Intereting Posts

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Limit of a Recursive Sequence
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isolated non-normal surface singularity
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difference between class, set , family and collection
surjective, but not injective linear transformation
Convergence of probability measures in total variation and limits of integrals
Has it been proved that odd perfect numbers cannot be triangular?
$L$-function, easiest way to see the following sum?
Eigenvalue problems for matrices over finite fields

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:

- A subgroup of $\operatorname{GL}_2(\mathbb{Z}_3)$
- The sum of the elements in a field of at least three elements is 0
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- Every radical ideal in a Noetherian ring is a finite intersection of primes
- The number of solutions of $x^n = e$ in a finite group is a multiple of n, whenever n divides the group order.

**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

Now for my question:

Can

CHbe 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.

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Nakayama Lemma.Let $N$ be a finitely generated $R$-module, and $J\subseteq R$. Suppose that $J$ is closed under addition and multiplication and $JN=N$. Then there is $a\in J$ such that $(1+a)N=0$. (Here by $JN$ we denote the subset of linear combinations of $N$ with coefficients in $J$.)

Cayley-Hamilton Theorem.Let $A$ be a commutative ring, $I$ an ideal of $A$, $M$ a finitely generated $A$-module, $\phi$ an $A$-module endomorphism of $M$ such that $\phi(M)\subseteq IM$. Then there are $n\ge 1$ and $a_i\in I^i$ such that $\phi^n+a_1\phi^{n-1}+\dots+a_n=0$.

**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 $a_i\in I^i$.

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