Consider the Cayley-Hamilton Theorem in the following form:
CH: Let $A$ be a commutative ring, $\mathfrak{a}$ an ideal of $A$, $M$ a finitely generated $A$-module, $\phi$ an $A$-module endomorphism of $M$ such that $\phi(M)\subseteq\mathfrak{a}M$. Then there are coefficients $a_i\in\mathfrak{a}$ such that $\phi^n+a_1\phi^{n-1}+\dots+a_n=0$.
This theorem can be proved by using elementary linear algebra in the context of rings. As a corollary, we find the following two versions of Nakayama's Lemma:
NAK1: Let $A$ be a commutative ring, $M$ a finitely generated $A$-module and $\mathfrak{a}\subseteq A$ an Ideal such that $\mathfrak{a}M=M$. Then there is an $x=1\mod\mathfrak{a}$ such that $xM=0$.
Proof. One just sets $\phi=\operatorname{id}$ and plugs in $x=1+a_1+\dots+a_n$.
It follows:
NAK2: Let $M$ be a finitely generated $A$-module, $\mathfrak{a}$ an ideal contained in the Jacobson radical of $A$. Then $\mathfrak{a}M=M$ implies $M=0$.
Proof. Indeed, $xM=0$ for an element $x\in 1+\mathfrak{a}\subseteq 1+J(A)$, which is a unit, hence $M=0$.
However, one can prove Nakayama's Lemma avoiding linear algebra:
Alternative proof of NAK2: Let $u_1,\dots,u_n$ be a generating system of $M$. $u_n\in M=\mathfrak{a}M$, so $u_n=a_1u_1+\dots+a_nu_n$. Subtracting, $(1-a_n)u_n=a_1u_1+\dots+a_{n-1}u_{n-1}$. But $1-a_n$ is a unit, since $a_n\in J(A)$, hence $u_n\in\langle u_1,\dots,u_{n-1}\rangle$. Iterating, we see that all $u_i$ have been zero.
Alternative proof of Nak1: Let $S=1+\mathfrak{a}$. Then $S^{-1}\mathfrak{a}\subseteq J(S^{-1}A)$. If $M=\mathfrak{a}M$, then $S^{-1}M=S^{-1}(\mathfrak{a}M)=(S^{-1}\mathfrak{a})(S^{-1}M)$, thus Nak2 implies $S^{-1}M=0$. As $M$ is finitely generated, there is an $x\in S$ such that $xM=0$.
Now for my question:
Can CH be deduced from Nakayama's Lemma, avoiding linear algebra, in particular the theory of determinants?
By the way, the arguments are taken from Atiyah-Macdonald, I did not find them myself.
Best Answer
Nakayama Lemma implies Cayley-Hamilton Theorem:
$M$ is an $A[X]$-module via $Xm=\varphi(m)$. Moreover, $M$ is also a finitely generated $A[X]$-module. By hypothesis $XM\subseteq IM$. Now consider the ring $A[X,X^{-1}]$, the localization of $A[X]$ with respect to the multiplicative set $S$ generated by $X$, and the finitely generated $A[X,X^{-1}]$-module $S^{-1}M$ (which we denote by $M[X^{-1}]$). The set $$J=\{a_1X^{-1}+\cdots+a_{r}X^{-r}:a_i\in I^i, r\ge1\}$$ is closed under addition and multiplication, and moreover $JM[X^{-1}]=M[X^{-1}]$: if $m\in M$ and since $XM\subseteq IM$ we have $Xm\in IM$. Then $Xm=a_1m_1+\cdots+a_nm_n$ with $a_j\in I$, and therefore $m=(a_1X^{-1})m_1+\cdots+(a_nX^{-1})m_n\in JM[X^{-1}]$.
Now by Nakayama Lemma (for $R=A[X,X^{-1}]$ and $N=M[X^{-1}]$) there is $p\ge 1$ such that $(1+a_1X^{-1}+\cdots+a_{p}X^{-p})M[X^{-1}]=0$. In particular, $(1+a_1X^{-1}+\cdots+a_{p}X^{-p})M=0$, that is, $\dfrac{(X^p+a_1X^{p-1}+\cdots+a_p)m}{X^p}=0$ for all $m\in M$. Since $M$ is finitely generated there is $s\ge0$ such that $X^s(X^p+a_1X^{p-1}+\cdots+a_p)M=0$. Now set $n=s+p$ and conclude that $\varphi^{n}+a_1\varphi^{n-1}+\cdots+a_n=0$ with $a_i\in I^i$.