Nakayama Lemma

~~啥都能水系列~~

茴字的四种写法之Nakayama lemma. 大定理的最后几个部分和一般的Nakayama lemma (finite 版本)差的有点远了, 但是还是抄上来了.

Nakayama Lemma

Lemma 1 (Existence of adjoint matrix). Let $R$ be a ring. Let $n \geq m$ . Let $A$ be an $n \times m$ matrix with coefficients in $R$ . Let $J \subset R$ be the ideal generated by the $m \times m$ minors of $A$ .

For any $f \in J$ there exists a $m \times n$ matrix $B$ such that $BA = f 1_{m \times m}$ .

Proof

Proof. For $I \subset \{1, \ldots, n\}$ with $|I| = m$ , we denote by $E_I$ the $m \times n$ matrix of the projection

R^{\oplus n} = \bigoplus\nolimits_{i \in \{1, \ldots, n\}} R\longrightarrow \bigoplus\nolimits_{i \in I} R

and set $A_I = E_I A$ , i.e., $A_I$ is the $m \times m$ matrix whose rows are the rows of $A$ with indices in $I$ . Let $B_I$ be the adjugate (transpose of cofactor) matrix to $A_I$ , i.e., such that $A_I B_I = B_I A_I = \det(A_I) 1_{m \times m}$ . The $m \times m$ minors of $A$ are the determinants $\det A_I$ for all the $I \subset \{1, \ldots, n\}$ with $|I| = m$ . If $f \in J$ then we can write $f = \sum c_I \det(A_I)$ for some $c_I \in R$ . Set $B = \sum c_I B_I E_I$ to see that the lemma holds. ◻

Now we give a large version of Nakayama lemma.

Theorem 1 (Nakayama Lemma). Let $R$ be a ring, $\mathfrak{R}$ is the Jacobson radical and $\mathfrak{a}$ is an ideal. $M$ is an $R$ -module.

If $\mathfrak{a}M=M$ and $M$ is finite generated, then there is $x\in R$ s.t. $x-1\in \mathfrak{a}$ and $xM=0$ .
If $\mathfrak{a}M=M$ and $M$ is finite generated, $\mathfrak{a}\subset \mathfrak{R}$ , then $M=0$ .
If $N,N'\subset M$ , $M=N+\mathfrak{a}N'$ , $N'$ is finite generated, then there is $x\in R$ s.t. $x-1\in \mathfrak{a}$ and $xM\subset N$ , $M_{x}=N_{x}$ .
If $N,N'\subset M$ , $M=N+\mathfrak{a}N'$ , $N'$ is finite generated, $\mathfrak{a}\subset \mathfrak{R}$ , then $M=N$ .
Let $f:N\to M$ be a homomorphism, $\bar{f}:N /\mathfrak{a}N\to M /\mathfrak{a}M$ is surjective. If $M$ is finite generated then there is $x\in R$ s.t. $x-1\in \mathfrak{a}$ and $f_{x}:N_{x}\to M_{x}$ is surjective.
Let $f:N\to M$ be a homomorphism, $\bar{f}:N /\mathfrak{a}N\to M /\mathfrak{a}M$ is surjective. If $M$ is finite generated and $\mathfrak{a}\subset\mathfrak{R}$ , then $f$ is surjective.
If $x_{1},\dots, x_{n}\in M$ generates $M /\mathfrak{a}M$ and $M$ is finite generated, then there’s $x\in R$ s.t. $x-1\in \mathfrak{a}$ and $x_{1},\dots, x_{n}$ generates $M_{x}$ as an $R_{x}$ -module.
If $x_{1},\dots, x_{n}\in M$ generates $M /\mathfrak{a}M$ , $M$ is finite generated and $\mathfrak{a}\subset\mathfrak{R}$ , then $M$ is generated by $x_{1},\dots, x_{n}$ .
If $\mathfrak{a}M=M$ , $\mathfrak{a}$ is nilpotent, then $M=0$ .
If $N,N'\subset M$ , $M=N+\mathfrak{a}N'$ , $\mathfrak{a}$ is nilpotent, then $M=N$ .
Let $f:N\to M$ be a homomorphism, $\bar{f}:N /\mathfrak{a}N\to M /\mathfrak{a}M$ is surjective. If $\mathfrak{a}$ is nilpotent, then $f$ is surjective.
If $\{x_{\alpha}\}_{\alpha\in A}$ is a set of elements in $M$ which generates $M /\mathfrak{a}M$ and $\mathfrak{a}$ is nilpotent, then $M$ is generated by $\{x_{\alpha}\}_{\alpha\in A}$ .

Proof

Proof.

