]> Commutative Algebra - Finitely Generated Modules and Free Modules

## Finitely Generated Modules and Free Modules

Let $R$ be a ring and $M$ an $R$-module. Note if $M\cong R$ as $R$-modules we may regard $M$ as a ring isomorphic to $R$: if $\theta :M\to R$ we may define multiplication by $xy=\theta \left(x\right)y$ for all $x,y\in M$.

$M$ is free if $M$ is isomorphic to an $R$-module of the form ${\oplus }_{i\in I}{M}_{i}$ where each ${M}_{i}\cong R$. Sometimes this is denoted ${R}^{\left(I\right)}$. A finitely generated free module is isomorphic to $R\oplus ...\oplus R$ where there are $n$ summands, and is written ${R}^{n}$. By convention ${R}^{0}$ is the zero module.

Proposition: $M$ is a finitely generated $R$-module $⇔$ $M$ is isomorphic to a quotient of ${R}^{n}$ for some $n\ge 0$.

Proof: ($⇒$) Suppose $M=⟨{x}_{1},...,{x}_{n}⟩$. Define $\varphi :{R}^{n}\to M$ by

$\left({a}_{1},...,{a}_{n}\right)↦{a}_{1}{x}_{1}+...+{a}_{n}{x}_{n}$

Then $\varphi$ is a surjective $R$-module homomorphism, thus $M\cong {R}^{n}/\mathrm{ker}\varphi$.

($⇐$) Suppose $M$ is isomorphic to a quotient of ${R}^{n}$ for some $n$. Then we have an onto module homomorphism $\varphi :{R}^{n}\to M$. Set ${e}_{i}=\left(0,...,0,1,0,...,0\right)$, (the $i$+++th coordinate is 1). Then the ${e}_{i}$ generate ${R}^{n}$ thus the $\varphi \left({e}_{i}\right)$ generate $M$. $\blacksquare$+.

Proposition: Let $M$ be a finitely generated $R$-module, $I◃R$ be an ideal, and $\varphi \in {\mathrm{Hom}}_{R}\left(M,M\right)$ be a homomorphism with $\varphi \left(M\right)\subset IM$. Then $\varphi$ is the root of a monic polynomial with nonleading coefficients from $I$.

Proof: Suppose $M=⟨{x}_{1},...,{x}_{n}⟩$. Now

$IM=\left\{\sum _{j=1}^{n}{b}_{j}{y}_{j}\mid m\ge 1,{b}_{j}\in I,{y}_{j}\in M\right\}$

Since every element of $M$ can be written as a linear combination of the ${x}_{i}$, and since $I$ is an ideal, we have

$IM=\left\{\sum _{i=1}^{n}{a}_{i}{x}_{i}\mid n\ge 1,{c}_{i}\in I\right\}$

For each $i$ write $\varphi \left({x}_{i}\right)={\sum }_{j=1}^{n}{a}_{ij}{x}_{j}$ for some ${a}_{ij}\in I$.

Now we can use the Cayley-Hamilton Theorem: let $A$ is the matrix $\left({a}_{ij}\right)$ and let $Chi\left(x\right)$ be its characteristic polynomial, that is $Chi\left(x\right)=\mid xI-A\mid$. Then by the Cayley-Hamilon Theorem $Chi\left(A\right)=0$ hence $Chi\left(\varphi \right)=0$.

Alternatively, we may write

$\sum _{j=1}^{n}\left({\delta }_{ij}\varphi -{a}_{ij}\right){x}_{j}=0$

where ${\delta }_{ij}$ is the Kronecker delta. By multiplying on the left by the adjoint of ${\delta }_{ij}\varphi -{a}_{ij}$ we see that $\mathrm{det}\left({\delta }_{ij}\varphi -{a}_{ij}\right)$ must map each ${x}_{i}$ to zero, hence is the zero endomorphism of $M$. The determinant is an equation of the required form. $\blacksquare$

Corollary: Let $M$ be a finitely generated $R$-module and $I$ be and ideal of $R$ with $IM=M$. Then for some $x\in 1+I$ we have $xM=0$.

Proof: Take $\varphi$ to be the identity in the previous theorem, thus we have ${\varphi }^{n}+{a}_{n-1}{\varphi }^{n-1}+...+{a}_{0}=0$ for some ${a}_{i}\in I$. Then set $x=1+{a}_{n-1}+...+{a}_{0}$.

Nakayama's Lemma: Let $M$ be a finitely generated $R$-module and $I◃R$ be an ideal such that $I\subset J\left(R\right)$ where $J\left(R\right)$ is the Jacobson radical of $R$. Then $IM=M⇒M=0$.

Proof: By the previous corollary $xM=0$ for some $x\in 1+J\left(R\right)$. Then $1-x\in J\left(R\right)$, thus $x=1-\left(1-x\right)$ is a unit in $R$, hence $M={x}^{-1}xM=0$.

Alternatively, suppose $M$ is nonzero. Then let ${u}_{1},...,{u}_{m}$ be a minimal set of generators of $M$. Then since ${u}_{n}\in IM$ we have ${u}_{n}={a}_{1}{u}_{1}+...+{a}_{n}{u}_{n}$ for some ${a}_{i}\in I$. Then

$\left(1-{a}_{n}\right){u}_{n}={a}_{1}{u}_{1}+...+{a}_{n-1}{u}_{n-1}$

and since ${a}_{n}\in I\subset J\left(R\right)$ we have that $1-{a}_{n}$ is a unit, implying that ${u}_{n}\in ⟨{u}_{1},...,{u}_{n-1}⟩$ contradicting the minimality of the set of generators. $\blacksquare$

Corollary: Let $M$ be a finitely generated $R$-module. Let $N$ be a submodule of $M$ and $I$ an ideal contained in $J\left(R\right)$. Then $M=IM+N⇒M=N$.

Proof: Note $I\left(M/N\right)=\left(IM+N\right)/N=M/N$ so by the lemma $M/N=0$.

Let $R$ be a local ring with maximal ideal $I$. Let $K=R/I$ be the residue field of $R$. Let $M$ be a finitely generated $R$-module. Note $I\subset \mathrm{Ann}\left(M/IM\right)$ so $M/I$ is can be viewed as an $R/I$-module, that is $M/I$ is a finite-dimensional $K$-vector space.

Proposition: Let ${x}_{1},...,{x}_{n}\in M$ be elements such that $\left\{{x}_{1}+IM,...,{x}_{n}+IM\right\}$ is a basis for the vector space $M/IM$. Then $M=⟨{x}_{1},...,{x}_{n}⟩$.

Proof: Let $N=⟨{x}_{1},...,{x}_{n}⟩$. Then the map $N\to M\to M/IM$ maps $N$ onto $M/IM$ thus $N+IM=M$ so by the above corollary $N=M$.