$.

  • $\begingroup$ Dear Servaes, thank you for your instructive answer! How can one prove that $\chi_f\in R[t]$ is monic? $\endgroup$ – Arrow Mar 25 at 19:56
  • $\begingroup$ 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. $\endgroup$ – Arrow Mar 26 at 11:09
  • $\begingroup$ 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.) $\endgroup$ – Arrow Mar 26 at 12:22
  • $\begingroup$ @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. $\endgroup$ – Servaes Mar 27 at 11:50
  • $\begingroup$ 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 $\chi_f$. $\endgroup$ – Servaes Mar 27 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.