Choose $y_{1},\dots, y_{m}$ as generators of $M$ . For each $i$ , we have $y_{i}=\sum_{j}z_{ij}y_{j}$ with $z_{ij}\in \mathfrak{a}$ . Thus $\sum_{j}(\delta_{ij}-z_{ij})y_{j}=0$ . Let $x$ be the determinant of $m\times m$ matrix $A=(\delta_{ij}-z_{ij})$ . Note that $x-1\in \mathfrak{a}$ . It’s easy to see $A\begin{pmatrix} y_{1}\\\vdots\\ y_{m} \end{pmatrix}=0$ . Let $B$ be the adjoint matrix of $A$ , $BA\begin{pmatrix} y_{1}\\\vdots\\ y_{m} \end{pmatrix}=\begin{pmatrix} xy_{1}\\\vdots\\ xy_{m} \end{pmatrix}=0$ , so $xy_{i}=0$ , $xM=0$ .
By 1, there’s $x\in R$ , $x-1\in\mathfrak{a}$ s.t. $xM=0$ . Since $x-1\in \mathfrak{R}$ , $1+(x-1)$ is a unit, there’s $x^{-1}\in R$ . Thus $M=x^{-1}xM=x^{-1}0=0$ .
If $N'$ is a finite generated module, then $M /N=\mathfrak{a}N' /N$ is also finite generated. $M /N=\mathfrak{a}(N' /N)\subset\mathfrak{a}(M /N)\subset M /N$ , so $\mathfrak{a}(M /N)=M /N$ . Applying 1, there’s $x\in R$ , $x-1\in \mathfrak{a}$ s.t. $x(M /N)=0$ . Thus $xM\subset N$ and $M_{x}\subset N_{x}$ . Since $N\subset M$ , $N_{x}\subset M_{x}$ . So $M_{x}=N_{x}$ .
Applying 2 to finite generated module $M/N$ , we have $M /N=0$ , i.e. $M=N$ .
Since $\bar{f}:N /\mathfrak{a}N\to M /\mathfrak{a}M$ is surjective, $M=\mathrm{Im}f+\mathfrak{a}M$ . Applying 3 to finite generated $M$ , we have $x\in R$ , $x-1\in\mathfrak{a}$ and $xM\subset\mathrm{Im}f$ , i.e. $f_{x}:N_{x}\to M_{x}$ is surjective.
Similarly, $M=\mathrm{Im}f+\mathfrak{a}M$ . Applying 4 to finite generated $M$ , we have $M=\mathrm{Im}f$ , i.e. $f$ is surjective.
Take $f:R^{n}\to M$ given by $f:(a_{1},\cdots, a_{n})\mapsto a_{1}x_{1}+\cdots+a_{n}x_{n}$ . $\bar{f}:R^{n} /\mathfrak{a}R^{n}\to M /\mathfrak{a}M$ given by $\bar{f}:(\bar{a_{1}},\cdots, \bar{a_{n}})\mapsto \bar{a_{1}}x_{1}+\cdots+ \bar{a_{n}}x_{n}$ is obviously surjective. Applying 5, there is $x\in R$ , $x-1\in\mathfrak{a}$ s.t. $f_{x}:R_{x}^{n}\to M_{x}$ surjective. Thus $x_{1},\dots, x_{n}$ generates $M_{x}$ over $R_{x}$ .
Take $f:R^{n}\to M$ given by $f:(a_{1},\cdots, a_{n})\mapsto a_{1}x_{1}+\cdots+a_{n}x_{n}$ . $\bar{f}:R^{n} /\mathfrak{a}R^{n}\to M /\mathfrak{a}M$ given by $\bar{f}:(\bar{a_{1}},\cdots, \bar{a_{n}})\mapsto \bar{a_{1}}x_{1}+\cdots+ \bar{a_{n}}x_{n}$ is obviously surjective. Applying 6, we obtain the surjectivity of $f$ . Thus $x_{1},\dots, x_{n}$ generates $M$ .
$M=\mathfrak{a}M$ so $M=\mathfrak{a}^{n}M$ for all $n\geq 0$ . Assume $\mathfrak{a}^{n}=0$ , then $M=0M=0$ .
Similar argument as above, we have $\mathfrak{a}(M /N)=M /N$ . Applying 9, we have $M /N=0$ , i.e. $M=N$ .
Similar as above argument, we have $M=\mathrm{Im}f+\mathfrak{a}M$ . Applying 10, we have $M=\mathrm{Im}f$ , i.e. $f$ is surjective.
Take $f:R^{A}\to M$ given by $f:(a_{\alpha})_{\alpha\in A}\mapsto \sum_{\alpha\in A}a_{\alpha}x_{\alpha}$ . Then $\bar{f}:R^{A} /\mathfrak{a}R^{A}\to M /\mathfrak{a}M$ is surjective. Applying 11, we have $f$ surjective. Thus $M$ is generated by $\{x_{\alpha}\}_{\alpha\in A}$ . ◻