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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?

Servaes
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Arrow
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  • 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
  • Would this be functorial enough: https://math.stackexchange.com/a/4701698/ ? – Alexey Jul 29 '23 at 15:13

1 Answers1

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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 $V$.

Servaes
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  • 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
  • Dear Servaes, thanks for your replies! Do you see an algebraic way to prove the characteristic polynomial is monic? Also, I don't understand why $f-t\in E$ is injective and why this implies the char. pol. is not a zero divisor. Assuming $(f-t)v=0$, how to proceed? – Arrow Apr 07 '19 at 21:16
  • @Arrow To see that $f-t\in E$ is injective, let $v\in S$ be such that $(f-t)v=0$. We can write $$v=\sum_{i=0}^nv_i\otimes t^i,$$ with $v_n\neq0$ (unless $v=0$), i.e. as a polynomial with coefficients in $V$. Then the leading coefficient of $(f-t)v$ equals $-v_n\neq0$, so $(f-t)v\neq0$. So $f-t$ is injective. To conclude that the characteristic polynomial is not a zero divisor, see this answer. (Take $R[t]$ in stead of $R$, and $S$ in stead of $M$, and $f-t$ in stead of $u$) – Servaes Apr 09 '19 at 12:35
  • I do not immediately see an algebraic way to prove that the characteristic polynomial is monic though. I don't doubt that there is an easy way, but I don't see it at the moment. – Servaes Apr 09 '19 at 12:45
  • Woah, there's a lot going on behind that link! I thought 'invertible iff invertible determinant' should be seen as a consequence of Cayley-Hamilton, but it seems I am wrong. Thank you for your patience! – Arrow Apr 10 '19 at 07:53